1. m, n are positive integers such that mn + m + n = 71, m2n + mn2 = 880, find m2 + n2.
2. The rectangle ABCD has AB = 4, BC = 3. The side AB is divided into 168 equal parts
by points P1, P2, ... , P167 (in that order with P1 next to A), and the side BC is divided into
168 equal parts by points Q167, Q166, ... , Q1 (in that order with Q1 next to C). The parallel
segments P1Q1, P2Q2, ... , P167Q167 are drawn. Similarly, 167 segments are drawn between
AD and DC, and finally the diagonal AC is drawn. Find the sum of the lengths of the 335
parallel segments.
3. Expand (1 + 0.2)1000 by the binomial theorem to get a0 + a1 + ... + a1000, where ai =
1000Ci (0.2)i. Which is the largest term?
4. How many real roots are there to (1/5) log2x = sin(5πx) ?
5. How many fractions m/n, written in lowest terms, satisfy 0 < m/n < 1 and mn = 20! ?
6. The real number x satisfies [x + 0.19] + [x + 0.20] + [x + 0.21] + ... + [x + 0.91] =
546. Find [100x].
7. Consider the equation x = √19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/x)))).
Let k be the sum of the absolute values of the roots. Find k2.
8. For how many reals b does x2 + bx + 6b have only integer roots?
9. If sec x + tan x = 22/7, find cosec x + cot x.
10. The letter string AAABBB is sent electronically. Each letter has 1/3 chance
(independently) of being received as the other letter. Find the probability that using the
ordinary text order the first three letters come rank strictly before the second three. (For
example, ABA ranks before BAA, but after AAB.)
11. 12 equal disks are arranged without overlapping, so that each disk covers part of a
circle radius 1 and between them they cover every point of the circle. Each disk touches
two others. (Note that the disks are not required to cover every point inside the circle.)
Find the total area of the disks.
12. ABCD is a rectangle. P, Q, R, S lie on the sides AB, BC, CD, DA respectively so
that PQ = QR = RS = SP. PB = 15, BQ = 20, PR = 30, QS = 40. Find the perimeter of
ABCD.
13. m red socks and n blue socks are in a drawer, where m + n ≤ 1991. If two socks are
taken out at random, the chance that they have the same color is 1/2. What is the largest
possible value of m?
14. A hexagon is inscribed in a circle. Five sides have length 81 and the other side has
length 31. Find the sum of the three diagonals from a vertex on the short side.
15. Let Sn be the minimum value of ∑ √((2k-1)2 + ak2) for positive reals a1, a2, ... , an
with sum 17. Find the values of n for which Sn is integral.
1. m, n are positive integers such that mn + m + n = 71, m2n + mn2 = 880, find m2 + n2.
2. The rectangle ABCD has AB = 4, BC = 3. The side AB is divided into 168 equal parts
by points P1, P2, ... , P167 (in that order with P1 next to A), and the side BC is divided into
168 equal parts by points Q167, Q166, ... , Q1 (in that order with Q1 next to C). The parallel
segments P1Q1, P2Q2, ... , P167Q167 are drawn. Similarly, 167 segments are drawn between
AD and DC, and finally the diagonal AC is drawn. Find the sum of the lengths of the 335
parallel segments.
3. Expand (1 + 0.2)1000 by the binomial theorem to get a0 + a1 + ... + a1000, where ai =
1000Ci (0.2)i. Which is the largest term?
4. How many real roots are there to (1/5) log2x = sin(5πx) ?
5. How many fractions m/n, written in lowest terms, satisfy 0 < m/n < 1 and mn = 20! ?
6. The real number x satisfies [x + 0.19] + [x + 0.20] + [x + 0.21] + ... + [x + 0.91] =
546. Find [100x].
7. Consider the equation x = √19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/(√19 + 91/x)))).
Let k be the sum of the absolute values of the roots. Find k2.
8. For how many reals b does x2 + bx + 6b have only integer roots?
9. If sec x + tan x = 22/7, find cosec x + cot x.
10. The letter string AAABBB is sent electronically. Each letter has 1/3 chance
(independently) of being received as the other letter. Find the probability that using the
ordinary text order the first three letters come rank strictly before the second three. (For
example, ABA ranks before BAA, but after AAB.)
11. 12 equal disks are arranged without overlapping, so that each disk covers part of a
circle radius 1 and between them they cover every point of the circle. Each disk touches
two others. (Note that the disks are not required to cover every point inside the circle.)
Find the total area of the disks.
12. ABCD is a rectangle. P, Q, R, S lie on the sides AB, BC, CD, DA respectively so
that PQ = QR = RS = SP. PB = 15, BQ = 20, PR = 30, QS = 40. Find the perimeter of
ABCD.
13. m red socks and n blue socks are in a drawer, where m + n ≤ 1991. If two socks are
taken out at random, the chance that they have the same color is 1/2. What is the largest
possible value of m?
14. A hexagon is inscribed in a circle. Five sides have length 81 and the other side has
length 31. Find the sum of the three diagonals from a vertex on the short side.
15. Let Sn be the minimum value of ∑ √((2k-1)2 + ak2) for positive reals a1, a2, ... , an
with sum 17. Find the values of n for which Sn is integral.
1. Find the sum of all positive rationals a/30 (in lowest terms) which are < 10.
2. How many positive integers > 9 have their digits strictly increasing from left to right?
3. At the start of a weekend a player has won the fraction 0.500 of the matches he has
played. After playing another four matches, three of which he wins, he has won more
than the fraction 0.503 of his matches. What is the largest number of matches he could
have won before the weekend?
4. The binomial coefficients nCm can be arranged in rows (with the nth row nC0, nC1,
... nCn) to form Pascal's triangle. In which row are there three consecutive entries in the
ratio 3 : 4 : 5?
5. Let S be the set of all rational numbers which can be written as 0.abcabcabcabc...
(where the integers a, b, c are not necessarily distinct). If the members of S are all written
in the form r/s in lowest terms, how many different numerators r are required?
6. How many pairs of consecutive integers in the sequence 1000, 1001, 1002, ... , 2000
can be added without a carry? (For example, 1004 and 1005, but not 1005 and 1006.)
7. ABCD is a tetrahedron. Area ABC = 120, area BCD = 80. BC = 10 and the faces ABC
and BCD meet at an angle of 30o. What is the volume of ABCD?
8. If A is the sequence a1, a2, a3, ... , define ΔA to be the sequence a2 - a1, a3 - a2, a4 - a3,
... . If Δ(ΔA) has all terms 1 and a19 = a92 = 0, find a1.
9. ABCD is a trapezoid with AB parallel to CD, AB = 92, BC = 50, CD = 19, DA = 70.
P is a point on the side AB such that a circle center P touches AD and BC. Find AP.
10. A is the region of the complex plane {z : z/40 and 40/w have real and imaginary
parts in (0, 1)}, where w is the complex conjugate of z (so if z = a + ib, then w = a - ib).
(Unfortunately, there does not appear to be any way of writing z with a bar over it in
HTML4). Find the area of A to the nearest integer.
11. L, L' are the lines through the origin that pass through the first quadrant (x, y > 0)
and make angles π/70 and π/54 respectively with the x-axis. Given any line M, the line
R(M) is obtained by reflecting M first in L and then in L'. Rn(M) is obtained by applying
R n times. If M is the line y = 19x/92, find the smallest n such that Rn(M) = M.
12. The game of Chomp is played with a 5 x 7 board. Each player alternately takes a bite
out of the board by removing a square any and any other squares above and/or to the left
of it. How many possible subsets of the 5 x 7 board (including the original board and the
empty set) can be obtained by a sequence of bites?
13. The triangle ABC has AB = 9 and BC/CA = 40/41. What is the largest possible area
for ABC?
14. ABC is a triangle. The points A', B', C' are on sides BC, CA, AB and AA', BB', CC'
meet at O. Also AO/A'O + BO/B'O + CO/C'O = 92. Find (AO/A'O)(BO/B'O)(CO/C'O).
15. How many integers n in {1, 2, 3, ... , 1992} are such that m! never ends in exactly n
zeros?
1
2
3
4
5
6
7
8
400
502
164
62
660
156
320
819
9
10
11
12
13
14
15
161/3
572
945
792
Outline solutions
1. Sum of such rationals < 1 is (1+7+11+13+17+19+23+29)/30 = 4. The rationals
between 1 and 2 are all 1 larger, so sum 12. Similarly, those between 2 and 3 have sum
20, so total sum = 4 + 12 + 20 + ... + 76 = 400.
2. Each of the 9 non-zero digits can be in the number once or not. The order of the digits
is then forced, so that gives 29 = 512 numbers, but the 1 zero-digit number and the 9 onedigit numbers are not allowed, giving 502 valid numbers.
3. Initially he won n out of 2n, then n+3 out of 2n+4 with n+3 > 0.503(2n+4), or 0.006n
< 0.988, n < 164.6. Thus largest n is 164.
4. Suppose the first coefficient is n!/(r!(n-r)!), then 4/3 = (n-r)/(r+1), 5/4 = (n-r-1)/(r+2),
so n = 62, r = 26.
5. We have 0.abcabcabc ... = abc/999. If abc is not divisible by 3 or 37, then this is in
lowest terms. There are 333 multiples of 3, 27 multiples of 37, and 9 multiples of both,
hence 999-333-27+9 = 648 which are neither. The 12 numbers which are multiples of 81
reduce to multiples of 3. There are no numbers which are multiples of 372, so we cannot
get numerators which are multiples of 37. Thus 648 + 12 = 660 in total.
6. For c = 0, 1, 2, 3 or 4 and denoting c+1 by c', we see that 1abc + 1abc' has no carry iff
a, b = 0, 1, 2, 3, 4 (total 125 possibilities). If c = 5, 6, 7, 8, then there must be a carry.
Suppose c = 9. Again a, b = 0, 1, 2, 3, 4 have no carry (25 possibilities). If b = 9, then a =
0, 1, 2, 3, 4, 9 have no carry (6 possibilities). Grand total 125 + 25 + 6.
7. The perpendicular from D to BC must have length 16 (then area BCD = ½16·10 =
80). Hence the perpendicular from D to ABC has length 16 sin 30o = 8. Hence vol =
8·120/3 = 320.
8. Examining small terms we soon guess that an = (n-1)a2 - (n-2)a1 + ½(n-1)(n-2), or
equivalently that an = ½n2 + an + b. But a19 = a92 = 0, so the polynomial must be (n-19)(n92)/2 which has value 9·91 = 819 at n = 1.
9. Extend AD, BC to meet at X. Then B lies on the bisector of ∠ AXB. But AP/PB =
AX/BX. By similar triangles AX/BX = DX/CX = AD/BC = 7/5. Hence AP = (7/12)92 =
161/3.
10. Re(z/40) ∈ (0, 1), Im(z/40) ∈ (0, 1) simply means that if z = a + ib, then 0 < a, b <
40. Now 40/(a-ib) = 40a/(a2+b2) + 40bi/(a2+b2). So we require 0 < a, 0 < b, a2+b2 > 40a,
or (a-20)2 + b2 > 202, and similarly a2 + (b-20)2 > 202.
So we want the area inside the square outside the two circles. The overlap has area
2(400π/4 - 400/2), so requd area = 1600 - 400π + (200π - 400) = 1200 - 200π = 200(6 - π)
= 200 x 2.858 = 571.6
11. The effect of the two rotations is to change the angle of the line M from θ to θ - π/35
- π/27 = θ - 62π/945. In other words the two reflections give a rotation of 62π/945.
Repeating n times gives a rotation of 62πn/945. We need this to be a multiple of π, so the
smallest n is 945.
12. Take any path along cell edges from the top left to the bottom right (moving an edge
at a time, always right or down). Each such path corresponds to a possible position. A
path is a string of 5 moves down and 7 moves right, so 12C5 = 792 possible paths.
13.
14.
15. There are always enough powers of 2, we are constrained by powers of 5. We jump
powers of 10, at multiples of 25. So 24! ends in 4 zeros (one 5 each from 5, 10, 15, 20),
but 25! ends in 6 zeros. Similarly, multiples of 125 skip two, multiples of 625 skip three
etc. So consider 7980! It has 1596 multiples of 5, including 319 multiples of 25, 63
multiples of 125, 12 multiples of 625, and 2 multiples of 3125. Hence total of
1596+319+63+12+2 = 1992. It has skipped 319+63+12+2 = 396.
1. How many even integers between 4000 and 7000 have all digits different?
2. Starting at the origin, an ant makes 40 moves. The nth move is a distance n2/2 units.
Its moves are successively due E, N, W, S, E, N ... . How far from the origin does it end
up?
3. In a fish contest one contestant caught 15 fish. The other contestants all caught less. an
contestants caught n fish, with a0 = 9, a1 = 5, a2 = 7, a3 = 23, a13 = 5, a14 = 2. Those who
caught 3 or more fish averaged 6 fish each. Those who caught 12 or fewer fish averaged
5 fish each. What was the total number of fish caught in the contest?
4. How many 4-tuples (a, b, c, d) satisfy 0 < a < b < c < d < 500, a + d = b + c, and bc ad = 93?
5. Let p0(x) = x3 + 313x2 - 77x - 8, and pn(x) = pn-1(x-n). What is the coefficient of x in
p20(x)?
6. What is the smallest positive integer that can be expressed as a sum of 9 consecutive
integers, and as a sum of 10 consecutive integers, and as a sum of 11 consecutive
integers?
7. Six numbers are drawn at random, without replacement, from the set {1, 2, 3, ... ,
1000}. Find the probability that a brick whose side lengths are the first three numbers can
be placed inside a box with side lengths the second three numbers with the sides of the
brick and the box parallel.
8. S has 6 elements. How many ways can we select two (possibly identical) subsets of S
whose union is S?
9. Given 2000 points on a circle. Add labels 1, 2, ... , 1993 as follows. Label any point 1.
Then count two points clockwise and label the point 2. Then count three points clockwise
and label the point 3, and so on. Some points may get more than one label. What is the
smallest label on the point labeled 1993?
10. A polyhedron has 32 faces, each of which has 3 or 5 sides. At each of it s V vertices
it has T triangles and P pentagons. What is the value of 100P + 10T + V? You may
assume Euler's formula (V + F = E + 2, where F is the number of faces and E the number
of edges).
11. A and B play a game repeatedly. In each game players toss a fair coin alternately.
The first to get a head wins. A starts in the first game, thereafter the loser starts the next
game. Find the probability that A wins the sixth game.
12. A = (0, 0), B = (0, 420), C = (560, 0). P1 is a point inside the triangle ABC. Pn is
chosen at random from the midpoints of Pn-1A, Pn-1B, and Pn-1C. If P7 is (14, 92), find the
coordinates of P1.
13. L, L' are straight lines 200 ft apart. A and A' start 200 feet apart, A on L and A' on L'.
A circular building 100 ft in diameter lies midway between the paths and the line joining
A and A' touches the building. They begin walking in the same direction (past the
building). A walks at 1 ft/sec, A' walks at 3 ft/sec. Find the amount of time before they
can see each other again.
14. R is a 6 x 8 rectangle. R' is another rectangle with one vertex on each side of R. R'
can be rotated slightly and still remain within R. Find the smallest perimeter that R' can
have.
15. The triangle ABC has AB = 1995, BC = 1993, CA = 1994. CX is an altitude. Find
the distance between the points at which the incircles of ACX and BCX touch CX.
1
2
3
4
5
6
7
728
580
943
870
763
495
1/4
9
10
11
12
13
14
15
118
250
364/729
(336,8)
160/3 sec
√448
332/665
8 365
Outline solutions
1. Suppose the first digit is even. Then there are 2 possibilities for the first digit, 4 for
the last, 8 for the second and 7 for the third, total 448. Similarly, if the first digit is odd,
then it must be 5 and there are 5·8·7 = 280 possibilities. Total 728.
2. Final x-coordinate = (12 - 32 + 52 - 72 + ... -392)/2 = -4(1 + 3 + 5 + ... + 19) = -400.
Final y-coordinate = (22 - 42 + 62 - ... + 402)/2 = -2(3 + 7 + 11 + ... + 39) = -420. So
distance = 20√(202 + 212) = 20·29 = 580.
3. N fish and M contestants. Then 6 fish average: 6(M - 21) = N - 19, and 5 fish
average: 5(M - 8) = N - 108. Hence N = 943.
4. Put b = a + n, c = a + n + m. Then d = a + 2n + m. So 93 = bc - ad = n(n + m). So n =
1, n+m = 93 or n = 3, n+m = 31. There are 405 tuples (a, a+1, a+93, a+94) and 465 tuples
(a, a+3, a+31, a+34)
5. p20(x) = p19(x-20) = p18(x-20-19) = ... = p0(x - (1 + 2 + ... + 20)) = p0(x-210). So
coefficient is 3·2102 - 313·2·210 - 77 = 763.
6. n + (n+1) + (n+2) + ... + (n+8) = 9(n+4), n + (n+1) + ... + (n+9) = 5(2n+9), n + (n+1)
+ ... + (n+10) = 11(n+5). So N must be divisible by 9, 5 and 11. Conversely, 5·9·11 = 495
works.
7. Let the numbers be a1 > a2 > a3 > a4 > a5 > a6. There are 20 ways of choosing 3 of
these for the dimensions of the brick, all equally likely. For the brick to fit, a1 must be a
box side, and a6 must be a brick side. If a2 is a box side, then the brick certainly fits,
giving 3 possibilities for the box (a1, a2, a3 or a1, a2, a4 or a1, a2, a5). If a2 is a brick side,
then a3 must be a box side, giving 2 possibilities for the box (a1, a3, a4 or a1, a3, a5).
8. Let the subsets be A and B. For each element we have three choices (A, B or both).
That gives each pair of subsets twice except for the case A = B = S. Hence (36 + 1)/2.
9. We must have n(n+1)/2 = 1993·1994/2 mod 2000, or (1993-n)(1994+n) = 0 mod
2000. Just one of 1993-n, 1994+n is odd, so either 1993-n or 1994+n is a multiple of 125
and the other is a multiple of 16. If 1993-n is a multiple of 125, then n = 118 mod 125
and 6 mod 16, and smallest is obviously 118. If 1993-n is a multiple of 16, then n = 9
mod 16 and 6 mod 125, so smallest is > 118.
10. V + 30 = E (Euler). VT + VP = 2E (counting edges). So V(T+P-2) = 60. Also, there
are VT/3 triangles and VP/5 pentagons, so V(T/3 + P/5) = 32. Hence 32V(T+P-2) =
32·60 = 60V(T/3 + P/5). Hence 3T + 5P = 16. Hence P + T = 2. Hence V = 30.
11. Chance A wins the first game is ½(1 + ¼ + ¼2 + ... ) = 2/3. Let pn be chance A wins
the nth game. Then it is an easy induction that pn = (1 + (-1)n+1/3n)/2
12. Working backwards we find that the requirement that Pn lies inside the triangle
forces P6 = (28, 184), P5 = (56, 368), P4 = (112, 316), P3 = (224, 212), P2 = (448, 4), P1 =
(336, 8).
13. We may assume A starts at (0, 0) and reaches X (t, 0) after t sec, and that A' starts at
(0, 200) and reaches X' (3t, 200) after t sec. The center of the building O is at (50, 100).
The line XX' is 100x - ty - 100t = 0, so O is a distance |100·50-100t-100t|/√(1002+t2)
from the line. We want this to be 50, so 15t2 = 800t.
14. Let R be ABCD, center O. Let R' be PQRS, as shown. Let ∠ ASP = θ. Then ∠ BRS =
∠ CQR = ∠ DPQ = θ. So ASP and CQR are similar, but PS = QR, so AP = CR. Hence PR
passes through O. But ∠ PQR = 90o, so PR is a diameter of the circle PQRS. Similarly for
QS. Hence O is the center of the circle PQRS.
We have AB + BC = (PS cos θ + RS sin θ) + (RS cos θ + QR sin θ), so perimeter R' =
28/(sin θ + cos θ) = 14√2/sin(θ+45o), which decreases as θ increases. Evidently the
rectangle shown can be rotated, but the other rectangle with the same circle (and the same
Q, S say) cannot. So the largest θ with a rotating rectangle occurs when P is the midpoint
of AD. That implies area R' = ½ area R = 24, so PQ·QR = 24, PQ2 + QR2 = 64. Hence
(PQ + QR)2 = 64+2·24 = 112.
15. We have AC = AU + CU = AT + CR = (AX - r) + (CX - r), where r is the inradius
of ACX. Put AX = x, CX = z, AC = b. Then 2r = x + z - b. Similarly, if r' is the inradius
of BCX and BX = y, then 2r' = y + z - a, so 2RS = |2r - 2r'| = |x - y - b + a| = |x - y - 1|.
Now x2 = b2 - z2, y2 = a2 - z2, so x2 - y2 = b2 - a2, and x - y = (b2 - a2)/c = 3987/1995.
Hence RS = 996/1995.
1. The sequence 3, 15, 24, 48, ... is those multiples of 3 which are one less than a square.
Find the remainder when the 1994th term is divided by 1000.
2. The large circle has diameter 40 and the small circle diameter 10. They touch at P. PQ
is a diameter of the small circle. ABCD is a square touching the small circle at Q. Find
AB.
3. The function f satisfies f(x) + f(x-1) = x2 for all x. If f(19) = 94, find the remainder
when f(94) is divided by 1000.
4. Find n such that [log21] + [log22] + [log23] + ... + [log2n] = 1994.
5. What is the largest prime factor of p(1) + p(2) + ... + p(999), where p(n) is the product
of the non-zero digits of n?
6. How many equilateral triangles of side 2/√3 are formed by the lines y = k, y = x√3 +
2k, y = -x√3 + 2k for k = -10, -9, ... , 9, 10?
7. For how many ordered pairs (a, b) do the equations ax + by = 1, x2 + y2 = 50 have (1)
at least one solution, and (2) all solutions integral?
8. Find ab if (0, 0), (a, 11), (b, 37) is an equilateral triangle.
9. A bag contains 12 tiles marked 1, 1, 2, 2, ... , 6, 6. A player draws tiles one at a time at
random and holds them. If he draws a tile matching a tile he already holds, then he
discards both. The game ends if he holds three unmatched tiles or if the bag is emptied.
Find the probability that the bag is emptied.
10. ABC is a triangle with ∠ C = 90o. CD is an altitude. BD = 293, and AC, AD, BC are
all integers. Find cos B.
11. Given 94 identical bricks, each 4 x 10 x 19, how many different heights of tower can
be built (assuming each brick adds 4, 10 or 19 to the height)?
12. A 24 x 52 field is fenced. An additional 1994 of fencing is available. It is desired to
divide the entire field into identical square (fenced) plots. What is the largest number that
can be obtained?
13. The equation x10 + (13x - 1)10 = 0 has 5 pairs of complex roots a1, b1, a2, b2, a3, b3, a4,
b4, a5, b5. Each pair ai, bi are complex conjugates. Find ∑ 1/(aibi).
14. AB and BC are mirrors of equal length. Light strikes BC at C and is reflected to AB.
After several reflections it starts to move away from B and emerges again from between
the mirrors. How many times is it reflected by AB or BC if ∠ b = 1.994o and ∠ a =
19.94o?
At each reflection the two angles x are equal:
15. ABC is a paper triangle with AB = 36, AC = 72 and ∠ B = 90o. Find the area of the
set of points P inside the triangle such that if creases are made by folding (and then
unfolding) each of A, B, C to P, then the creases do not overlap.
1
2
3
4
5
6
7
8
63
8+4√19
561
312
103
660
72
315
9
10
11
12
13
14
15
9/385
29/421
465
702
850
71
270π - 324√3
Outline solutions
1. The terms are (3·1-1)2-1,(3·1+1)2-1,(3·2-1)2-1,(3·2+1)2-1, ... . So 1994th term is
(3·997+1)2-1 = (-8)2-1 = 63 mod 1000.
2. Let AB=x. Then considering triangle with hypoteneuse AO (where O is the center of
the large circle): 202=(x/2)2+(x-10)2, so x=8+4√19.
3. f(94)=942-f(93)=942-932+f(92)=942-932+922-f(91)= ... = (942-932) + (922-912) +...+
(222-212)+ 202-f(19) = 94+93+...+21+400-94 = 115·37+306 = 4561.
4. [log21]=0, [log22]=[log23]=1, [log24]=[log25]=[log26]=[log27]=2 etc. SO sum of first
(1+2+4+8+16+32+64+128)=255 terms is (1·0+2·1+4·2+8·3+16·4+32·5+64·6+128·7) =
2+8+24+64+160+384+896=1538. The next two thousand terms are all 8. We need
another 1994-1538=8·57, so another 57 terms. Hence n =312.
5. If we expand (1+1+2+3+...+9)3 we get the required sum, except that it also includes
1·1·1 for 000. Thus we get 463-1 = 45·2163 = 45(3·7·103), largest prime 103.
6. Between the lines y = 10 and y = -10 we have lines parallel to them, but outside them
we do not. Similarly for the lines y = x√3 ± 20, and for the lines y = -x√3 ± 20. Thus the
area where triangles are formed is the hexagon bounded by these six lines. It has long
diagonal length 40/√3 from -20/√3 (the intersection of y = 0, y = x√3 + 20 and y = -x√3 20) to 20/√3 (the intersection of y = 0, y = x√3 - 20 and y = -x√3 + 20). So we can regard
it as made up of 6 equilateral triangles side 20/√3. Each of these is divided into
equilateral triangles side 2/√3. Each has side 1/10 of the large triangle, so area 1/100, so
there are 100 of them, or 600 in all. But there is a trap. There is a line of triangles outside
each edge of the hexagon (with bases on the hexagon). Each edge has 10 triangles, so 60
in all.
7. There are two ways of writing 50 as a sum of two squares, namely 52 + 52 and 72 + 12.
So there are 12 lattice points on the circle x2 + y2 = 50, namely 4 points (±5, ±5), 4 points
(±1, ±7) and 4 points (±7, ±1). As lines ax + by = 1 we can take the tangent at each point
(12 lines) and the line joining any pair which are not diametrically opposite. There are
12·10/2 = 60 such pairs. So 72 lines in all.
8. Using a2+112=b2+372=(a-b)2+262 is a poor approach. It can be done (take h=a+b,
k=a-b, then hk = a2-b2 = 1248, so we get a and b in terms of k. Then get a quartic in k
which factorises easily with only one positive real root). Better is to use angles. Put O =
(0,0), A = (a,11), B = (b,37). Take angle between OA and x-axis to be x, and the side of
the triangle to be t. Then 37/t = sin(x+60o) = sin x cos 60o + cos x sin 60o = 11/2r +
a√3/2r. Hence a = 21√3. Similarly, if y is the angle between OB and the y-axis, then
considering sin(y+60o), we get b = 5√3.
9. Let pn be the prob for n pairs of tiles. We calculate the prob. that first 3 tiles contain a
pair. No. of ways of choosing the pair type is n, no. of ways of choosing odd tile is 2n-2,
so n(2n-2) in total. No. of ways of choosing 3 tiles (order irrelevant) is 2nC3 (binomial
coeff (2n)!/3!(2n-3)! ). So prob. n(2n-2)(6/2n·2n-1·2n-2) = 3/(2n-1). At that point we
have picked one tile from the remaining n-1 pairs. So the prob. of emptying the bag is pn1. Thus pn = (3/2n-1) pn-1. Obviously p2 = 1, so p6 = (3/11)(3/9)(3/7)(3/5) = 9/385.
10. BD/BC = BC/AB (similar triangles), so AB·BD = BC2. But AB and BC are integers,
so for some n we have AB = 29n2, BC = 292n. Also AD/AC = AC/AB, so AC2 = AB·AD
= 29n229(n2-292). But AC is an integer, so n2 - 292 = m2 for some integer m. Hence (nm)(n+m) = 292. Hence n+m = 292, n-m = 1, so n = 421. Hence cos B = BD/BC = 29/n =
29/421.
11. Suppose x bricks are oriented to add 4 to the height, y to add 19 and z to add 10.
Since 5·10 = 3·4 + 2·19 we can take z = 0, 1, 2, 3, or 4. Also we must have y = 94-x-z, so
height = 4x+10z+19(94-x-z) = 1786-15x-9z. If z=0, height = 1786-15x = 1 mod 5; if z=1,
height = 1777-15x = 2 mod 5; if z=2, height = 1768-15x = 3 mod 5; if z=3, height =
1759-15x = 4 mod 5; if z=4, height = 1750-15x = 0 mod 5. So these values of x,z all give
different heights. We can check that 1750 > 15·94 = 1510, so for z=0 there are 95
possible values of x (0, 1, ... , 94), for z=1, 94 etc. Hence in total 95+94+93+92+91 = 465
possible heights.
12. m long fences, n short fences, so 52m+24n ≤ 1994. The small plots must be square,
so 52/(n+1) = 24/(m+1). 52 is divisible by 13 and 24 is not, so n+1 must be a multiple of
13. n = 38 gives m = 17 and 1796 of fencing. The next highest possibility gives n = 51, m
= 23 and 2420 of fencing. So n = 38, m = 17 and 39·18 = 702 plots.
13. A root of the polynomial satisfies (13 - 1/x)10 = -1. So 13 - 1/ak = eiθk, where θ1 =
18o, θ2 = 54o, θ3 = 90o, θ4 = 126o, θ5 = 162o. Hence 1/akbk = (13 - eiθk)(13 - e-iθk) = 170 26 cos θk. So ∑ 1/akbk = 850 - 26(cos 18o + cos 54o + cos 90o + cos 126o + cos 162o). But
cos 90o = 0 and cos 162o = - cos 18o, cos 54o = - cos 126o, so ∑ 1/akbk = 850.
14.
It is convenient to have a series of phantom mirrors with angle b between each. Then we
can regard the path as a straight line. It stops reflecting when the distance from B exceeds
BC. After n+1 reflections it is at a mirror BD with ∠ CBD = nb. The distance exceeds BC
when ∠ CBD > 180o-2a. Now 2a/b = 20 and 90·1.994 < 180o, and 91·1.994 > 180o, so we
get 1+90-20 = 71 reflections.
15.
The folds to bring A and B to P are the perpendicular bisectors of AP and BP, which
meet at O. We want O to lie below AB. Note that O is the circumcenter of ABP, so
∠ APB = ½∠ AOB. We want ∠ AOB > 180o, so ∠ APB > 90o, so P must lie inside the
semicircle with AB as diameter.
Similarly P must lie inside the semicircles on AC and BC. The semicircle on AC does not
impose any constraint since B lies on it and so the entire triangle lies inside it. Thus we
want the area of the intersection of the semicircles on AB and BC. The upper arc BC has
radius 18 and subtends an angle 120o, so the area between it and the chord BD is
(1/3)π182 - 182cos 30o sin 30o = 108π - 81√3. Similarly, the other arc has radius 18√3 and
angle 60o, so area (1/6)π3·182 - 243√3. So total area 270π - 324√3.
1. Starting with a unit square, a sequence of square is generated. Each square in the
sequence has half the side-length of its predecessor and two of its sides bisected by its
predecessor's sides as shown. Find the total area enclosed by the first five squares in the
sequence.
2. Find the product of the positive roots of √1995 xlog1995x = x2.
3. A object moves in a sequence of unit steps. Each step is N, S, E or W with equal
probability. It starts at the origin. Find the probability that it reaches (2, 2) in less than 7
steps.
4. Three circles radius 3, 6, 9 touch as shown. Find the length of the chord of the large
circle that touches the other two.
5. Find b if x4 + ax3 + bx2 + cx + d has 4 non-real roots, two with sum 3 + 4i and the
other two with product 13 + i.
6. How many positive divisors of n2 are less than n but do not divide n, if n = 231319?
7. Find (1 - sin t)(1 - cos t) if (1 + sin t)(1 + cos t) = 5/4.
8. How many ordered pairs of positive integers x, y have y < x ≤ 100 and x/y and
(x+1)/(y+1) integers?
9. ABC is isosceles as shown with the altitude AM = 11. AD = 10 and ∠ BDC = 3
∠ BAC. Find the perimeter of ABC.
10. What is the largest positive integer that cannot be written as 42a + b, where a and b
are positive integers and b is composite?
11. A rectangular block a x 1995 x c, with a ≤ 1995 ≤ c is cut into two non-empty parts
by a plane parallel to one of the faces, so that one of the parts is similar to the original.
How many possibilities are there for (a, c)?
12. OABCD is a pyramid, with ABCD a square, OA = OB = OC = OD, and ∠ AOB =
45o. Find cos θ, where θ is the angle between two adjacent triangular faces.
13. Find ∑11995 1/f(k), where f(k) is the closest integer to k¼.
14. O is the center of the circle. AC = BD = 78, OA = 42, OX = 18. Find the area of the
shaded area.
15. A fair coin is tossed repeatedly. Find the probability of obtaining five consecutive
heads before two consecutive tails.
1
2
3
4
5
6
7
8
1279/1024
19952
3/64
√224
51
589
13/4 - √10
85
9
10
11
12
13
14
15
11+11√5
215
40
-3 + 2√2
400
294π - 81√3
3/34
Outline solutions
1. Square area 1, ¼, ¼2 etc. ¾ of each square does not overlap earlier squares. So total
area = 1 + ¾(¼ + ¼2 + ¼3 + ¼4) = 1 + ¾(64/256 + 16/256 + 4/256 + 1/256) = 1 +
255/1024.
2. Take logs (all to base 1995). We get ½ + log2x = 2 log x. So log x = 1 + √½ or 1 - √½.
Hence x = 19951+√½ or 19951-√½, both positive reals, product 19952.
3. It needs either 4 steps (N, N, E, E in some order) or 6 steps (also a pair of steps in
opposite directions). The first has prob 6/44. The second has NNEENS or NNEEEW in
some order. There are 6!/3!2!1! = 60 possibilities for NNEENS. But there is a trap. In
some of those 60 we already reach (2, 2) on step 4, so we are double-counting. In fact it is
easy to see there are 12 such possibilities. So we have 48 possibilities. Similarly for
NNEEEW, so 96 in all, prob 96/46 = 6/44, so total prob 3/64.
4. XAC and XBD are similar, so XA = 9. Let O be the center of the large circle. Then
XOM is also similar and XO = 15. Hence OM = (15/9) AC = 5. Hence if the chord length
is 2x, we have x2 + 52 = 92, so x = √56.
5. Let roots with sum 3 + 4i be α, β. Their sum is not real, so they are not complex
conjugates. But the roots occur in complex conjugate pairs (since the polynomial is real),
so the other roots γ, δ must be the complex conjugates of α and β. Hence γ + δ = 3 - 4i
and αβ = 13 - i. Hence b = (α + β)(γ + δ) + αβ + γδ = 25 + 26 = 51.
6. n2 has 63·39 factors. Apart from the factor n, they form pairs d, n2/d, with one factor
of each pair < n. Hence there are (63·39-1)/2 = 1228 factors of n2 less than n. There are
(32·20-1) = 639 factors of n less than n, all of which are obviously factors of n2. Hence
there are 1228 - 639 = 589.
7. Put k = (1 - sin t)(1 - cos t). Then 5k/4 = cos2t sin2t. But 5/4 + k = 2 + 2 sin t cos t, so
5k = (k - 3/4)2. Solving k = 13/4 ± √10. But k ≤ 2, so k = 13/4 - √10.
8. x/y - (x+1)/(y+1) = (x-y)/(y(y+1)) must also be an integer, so x must be y plus a
multiple of y(y+1). Thus we get:
y = 1, x = 3, 5, 7, ... , 99 (49 solutions)
y = 2, x = 8, 14, ... , 98 (16 solutions)
y = 3, x = 15, 27, ... , 99 (8 solutions)
y = 4, x = 24, 44, 64, 84 (4 solutions)
y = 5, x = 35, 65, 95 (3 solutions)
y = 6, x = 48, 90 (2 solutions)
y = 7, x = 63
y = 8, x = 80
y = 9, x = 99
9. Let tan BAM = x. Then 11x = tan BDM = (3x - x3)/(1 - 3x2), so x = 0, ½ or -½. Hence
x = ½, and BM = 11/2. Hence AB = (11/2)√5 and perimeter = 11 + 11√5.
10. We need only consider 42a + 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41. Note
that 84+1, 42+2, 42+3, 42+7, 84+11, 42+13, 126+17, 126+19, 42+23, 126+29, 84+31,
84+37, 84+41 are composite, so the tricky case is 42n + 5. We have to go to 5·42 + 5 to
get a composite number. (So 6·42 + 5 = 42 + (5·42+5), but 215 = 5·42 + 5 cannot be
written in the required form).
11. We must cut the longest edges, so the similar piece has dimensions a x 1995 x k for
some 1 ≤ k < c. The shortest edge of this piece cannot be a, so it must be k. Thus a x 1995
x c and k x a x 1995 are similar. Hence c = 19952/a, k = a2/1995. Now 1995 = 3·5·7·19,
so 19952 has 34 factors, of which (34-1)/2 = 40 are < 1995.
12. Let AX be the perpendicular from A to OB. Let AX = 1, then OX = 1, OA = √2, so
BX = √2 - 1. Hence AB2 = 1 + 3 - 2√2, and AC2 = 8 - 4√2. But AC2 = AX2 + CX2 2·AX·CX cos θ, so cos θ = -3 + 2√2.
13. f(k) must take one of the values 1, 2, ... , 7. The boundary values are (n + ½)4 = n4 +
2n3 + ½(3n2 + n) + 1/16, giving 5 1/16, 39 1/16, 150 1/16, 410 1/16, 915 1/16, 1785 1/16.
So the sum is 5 + 34/2 + 111/3 + 260/4 + 505/5 + 870/6 + 210/7 = 400.
14. OY2 = OA2 - AY2, so OY = 9√9. XY2 = OX2 - OY2, so XY = 9 and ∠ OXY = 60o.
Hence DX = AX = AY + XY = 48, BX = CX = 30. Also ∠ AXB = 180o - 2·60o = 60o. So
area triangle AXB = ½AX·BX sin 60o = 360√3. ∠ ACB = ∠ XBC and their sum is ∠ AXB
= 60o, so ∠ ACB = 30o. Hence ∠ AOB = 60o. So area sector OAB = 422π/6 and area
between AB and minor arc AB = 294π - 422(√3)/4. Hence required area = 294π - 81√3.
15. Let p = prob that first toss is H and 5H occurs before 2T, q = prob that first toss is T
and 5H occurs before 2T. We have q = ½p, because the T must be followed by an H.
Also p = q/2 + q/4 + q/8 + q/16 + 1/32 (there are either 1, 2, 3, 4, or 5 Hs in the initial
run) = (15/16)q + 1/32 = (15/32)p + 1/32. So p = 1/17, q = 1/34.
1. The square below is magic. It has a number in each cell. The sums of each row and
column and of the two main diagonals are all equal. Find x.
2. For how many positive integers n < 1000 is [log2n] positive and even?
3. Find the smallest positive integer n for which (xy - 3x - 7y - 21)n has at least 1996
terms.
4. A wooden unit cube rests on a horizontal surface. A point light source a distance x
above an upper vertex casts a shadow of the cube on the surface. The area of the
shadow (excluding the part under the cube) is 48. Find x.
5. The roots of x3 + 3x2 + 4x - 11 = 0 are a, b, c. The equation with roots a+b, b+c,
c+a is x3 + rx2 + sx + t = 0. Find t.
6. In a tournament with 5 teams each team plays every other team once. Each game
ends in a win for one of the two teams. Each team has ½ chance of winning each
game. Find the probability that no team wins all its games or loses all its games.
7. 2 cells of a 7 x 7 board are painted black and the rest white. How many different
boards can be produced (boards which can be rotated into each other do not count as
different).
8. The harmonic mean of a, b > 0 is 2ab/(a + b). How many ordered pairs m, n of
positive integer with m < n have harmonic mean 620?
9. There is a line of lockers numbered 1 to 1024, initially all closed. A man walks
down the line, opens 1, then alternately skips and opens each closed locker (so he
opens 1, 3, 5, ... , 1023). At the end of the line he walks back, opens the first closed
locker, then alternately skips and opens each closed locker (so he opens 1024, skips
1022 and so on). He continues to walk up and down the line until all the lockers are
open. Which locker is opened last?
10. Find the smallest positive integer n such that tan 19no = (cos 96o + sin 96o)/(cos
96o - sin 96o).
11. Let the product of the roots of z6 + z4 + z3 + z2 + 1 = 0 with positive imaginary
part be r(cos θo + i sin θo). Find θ.
12. Find the average value of |a1 - a2| + |a3 - a4| + |a5 - a6| + |a7 - a8| + |a9 - a10| for all
permutations a1, a2, ... , a10 of 1, 2, ... , 10.
13. AB = √30, BC = √15, CA = √6. M is the midpoint of BC. ∠ ADB = 90o. Find area
ADB/area ABC.
14. A 150 x 324 x 375 block is made up of unit cubes. Find the number of cubes
whose interior is cut by a long diagonal of the block.
15. ABCD is a parallelogram. ∠ BAC = ∠ CBD = 2 ∠ DBA. Find ∠ ACB/∠ AOB,
where O is the intersection of the diagonals.
1
2
3
4
5
6
7
8
200
340
44
1/6
23
17/32
300
799
9
10
11
12
13
14
15
342
159
276
55/3
27/38
768
7/9
Outline solutions
1. We find successively, a = 114, b = x - 95, c = 191, d = x - 190. Then 115 = b + d = 2x
- 285, so x = 200.
2. [log2n] = 2 for n = 4, 5, 6, 7; 4 for n = 16, 17, ... , 31; 6 for n = 64, 65, ... , 127; 8 for n
= 256, 257, ... , 511. So 4 + 16 + 64 + 256 = 340.
3. There are potentially (n+1)2 terms, namely xayb for 0 ≤ a, b ≤ n. So to get at least
1996 we need n = 44. The only problem is whether some of these terms have zero
coefficient. But if (xy)axbyc = xyAxByC, then a+c = A+C. Signs are (-1)b(-1)n-a-b-c = (-1)n-ac
, and (-1)n-A-C, so they are equal. So no terms vanish.
4. By similar triangles the total shadow (including the part under the cube) is a square
with side (1+x)/x. So (1+x)2/x2 = 49, or (8x+1)(6x-1) = 0.
5. a+b+c = -3, so the required equation has roots -3-a, -3-b, -3-c. If y = -3-x, then (-3-y)3
+ 3(-3-y)2 + 4(-3-y) - 11 = 0, or y3 + ... + 27-27+12+11, so t = 23.
6. Prob team A wins all 4 games is 1/16. Two teams cannot both win all their games, so
prob some team wins all 4 games is 5/16. Similarly prob some team loses all 4. But these
overlap. Prob team A wins all 4 games, and team B loses all 4 games is 1/128 (7 games
determined). So prob some team wins all 4 and some team loses all 4 is 20/128 = 5/32.
So prob some team wins all 4 or some team loses all 4 is 10/16 - 5/32 = 15/32. Hence
requd prob is 17/32.
7. There are 49·48/2 = 1176 ways of choosing 2 cells. In 24 cases the two cells are
diametrically opposite. In the other 1152 they are not. The 1152 form groups of 4 related
by rotation, so 288 distinct. The 24 form 12 groups of 2. So total 300.
8. We have mn = 219320m + 219320n, or (m - 219320)(n - 219320) = 238340. Now 238340 has
39·41 factors. Leaving aside the square root 219320 which does not give a solution, the
factors form 799 pairs (one < square root and one >). Each pair gives a solution m, n.
9. Write the labels in binary. 1st pass, open those ending in 1, leaving those ending in 0
still closed. 2nd pass, open those ending in 00, leaving -10 closed. 3rd pass open 010,
leaving -110 closed. 4th pass open -1110, leaving -0110 closed. 5th pass, open -00110,
leaving -10110. 6th pass open -110110, leaving -010110. Continuing, we leave the single
locker 0101010110 = 342.
10. (cos 96o + sin 96o)/(cos 96o - sin 96o) = (1 + tan 96o)/(1 - tan 96o). Now we need a
trick, which is that tan 45o = 1, so we can write the expression as (tan 45o + tan 96o)/(1 tan 45otan 96o) = tan 141o. Now tan 19no = tan 141o iff 19n = 141 mod 180. We have 180
= 9·19+9, 19 = 2·9 + 1. So 1 = 19 - 2·9 = 19 - 2(180 - 9·19) = 19·19 - 2·180. So 141 =
(19·141)19 - 282·180, or 141 = 19(19·141) mod 180. But 19·141 = 14·180 + 159, so 141
= 19·159 mod 180.
11. We need the factorisation: z6 + z4 + z3 + z2 + 1 = (z4 + z3 + z2 + z + 1)(z2 - z + 1). So
the roots are ei72o, ei144o, ei216o, ei288o, ei60o, ei300o. The roots with positive imaginary part
are ei72o, ei144o, ei60o, with product ei276o.
12. By symmetry the average of |ai - aj| is independent of i, j (provided they are
unequal). Suppose ai = k. Then the average of |k - aj| is (k-1 + k-2 + ... + 1 + 1 + 2 + ... +
10-k)/9 = (k2 -11k + 55)/9. So average of |ai - aj| is (1/10) ∑ (k2 -11k + 55)/9. Hence requd
average is ∑ (k2 -11k + 55)/18 = (385 - 605 + 550)/18 = 55/3.
13. We have area ADB/area ABC = area ADB/(2 area ABM) = AD/(2 AM).
14.
15.
1. How many of 1, 2, 3, ... , 1000 can be written as the difference of the squares of
two non-negative integers?
2. The 9 horizontal and 9 vertical lines on an 8 x 8 chessboard form r rectangles
including s squares. Find s/r in lowest terms.
3. M is a 2-digit number ab, and N is a 3-digit number cde. We have 9·M·N = abcde.
Find M, N.
4. Circles radii 5, 5, 8, k are mutually externally tangent. Find k.
5. The closest approximation to r = 0.abcd (where any of a, b, c, d may be zero) of
the form 1/n or 2/n is 2/7. How many possible values are there for r?
6. A1A2...An is a regular polygon. An equilateral triangle A1BA2 is constructed
outside the polygon. What is the largest n for which BA1An can be consecutive
vertices of a regular polygon?
7. A car travels at 2/3 mile/min due east. A circular storm starts with its center 110
miles due north of the car and travels southeast at 1/√2 miles/min. The car enters the
storm circle at time t1 mins and leaves it at t2. Find (t1 + t2)/2.
8. How many 4 x 4 arrays of 1s and -1s are there with all rows and all columns
having zero sum?
9. The real number x has 2 < x2 < 3 and the fractional parts of 1/x and x2 are the
same. Find x12 - 144/x.
10. A card can be red, blue or green, have light, medium or dark shade, and show a
circle, square or triangle. There are 27 cards, one for each possible combination. How
many possible 3-card subsets are there such that for each of the three characteristics
(color, shade, shape) the cards in the subset are all the same or all different?
11. Find [100(cos 1o + cos 2o + ... + cos 44o)/(sin 1o + sin 2o + ... + sin 44o)].
12. a, b, c, d are non-zero reals and f(x) = (ax + b)/(cx + d). We have f(19) = 19, f(97)
= 97 and f(f(x)) = x for all x (except -d/c). Find the unique y not in the range of f.
13. Let S = {(x, y) : | ||x| - 2| - 1| + | ||y| - 2| - 1| = 1. If S is made out of wire, what is
the total length of wire is required?
14. v, w are roots of z1997 = 1 chosen at random. Find the probability that |v + w| >=
√(2 + √3).
15. Find the area of the largest equilateral triangle that can be inscribed in a rectangle
with sides 10 and 11.
1
2
3
4
5
6
7
8
750
17/108
14,112
8/9
417
42
198
90
9
10
11
12
13
14
15
233
117
241
58
64√2
83/499
221√3 - 330
Outline solutions
1. 2n+1 = (n+1)2 - n2 and 4n = (n+1)2 - (n-1)2. Odd squares = 1 mod 4 and even squares
= 0 mod 4, so we cannot get 2 mod 4 by the difference of two squares. Thus we can get
all except 2 = 1·4-2, 6=2·4-2, 10=3·4-2, ... , 998=250·4-2.
2. A rectangle is uniquely defined by two opposite vertices. We can choose one vertex
in 81 ways. That rules out 17 in the same row and column, so we can choose second in 64
ways. So 81·64, but that gives every rectangle four times, so 81·16 rectangles. There is 1
8 x 8 square, 22 7 x 7 squares, 32 6 x 6 squares, ... , 82 1 x 1 squares, total 8·9·17/6 =
12·17 squares. So s/r = 12*middot;17/(81·16) = 17/108.
3. Put f(M, N) for the number abcde = 1000M + N. We have 9·M·111 < f(M, 111). Also
if N >= 123, then 9·M·N >= 9·M·123 = 1000M + 107M >= 1000M + 1070 > f(M,N), so
N must be 112, 113, ... , or 122. Now bootstrap. If N >= 113, then 9·M·N >= 1000M +
17M >= 1000M + 170 > f(M,N). So the only possibility is N = 112. Hence 1000M + N =
1008 M, so M = N/8 = 14.
4. Take triangle formed by centers of circles radii 5, 5, 8, and take center of 4th circle.
Altitude is 12 (5, 12, 13 triangle), so small right-angled triangle has sides 5+k, 4-k, 5.
Hence k = 8/9.
5. Closest fractions 2/n to 2/7 are 2/6 and 2/8. Similarly, closest of form 1/n are 1/3 and
1/4. So r must lie in [(1/4 + 2/7)/2, (1/3 + 2/7)/2] = [15/56, 13/42]. But 15/56 = 0.26786,
13/42 = 0.30952. Thus r = 0.2679, 0.2680, ... , 0.3095, giving 417 possibilities.
6. Angle AnA1A2 = 180o(1 - 2/n), so BA1An = 120o + 360o/n. For n = 42, that gives 128
4/7 = 180o(1 - 2/7). For n > 42, we have 120o < angle BA1An < 128 4/7, in other words
the angle is between that of a hexagon and a heptagon. So 42 is the largest.
7. At t mins, car is at (2t/3, 0), storm center (t/2, 110 - t/2), so dist2 = 10t2/36 - 110t +
1102. We want roots to dist2 = 512. They have sum 110·36/10 = 396.
8. 4C2 = 6 ways of choosing the first row, then 3 ways of choosing the first column. Let
A be top left square, B square in top row with same value, C square in first col with same
value. Let D be square that completes rectangle. If D has same value as A, B, C then all
values fixed. If not, then it is easy to check there are 4 ways of filling the remaining
squares. So 6·3·5 in total.
9. Since 2 < x2 < 3, we have 1 < x < 2, so 1/2 < 1/x < 1 and the fractional part of 1/x is
just 1/x. The fractional part of x2 is x2 - 2, so x3 - 2x - 1 = 0. Taking out the factor x + 1,
and solving x2 - x - 1 = 0, we get x = (√5 + 1)/2. Now using x2 = 1+x, we get
successively x4 = 2+3x, x8 = 13+21x, x12 = 89+144x, 1/x = x-1, so x12 - 144/x = 233.
10. The colors must be RRR, BBB, GGG or RBG. Similarly for the other
characteristics. Suppose the colors are RBG. Suppose the shades are LLL. Then the
shapes can be SSS, CCC, TTT (1 possibility each), or SCT (6 possibilities), total 9,
Similarly for shades MMM, DDD. If the shades are LMD, there are 6 possible orders and
9 possibilities for the shapes in each case, total 54. So 9+9+9+54 = 81 possibilities for
RBG. Suppose the colors are RRR. If the shades are LLL, then the shapes must be SCT
(1 possibility). Similarly for MMM and DDD. If the shades are LMD, then there are
1+1+1+6 possibilities, so 12 in all for RRR. Similarly for BBB and GGG. Hence
81+12+12+12 = 117 in all.
11. 2 cos n sin(1/2) = sin(n + 1/2) - sin(n - 1/2), so the numerator is 50(sin 44.5 - sin
0.5)/sin 0.5. Similarly the denominator is (cos 0.5 - cos 44.5)/(2 sin 0.5), so we get
100(sin 44.5 - sin 0.5)/(cos 0.5 - cos 44.5). Evidently this is roughly 100 (1/√2)(1 - 1/√2)
= 100(√2 + 1) = 241.42. Using sin 44.5 = sin 45 cos 0.5 - sin 0.5 cos 45 etc we get
100(cos 0.5 - (1 + √2)sin 0.5)/( (√2 - 1)cos 0.5 - sin 0.5) which is exactly 100(√2 + 1).
12. f(f(x)) = x simplifies to (a+d)(cx2 + (d-a)x - b) = 0, so d = -a. There is no loss of
generality in taking c = 1. Then f(19) = 19, f(97) = 97 give 38a + b = 361, 194a + b =
9409. Solving, a = 58, b = -1843. The unattained value is a/c.
13. We consider first the first quadrant. The first graph shows | |x| - 2| - 1|. So we need
only consider 0 <= x <= 4. The graph for the second term (in y) is the same, so it is not
hard to see that we get the pattern for x, y shown in the second graph. If (x, y) belongs to
S, then so does (-x, y), (-x, -y), (x, -y). Hence total length = 64√2.
14. Let the angle between the two vectors be 2x.Then length |v+w| = 2 cos x, so require
2 cos2x >= 1 + (√3)/2, or cos 2x >= (√3)/2. Thus the angle between them must not exceed
30o. Since 166 < 1997/12 < 167, there are 2·166 either side of a given vector. Thus prob
= (2·166)/1996 = 83/499.
15. It is straightforward to get a quartic in x. Obviously 100 + y2 = 121 + x2. Then either
use the third side or use the angles: y/10 = (tan 30o - x/11)/(1 + tan 30o (x/11)). Hence x4
+ 22√3 x3 + 84x2 + 2662√3 x - 4477 = 0. This factorises (x2 + 22√3 x - 37)(x2 + 121).
Hence x = 20 - 11√3. So area = (20 - 11√3)2(√3)/4.
1. For how many k is lcm(66, 88, k) = 1212?
2. How many ordered pairs of positive integers m, n satisfy m ≤ 2n ≤ 60, n ≤ 2m ≤ 60?
3. The graph of y2 + 2xy + 40|x| = 400 divides the plane into regions. Find the area of the
bounded region.
4. Nine tiles labeled 1, 2, 3, ... , 9 are randomly divided between three players, three tiles
each. Find the probability that the sum of each player's tiles is odd.
5. Find |A19 + A20 + ... + A98|, where An = ½n(n-1) cos(n(n-1)½π).
6. ABCD is a parallelogram. P is a point on the ray DA such that PQ = 735, QR = 112.
Find RC.
7. Find the number of ordered 4-tuples (a, b, c, d) of odd positive integers with sum 98.
8. The sequence 1000, n, 1000-n, n-(1000-n), ... terminates with the first negative term
(the n+2th term is the nth term minus the n+1th term). What positive integer n maximises
the length of the sequence?
9. Two people arrive at a cafe independently at random times between 9am and 10am
and each stay for m minutes. What is m if there is a 40% chance that they are in the cafe
together at some moment.
10. 8 sphere radius 100 rest on a table with their centers at the vertices of a regular
octagon and each sphere touching its two neighbors. A sphere is placed in the center so
that it touches the table and each of the 8 spheres. Find its radius.
11. A cube has side 20. Two adjacent sides are UVWX and U'VWX'. A lies on UV a
distance 15 from V, and F lies on VW a distance 15 from V. E lies on WX' a distance 10
from W. Find the area of intersection of the cube and the plane through A, F, E.
12. ABC is equilateral, D, E, F are the midpoints of its sides. P, Q, R lie on EF, FD, DE
respectively such that A, P, R are collinear, B, Q, P, are collinear, and C, R, Q are
collinear. Find area ABC/area PQR.
13. Let A be any set of positive integers, so the elements of A are a1 < a2 < ... < an. Let
f(A) = ∑ ak ik. Let Sn = ∑ f(A), where the sum is taken over all non-empty subsets A of
{1, 2, ... , n}. Given that S8 = -176-64i, find S9.
14. An a x b x c box has half the volume of an (a+2) x (b+2) x (c+2) box, where a ≤ b ≤
c. What is the largest possible c?
15. D is the set of all 780 dominos [m,n] with 1≤m<n≤40 (note that unlike the familiar
case we cannot have m = n). Each domino [m,n] may be placed in a line as [m,n] or
[n,m]. What is the longest possible line of dominos such that if [a,b][c,d] are adjacent
then b = c?
1
2
3
4
5
6
7
8
25
480
800
3/14
40
308
19600
618
9
10
11
12
13
14
15
60-12√15
100+50√2
525
7+3√5
-352+16i
130 (3,11)
761
Outline solutions
1. 1212 = 224312, 66 = 2636, 88 = 224. We must have k = 2a3b (since 2, 3 are the only
primes dividing the lcm). a can be any of 0, 1, ... , 24, and b must be 12.
2. There are 302 points (m, n) with 1 ≤ m, n ≤ 60/2. We must exclude those with 2m < n
or 2n < m. By symmetry, there are the same number with 2m < n and 2n < m and none
satisfy both conditions. The number of points satisfying 2m < n is 0 for n = 1, 2; 1 for n =
3, 4; 2 for n = 5, 6; ... ; 14 for n = 29, 30. So total 2(1 + 2 + ... + 14) = 210. Hence requd
no. is 900 - 210 - 210.
3. For x ≥ 0, the equation is (y+20)(y+2x-20) = 0. For x ≤ 0, the equation is (y20)(y+2x+20).
So the bounded region is the parallelogram shown, area 20·40 = 800.
4. There are 5 odd tiles and 3 even tiles. Each player must get an odd number of odd
tiles, so one player must get 3 odd tiles, and the other two 1 odd tile each. There are 9C3
= 84 ways for a player to get 3 tiles from 9 and 5C3 = 10 ways to get 3 odd. Hence prob
5/42 for a given player to get all odd. There remain 6 tiles, 2 odd. So the prob the next
player has prob (1/3)(4/5)(3/4) x 3 = 3/5 of getting just 1 odd (and then the last player
must get 1 odd). Hence prob. (5/42)(3/5) = 1/14. But any of the three players can get all
odd (and these are disjoint poss), so total prob 3/14.
5. ½n(n-1) is odd for n = 2, 3 mod 4, even for n = 0, 1 mod 4. Thus A4n-1 + A4n = -½(4n1)(4n-2) + ½4n(4n-1) = 4n-1, A4n+1 + A4n+2 = -(4n+1), and A4n-1 + A4n + A4n+1 + A4n+2 = 2. Hence A19 + ... A98 = -20·2 = -40.
6. By similar triangles 735/(RC+112) = AP/BC, 847/RC = PD/BC = 1 + AP/BC = 1 +
735/(RC+112). hence RC2 = 847·112, RC = 308.
7. There are n ordered pairs summing to 2n. Consider triples summing to 2n+1: if 1st
number is 1, then n; if 1st is 3, then n-1 etc, so n(n+1)/2. Now consider 4-tuples summing
to 2n+2. If 1st number is 1, then n(n+1)/2; if 1st is 3, then (n-1)n/2 etc. So total (n2 + (n1)2 + ... + 12 + n + n-1 + ... + 1)/2 = n(n+1)(n+2)/6. Hence 48·49·50/6 = 19600.
8. The terms are 1000, n, 1000-n, 2n-1000, 2000-3n, 5n-3000, 5000-8n, 13n-8000,
13000-21n, 34n-21000, 34000-55n, 89n-55000. For the 3rd term to be non-neg, we need
n ≤1000, for the 4th n ≥500, for the 5th n < 667, for the 6th n ≥ 600, for the 7th n ≤ 625,
for the 8th n > 615, for the 9th n ≤ 619, for the 10th n > 617, for the 11th n < 619, for the
12th n > 617. Thus for n ≤ 617 the 10th term is negative, and for n ≥ 619 the 11th term is
negative. But for n = 618 at least the first 12 terms are positive.
9. The x-axis give the arrival time of one (in minutes after 9am), the y-axis the arrival
time of the other. The arrival times can differ by up to m mins, so the shaded band gives
the arrival pairs (x, y) for which they meet. Thus we need (60-m)2/602 = 0.6, giving m2 120m + 1440 = 0, roots 60 ± √2160. But m < 60, so m = 60 - 12√15.
10. The centers of the 8 spheres form an octagon. Let the distance of the center O of this
octagon from each sphere-center be d. Let the central sphere have radius r. Then the
triangle shown has sides r+100, d, r-100, so d2 = 400r. Two adjacent sphere centers and O
form a triangle with sides d, d, 200, and angle 45o opp the 200, so by the cosine rule 2002
= 2d2(1 - 1/√2), or d2 = 2002/(2-√2). Thus r = 100/(2-√2) = 50(2+√2).
11. We are given the points A, F, E below. Let B be the midpoint of its edge. Then BE
is parallel to AF, so B also belongs to the plane. Hence the center of the cube, O, belongs
to the plane, and so CD, obtained by rotating AF about O belongs to the plane.
AF = 15√2. BE = 20√2, so AC/2 = √(AB2 + ((5√2)/2)2) = √(125 + 25/2) = 15/√2. So AC
= 15√2. Hence area hexagon = AC·AF + AC(5/√2) = 450 + 75 = 525.
12. Compare the figures APRE and CRQD. If PE > RD, then RE > DQ, so QF > RD.
Similarly that implies QF > PE, and that implies PE < RD. Contradiction. Similarly if PE
< RD. So PE = RD. Similarly PE = QF. Put PF = QD = RE = x, AF = 1. Then BQD, PQF
are similar, so x/1 = (1-x)/x or x2 + x - 1 = 0, so x = (√5 - 1)/2 (we discard the negative
root). By cosine rule PQ2 = x2 + (1-x)2 - x(1-x) = 3x2 - 3x + 1 = 3(x2+x-1)+4-6x = 7-3√5.
Area ABC/area PQR = 22/x2 = 7 + 3√5.
13. There are 8Ck subsets A of {1, 2, ... , 8} with k elements. If we adjoin 9 to one of
these we add 9ik+1 to f(A). So adjoining 9 to each of the subsets of {1, 2, ... , 8} increases
the sum by ∑ 8Ck 9ik+1 = 9i (1+i)8. The subsets of {1, 2, ... , 9} are the subsets of {1, 2, ...
, 8} and the subsets of {1, 2, ... , 8} with 9 adjoined to each. Hence S9 = 2S8 + 9i(1+i)8.
Note that (1+i)2 = 2i, so (1+i)8 = 16. Hence S9 = -352 - 128i + 144i = -352 + 16i.
14. c = 2(a+2)(b+2)/(2ab-(a+2)(b+2)) = 2(a+2)(b+2)/(ab-2a-2b-4) = 2 +
(8a+8b+16)/(ab-2a-2b-4). So (c-2)/8 = (a+b+2)/(ab-2a-2b-4). To make progress we need
to realise that it helps to put c'=c-2, b'=b-2, a'=a-2, so c'/8 = (a'+b'+6)/(a'b'-8). Now we
cannot have a or b = 1 or 2, because (1+2)/1 and (2+2)/2 are already ≥ 2 (so multiplying
by (c+2)/c gives > 2). So a', b' ≥ 1, so a'+b' ≤ a'b'+1. Hence c'/8 ≤ (a'b'+7)/(a'b'-8) = 1 +
15/(a'b'-8) ≤ 1 + 15. Hence c ≤ 130 and it is easy to check this is realised.
15. There are 39 elements [m, 40]. Unless 39 appears as the first or last number in the
line, it must occur an even number of times. Similarly for the other numbers. So 38
numbers and hence 19 dominos [m, n] cannot appear. So the sequence cannot be longer
than 780 - 19 = 761. We claim that a sequence of 1 + (4 + 8 + ... + 4n-4) with first
number 1 and last number 2 is possible for the set of dominos with largest number 2n.
Induction. n = 1. [1,2] gives 1. [1 2n-1][2n-1 2][2 2n][2n 3][3 2n-1][2n-1 4] ... [2n-2
2n][2n 1] prefixed to the line for 2n-2 gives 4n-4 more dominos than the line for 2n-2.
(The extra dominos are the pairs [2n-1 a] and [2n a] for a = 1, 2, ... , 2n-2.) Putting n = 40
shows that 761 is realised.
1. Find the smallest a5, such that a1, a2, a3, a4, a5 is a strictly increasing arithmetic
progression with all terms prime.
2. A line through the origin divides the parallelogram with vertices (10, 45), (10, 114),
(28, 153), (28, 84) into two congruent pieces. Find its slope.
3. Find the sum of all positive integers n for which n2 - 19n + 99 is a perfect square.
4. Two squares side 1 are placed so that their centers coincide. The area inside both
squares is an octagon. One side of the octagon is 43/99. Find its area.
5. For any positive integer n, let t(n) be the (non-negative) difference between the digit
sums of n and n+2. For example t(199) = |19 - 3| = 16. How many possible values t(n) are
less than 2000?
6. A map T takes a point (x, y) in the first quadrant to the point (√x, √y). Q is the
quadrilateral with vertices (900, 300), (1800, 600), (600, 1800), (300, 900). Find the
greatest integer not exceeding the area of T(Q).
7. A rotary switch has four positions A, B, C, D and can only be turned one way, so that
it can be turned from A to B, from B to C, from C to D, or from D to A. A group of 1000
switches are all at position A. Each switch has a unique label 2a3b5c, where a, b, c = 0, 1,
2, ... , or 9. A 1000 step process is now carried out. At each step a different switch S is
taken and all switches whose labels divide the label of S are turned one place. For
example, if S was 2·3·5, then the 8 switches with labels 1, 2, 3, 5, 6, 10, 15, 30 would
each be turned one place. How many switches are in position A after the process has been
completed?
8. T is the region of the plane x + y + z = 1 with x,y,z ≥0. S is the set of points (a, b, c) in
T such that just two of the following three inequalities hold: a ≤ 1/2, b ≤ 1/3, c ≤ 1/6. Find
area S/area T.
9. f is a complex-valued function on the complex numbers such that function f(z) = (a +
bi)z, where a and b are real and |a + ib| = 8. It has the property that f(z) is always
equidistant from 0 and z. Find b.
10. S is a set of 10 points in the plane, no three collinear. There are 45 segments joining
two points of S. Four distinct segments are chosen at random from the 45. Find the
probability that three of these segments form a triangle (so they all involve two from the
same three points in S).
11. Find sin 5o + sin 10o + sin 15o + ... + sin 175o. You may express the answer as
tan(a/b).
12. The incircle of ABC touches AB at P and has radius 21. If AP = 23 and PB = 27, find
the perimeter of ABC.
13. 40 teams play a tournament. Each team plays every other team just once. Each game
results in a win for one team. If each team has a 50% chance of winning each game, find
the probability that at the end of the tournament every team has won a different number
of games.
14. P lies inside the triangle ABC, and angle PAB = angle PBC = angle PCA. If AB =
13, BC = 14, CA = 15, find tan PAB.
15. A paper triangle has vertices (0, 0), (34, 0), (16, 24). The midpoint triangle has as its
vertices the midpoints of the sides. The paper triangle is folded along the sides of its
midpoint triangle to form a pyramid. What is the volume of the pyramid?
1
2
3
4
5
6
7
8
29
99/19
38
86/99
223
314
650
11/18
9 (√255)/2.
10 16/473
11 tan(175o/2)
12 345
13 40!/2780
14 168/295
15 408
Outline solutions
1. Best is 5,11,17,23,29
2. The line through (0,0) and (10,45+x) must pass through (28,153-x), so x = 270/38.
Hence slope (45 + 270/38)/10 = 99/19.
3. We have n2 - 19n + 99 > n2 - 20n + 100 = (n-10)2 for n ≥ 1. Also < (n-9)2 = n2 - 18n +
81 for n > 18. So cannot be a square for n > 18. Also if f(n) = n2 - 19n + 99, then f(19-n)
= f(n). So we just have to check the values 1, 2, ... , 9. We find n = 1,9 and hence also
10,18 work. Sum 38.
4. Let the sides of the triangle be x, y, √(x2+y2). It is clear from the diagram that the
perimeter of the triangle is the side of the square 1. Solving, y = (1-2x)/(2-2x) and
octagon side = 1-x-y = (2x2-2x+1)/(2-2x) = 43/99. Hence 198x2 - 112x + 13 = 0. Product
of roots is 13/198. So area octagon = area square - 4 x area triangle = 1 - 2xy = 1 - 26/198
= 86/99.
5. If there are k carries in going from n to n+2, then the digit sum of n+2 is 9k-2 smaller
than that of n. So possible values of t(n) are 2 and 7+9k, for k = 0, 1, 2, ... . Largest <
2000 is k = 221: 7 + 221·9 = 1996. Hence 223 values < 2000.
6. Side joining (900,300) and (1800,600) lies along line y=x/3. So corresponding side of
T(Q) is also a straight line, namely y = x/√3, which makes angle 30o with x-axis.
Reflecting in y = x, we see that side joining (300,900) and (600,1800) goes to line along
y = 3x, which makes angles 30o with y-axis. The vertices (1800,600) and (600,1800) lie
on x + y = 2400, so corresponding side of T(Q) is arc of circle radius √2400. Similarly
4th side is arc of circle radius √1200. Angle between y=x/3 and y=3x is 30o, so area =
(π/12)(2400 - 1200) = π/100.
7. 2a3b5c divides 2A3B5C iff a ≤ A, b ≤ B and c ≤ C. So there are (10-a)(10-b)(10-c)
multiples of 2a3b5c. Thus it ends up in position A iff 4 divides (10-a)(10-b)(10-c). So we
have: a = 2,6, any b, c, 200 poss; a = 0,4,8 b,c not both odd 3(100-25) = 225 poss; a odd,
b = 2,5, any c, 100 poss; a odd, b odd, c 2 or 6, 50 poss; or a odd, b 0,4,8, c even 75 poss.
Total 650.
8. T is an equilateral triangle as shown. S is the yellow region. The three small
equilateral triangles that make up T-S have area T/22, T/32, T/62, total (14/36) area T, so
area S/area T = 11/18.
9. f(z) is equidistant from 0 and z iff a = 1/2. So need 1/4 + b2 = 65, b=(√255)/2.
10. There are 10C3 ways of choosing a triangle, then 42 ways of choosing a 4th
segment, whereas there are 45C4 ways of choosing 4 segments. Hence prob =
(10·9·8·42·24)/(6·45·44·43·42) = 16/473.
11. 2 sin 5 sin 5k = cos(k-1)5 - cos(k+1)5. So (2 sin 5) sum = cos 0 + cos 5 - cos 175 cos 180 = 2 + 2 cos 5. Hence sum = (1+cos 5)/sin 5. We still have to put it into the
required form. cos 2x = 2 cos2x - 1, and sin 2x = 2 sin x cos x, so sum = cot(5/2) = tan
87.5o = tan(175/2).
12. Put r = inradius, s = semiperimeter. So s-a = 46, s-b = 54, c = 50. Area = rs = √(s(sa)(s-b)(s-c)). So s = 345/2, perimeter = 345.
13. Team totals must be 0, 1, 2, ... , 39. So we must be able to order the teams as T1, T2,
... , T40, so that Ti loses to Tj for i < j. In other words, this order uniquely determines the
result of every game. There are 40! such orders and 780 games, so 2780 possible outcomes
for the games. Hence prob = 40!/2780.
14. P is a Brocard point. We have cot PAB = cot A + cot B + cot C (*). Using cosine
formula we find cos A = 33/65, so sin A = 56/65 and cot A = 33/65. Similarly, cos C =
3/5. We recognize 3,4,5 triangle, so cot C = 3/4. Finally cos B = 5/13. We recognize
5,12,13 triangle, so cot B = 5/12. Hence cot PAB = 295/168.
To prove (*) take D so that BDA is similar to ABC. Then P lies on CD. So cot PCA =
CK/DK = CA/DK + AK/DK. If the altitude from B is BE, then CA/DK = CE/BE +
EA/BE = cot C + cot A, whilst AK/DK = cot B.
15. The sides of the midpoint triangle are 15, 17, 4√13. Its vertices are A (17,0,0), B
(8,12,0), C(25,12,0). It is not hard to check that the 4th vertex is D (16,12,12). Then DA
= 17, DB = 4√13, DC = 15 as required. Using Heron or considering the altitude from A to
BC, we find that the midpoint triangle has area 102. Hence vol = (1/3) 12·102 = 408.
1. Find the smallest positive integer n such that if 10n = M·N, where M and N are
positive integers, then at least one of M and N must contain the digit 0.
2. m, n are integers with 0 < n < m. A is the point (m, n). B is the reflection of A in the
line y = x. C is the reflection of B in the y-axis, D is the reflection of D in the x-axis, and
E is the reflection of D in the y-axis. The area of the pentagon ABCDE is 451. Find u + v.
3. m, n are relatively prime positive integers. The coefficients of x2 and x3 in the
expansion of (mx + b)2000 are equal. Find m + n.
4. The figure shows a rectangle divided into 9 squares. The squares have integral sides
and adjacent sides of the rectangle are coprime. Find the perimeter of the rectangle.
5. Two boxes contain between them 25 marbles. All the marbles are black or white. One
marble is taken at random from each box. The probability that both marbles are black is
27/50. If the probability that both marbles are white is m/n, where m and n are relatively
prime, find m + n.
6. How many pairs of positive integers m, n have n < m < 1000000 and their arithmetic
mean equal to their geometric mean plus 2?
7. x, y, z are positive reals such that xyz = 1, x + 1/z = 5, y + 1/x = 29. Find z + 1/y.
8. A sealed conical vessel is in the shape of a right circular cone with height 12, and base
radius 5. The vessel contains some liquid. When it is held point down with the base
horizontal the liquid is 9 deep. How deep is it when the container is held point up and
base horizontal?
9. Find the real solutions to: log10(2000xy) - log10x log10y = 4, log10(2yz) - log10y log10z
= 1, log10zx - log10z log10x = 0.
10. The sequence x1, x2, ... , x100 has the property that, for each k, xk is k less than the
sum of the other 99 numbers. Find x50.
11. Find [S/10], where S is the sum of all numbers m/n, where m and n are relatively
prime positive divisors of 1000.
12. The real-valued function f on the reals satisfies f(x) = f(398 - x) = f(2158 - x) =
f(3214 - x). What is the largest number of distinct values that can appear in f(0), f(1),
f(2), ... , f(999)?
13. A fire truck is at the intersection of two straight highways in the desert. It can travel
at 50mph on the highway and at 14mph over the desert. Find the area it can reach in 6
mins.
14. Triangle ABC has AB = AC. P lies on AC, and Q lies on AB. We have AP = PQ =
QB = BC. Find angle ACB/angle APQ.
15. There are cards labeled from 1 to 2000. The cards are shuffled and placed in a pile.
The top card is placed on the table, then the next card at the bottom of the pile. Then the
next card is placed on the table to the right of the first card, and the next card is placed at
the bottom of the pile. This process is continued until all the cards are on the table. The
final order (from left to right) is 1, 2, 3, ... , 2000. In the original pile, how many cards
were above card 1999?
1
2
3
4
5
6
7
8
8
21
667
260
26
997
1/4
12 - 3·37(1/3)
9
10
11
12
13
14
15
(1,5,1),(100,20,100)
75/98
248
177
700/31
4/7
927
Outline solutions
1. We need 2n or 5n to have a 0. It is easy to check the first is 58 = 390625.
2. BCDE is a rectangle sides 2u, 2v, so area 4uv. The triangle ABE has base BC length
2u and height u-v, so area u2 - uv. Hence u(u+3v) = 451 = 11·41. Hence u = 11, v = 10.
3. We must have (1998/3)(m/n) = 1, or m/n = 1/666. Hence m = 1, n = 666.
4. Let the sides of the squares be a, b, ... , i (in ascending order of size - a is not marked
in the diagram). We have a + b = c (1), a + c = d (2), c + d = e (3), d + e = f (4), b + c + e
= g (5), b + g = h (6), a + d + f = i (7), f + i = g + h (8). Using (1) - (7) we can solve in
terms of a, b, then (8) gives 5a = 2b, so a = 2, b = 5 (must have no common factor).
Hence f = 25, h = 33, i = 36, sides are 61, 69.
5. Suppose one box has n marbles, b of which are black, and the other 25 - n marbles, B
of which are black. So bB/(n(25-n)) = 27/50. n(25-n) must be a multiple of 5, so n must
be a multiple of 5. Hence n = 5, 10, 15 or 20. Without loss of generality, n = 5 or 10 (just
swap boxes). If n = 5, then bB = 54. So b divides 54 and <= 5, so b = 1, 2, or 3 giving B =
54, 27, 18. Thus b = 3, B = 18, and prob both white = (2/5)(1/10) = 1/25. If n = 10, then
bB = 81, so b = B = 9. Hence prob both white = (1/10)(2/5) = 1/25.
6. We have (m+n)/2 = √(mn) + 2, so mn = m2/4 + n2/4 + mn/2 - 2m - 2n + 4. Hence
(m+n) = 2 + 2( (m-n)/4 )2. So m-n must be a multiple of 4. 4 is too small, it gives m = 4, n
= 0. 999 is too big, it gives m = 1000000. But the 997 values 4·2, 4·3, ... , 4·998 all work.
7. 1/z = 5-x, y = 29 - 1/x. Hence x(29 - 1/x) = (5 - x), so x = 1/5, y = 24, z = 5/24 and z +
1/y = 1/4.
8. When point down the liquid dimensions are 3/4 those of the container, so its volume
is 27/64 that of the container. Hence the empty space has volume 37/64 that of the
container. When point up, the empty space occupies a cone, whose height must be
(37/64)1/3 that of the container. Hence depth of liquid is 12(1 - (37/64)1/3).
9. Since log(2000xy) = 3 + log(2xy), the first two equations give log y = 1 or log x = log
z, ie y = 10 or x = z. But y = 10, does not satisfy the first equation, so x = z. The last
equation gives 2 log x - log x log x = 0, so x = 1 or 100. If x = z = 1, then the second
equation gives y = 5. If x = z = 100, the second equation gives y = 20. So the solutions
are (1, 5, 1) and (100, 20, 100).
10. Let s = x1 + x2 + ... + x100. Then 2xk = s - k. Summing, s = 50s - (1 + 2 + ... + 100)/2,
so s = 25·101/49.
11. We have (1+2+4+8)(1+ 1/5 + 1/25 + 1/125) + (1+5+25+125)(1 + 1/2 + 1/4 + 1/8) -1
+ (1/10 + 1/20 + 1/40 + 1/50 + 1/100 + 1/200 + 1/250 + 1/500 + 1/1000) + (10/1 + 20/1 +
40/1 + 50/1 + 100/1 + 200/1 + 250/1 + 500/1 + 1000/1) = 2480.427.
12. From the first two, f(x) = f(1760+x). From the second two, f(x) = f(1056+x).
gcd(1056, 1760) = 352, so f(x) = f(x+352). Then f(x) = f(398-x) gives f(x) = f(46-x).
Conversely, it is easy to check that f(x) = f(46-x) and f(x) = f(352+x) imply the relations
given. Evidently, f(x) = f(352+x) means the values repeat after any block of 352. Take
that block centered on x = 23. We can choose separately the values at x = 23-176, 23175, ... , 23-1, 23. Then the reflection determines the values at 23+1, 23+2, ... , 23+176.
The period then determines all other values. So we get at most 177 different values. If we
take them all different, they certainly appear in x = 0, 1, ... , 999, which contains two
complete periods.
13. The fire truck can drive along the axis and then head across the prairie in a straight
line. If it starts straight across the prairie it can reach anywhere in a circle radius 1.4. If it
drives 5 along the road, the circle has radius 0. For points in between the circle is scaled
proportionately. So the envelope is the four-pointed star. The part in the first quadrant is
the difference between an isosceles right-angled triangle hypoteneuse 5√2 and an obtuseangled triangle with the same base. The height of the second triangle is 5/√2 tan(45o - x),
where tan x = 7/24 (the right-angled triangle has sides 5, 1.4 and hence 4.8 - it is a 7, 24,
25 triangle). So tan(45o - x) = 17/31. Hence total area = 700/31.
14. AQ = 2 cos x, so AB = (1 + 2 cos x), so (1 + 2 cos x) sin x/2 = 1/2. But sin 3x/2 - sin
x/2 = 2 cos x sin x/2, so (1 + 2 cos x) sin x/2 = sin 3x/2. Hence 3x/2 = 30o, so x = 20o.
Hence angle APQ = 140o, and angle ACB = 80o, so ratio = 4/7.
15. Work backwards. It is easy to see that shortly before the end the pile is 1998, 2000,
1999. Move 1999 to the top and put 1997 above it, giving 1997, 1999, 1998, 2000. Move
2000 to the top and put 1996 above it, giving 1996, 2000, 1997, 1999, 1998. Move 1998
to the top and put 1995 above it, giving 1995, 1998, 1996, 2000, 1997, 1999. Now denote
the state by (n, m), where the pile has n cards and there are m below the 1999. We see
that (n, m) is preceded by (n+1, m-1) for m > 0, and (n, 0) is preceded by (n+1, n-1).
Thus from (3, 0) we went to (4, 2), (5, 1), (6, 0), (7, 5), ... , (12, 0), ... , (24, 0), ... , (48, 0),
... , (96, 0), ... , (192, 0), ..., (384, 0), ... , (768, 0), ... , (1536, 0), (1537, 1535), (1538,
1534), (1539, 1533), ... , (2000, 1072). Hence at the start there are 2000 - 1073 = 927
cards above 1999.
1. Find 2/log4(20006) + 3/log5(20006).
2. How many lattice points lie on the hyperbola x2 - y2 = 20002?
3. A deck of 40 cards has four each of cards marked 1, 2, 3, ... 10. Two cards with the
same number are removed from the deck. Find the probability that two cards randomly
selected from the remaining 38 have the same number as each other.
4. What is the smallest positive integer with 12 positive even divisors and 6 positive odd
divisors?
5. You have 8 different rings. Let n be the number of possible arrangements of 5 rings on
the four fingers of one hand (each finger has zero or more rings, and the order matters).
Find the three leftmost non-zero digits of n.
6. A trapezoid ABCD has AB parallel to DC, and DC = AB + 100. The line joining the
midpoints of AD and BC divides the trapezoid into two regions with areas in the ratio 2 :
3. Find the length of the segment parallel to DC that joins AD and BC and divides the
trapezoid into two regions of equal area.
7. Find 1/(2! 17!) + 1/(3! 16!) + ... + 1/(9! 10!).
8. The trapezoid ABCD has AB parallel to DC, BC perpendicular to AB, and AC
perpendicular to BD. Also AB = √11, AD = √1001. Find BC.
9. z is a complex number such that z + 1/z = 2 cos 3o. Find [z2000 + 1/z2000] + 1.
10. A circle radius r is inscribed in ABCD. It touches AB at P and CD at Q. AP = 19, PB
= 26, CQ = 37, QD = 23. Find r.
11. The trapezoid ABCD has AB and DC parallel, and AD = BC. A, D have coordinates
(20,100), (21,107) respectively. No side is vertical or horizontal, and AD is not parallel to
BC. B and C have integer coordinates. Find the possible slopes of AB.
12. A, B, C lie on a sphere center O radius 20. AB = 13, BC = 14, CA = 15. `Find the
distance of O from the triangle ABC.
13. The equation 2000x6 + 100x5 + 10x3 + x - 2 = 0 has just two real roots. Find them.
14. Every positive integer k has a unique factorial expansion k = a1 1! + a2 2! + ... + am
m!, where m+1 > am > 0, and i+1 > ai ≥ 0. Given that 16! - 32! + 48! - 64! + ... + 1968! 1984! + 2000! = a1 1! + a2 2! + ... + an n!, find a1 - a2 + a3 - a4 + ... + (-1)j+1 aj.
15. Find the least positive integer n such that 1/(sin 45o sin 46o) + 1/(sin 47o sin 48o) + ...
+ 1/(sin 133o sin 134o) = 2/sin no.
1
2
3
4
5
6
1/6
98
55/703
180
376
25√29
9
10
11
12
13
14
0
-3,-4/3,-1,-1/2,1/3,3/4,1,2
(15√95)/8
(-1±√161)/40
495
7 (218-20)/19!
8 √110
15 1
Outline solutions
1. Put log45 = k. Then log54 = 1/k. So log42000 = 2+3k, log52000 = 3+2/k. Hence expr =
1/(3(2+3k)) + 1/(2(3+2/k)) = 1/6.
2. (x+y)(x-y)=2856. There are two solutions with y = 0. None with x = 0. Suppose x, y
are positive. Then x+y and x-y have same parity, so must both be even. The no. of factors
of 2000/4 is (6+1)(6+1) = 49. There is one factor 2453 and 24 pairs (2a,2b) with a, b
unequal and 4ab = 2000. Each of these gives a lattice point (x,y). But each of these gives
4 lattice points (±x,±y). So 2 + 4·24=98 in all.
3. prob cards also match removed cards = (2/38)(1/37), prob match each other but not
removed cards = (36/38)(3/37). Total prob = 110/(38.37) = 55/703.
4. Smallest no. with 6 odd divisors is 325. To get additionally 12 even, we must multiply
by 22.
5. The no. of rings may be: (A) 5000, (B) 4100, (C) 3200, (D) 3110, (E) 2210, (F) 2111.
There are 4 choices each for (A), (B), 12 each for (B), (C), (D). Then there are 8·7·6·5·4
= 6720 ways of choosing the rings. Hence 56·6720 = 376320.
6. Midpoint line has length AB+50. So (AB+75/(AB+25) = 3/2. Hence AB = 75.
Suppose distance between AB and CD is h and between requd line and AB is kh. Then
length = AB + 100k. So k(AB +50k) = (1/2)(AB+50) or 2k(75+50k) = 125, or 4k2+6k5=0, k=(√29-3)/4, so length = 75+25√29-75 = 25√29.
7. Put k = 19C2 + 19C3 + ... + 19C9. Then 2k+2(1+19) = 219. So k = 218-20 and sum =
(218-20)/19!
8. Put BC = x. Then AC = √(x2+11). ABC and BCD are similar. So BD/BC = AC/AB,
so BD = x(√(x2+11))/√11. Also cos ABD = cos ACB = x/√(x2+11). Applying cosine rule
to ABD, we get 1001 = AD2 = AB2 + BD2 - 2AB·BD cos ABD = 11 + x2(x2+11)/11 2x2. Hence (x2+99)(x2-110) = 0, so x = √110.
9. z = e±iθ, where θ = 3o. So z + 1/z = 2 cos 6000o = 2 cos 240o = -1.
10.
11. It is convenient to translate the axes so that A is at the origin and D is (1,7). Take X
on CD with AX parallel to BC. Then ADX is isosceles and X also has integer
coordinates. But the only integer solutions to x2 + y2 = 50 are (x,y) = (±1,±7), (±7,±1),
(±5,±5), so X must be one of these. We cannot have (1,7) (X cannot be D). (-1,7) makes
AB horizontal. (1,-7) makes AB vertical. (-1,-7) makes A,B,C,D collinear.
(7,1) is fine eg B (-1,1), C (8,0) and slope -1. (7,-1) is fine, eg B (3,-4), C (10,-5) and
slope -4/3. Similarly, (-7,1) slope 3/4 and (-7,-1) slope 1.
(5,5) is fine slope -1/2, (5,-5) slope -3, (-5,5) slope 1/3, (-5,-5) slope 2.
12. By Heron area = 84. Hence altitude from A is 12. Hence sin B = 12/13. If
circumradius is R, then 15 = 2R sin B, so R = 65/8. Hence square of requd dist = 202 R2 = (15√95)/8.
13. Factorises as (20x2 + x - 2)(100x4 + 10x2 + 1). First factor has roots (-1
±√(1+160))/40.
14. We have (n+1)! = n·n! + n! = n·n! + (n-1)(n-1)! + (n-1)! etc. Hence 2000! - 1984! =
1999·1999! + 1998·1998! + ... + 1984·1984!, and 1999 - 1998 + 1997 - 1996 + ... + 1985
- 1984 = 8. Thus the 62 pairs (2000! - 1984!) + (1968! - 1952!) + ... + (48! - 32!) give
8·62 = 496. Finally + 1·16! gives -1. Total 495
15.
1. Find the sum of all positive two-digit numbers that are divisible by both their digits.
2. Given a finite set A of reals let m(A) denote the mean of its elements. S is such that
m(S∪{1}) = m(S) - 13 and m(S∪{2001}) = m(S) + 27. Find m(S).
3. Find the sum of the roots of the polynomial x2001 + (½ - x)2001.
4. The triangle ABC has ∠ A = 60o, ∠ B = 45o. The bisector of ∠ A meets BC at T where
AT = 24. Find area ABC.
5. An equilateral triangle is inscribed in the ellipse x2 + 4y2 = 4, with one vertex at (0,1)
and the corresponding altitude along the y-axis. Find its side length.
6. A fair die is rolled four times. Find the probability that each number is no smaller than
the preceding number.
7. A triangle has sides 20, 21, 22. The line through the incenter parallel to the shortest
side meets the other two sides at X and Y. Find XY.
8. A number n is called a double if its base-7 digits form the base-10 number 2n. For
example, 51 is 102 in base 7. What is the largest double?
9. ABC is a triangle with AB = 13, BC = 15, CA = 17. Points D, E, F on AB, BC, CA
respectively are such that AD/AB = α, BE/BC = β, CF/CA = γ, where α + β + γ = 2/3,
and α2 + β2 + γ2 = 2/5. Find area DEF/area ABC.
10. S is the array of lattice points (x, y, z) with x = 0, 1 or 2, y = 0, 1, 2, or 3 and z = 0, 1,
2, 3 or 4. Two distinct points are chosen from S at random. Find the probability that their
midpoint is in S.
11. 5N points form an array of 5 rows and N columns. The points are numbered left to
right, top to bottom (so the first row is 1, 2, ... , N, the second row N+1, ... , 2N, and so
on). Five points, P1, P2, ... , P5 are chosen, P1 in the first row, P2 in the second row and so
on. Pi has number xi. The points are now renumbered top to bottom, left to right (so the
first column is 1, 2, 3, 4, 5 the second column 6, 7, 8, 9, 10 and so on). Pi now has
number yi. We find that x1 = y2, x2 = y1, x3 = y4, x4 = y5, x5 = y3. Find the smallest
possible value of N.
12. Find the inradius of the tetrahedron vertices (6,0,0), (0,4,0), (0,0,2) and (0,0,0).
13. The chord of an arc of ∠ d (where d < 120o) is 22. The chord of an arc of ∠ 2d is
x+20, and the chord of an arc of ∠ 3d is x. Find x.
14. How many different 19-digit binary sequences do not contain the subsequences 11 or
000?
15. The labels 1, 2, ... , 8 are randomly placed on the faces of an octahedron (one per
face). Find the probability that no two adjacent faces (sharing an edge) have adjacent
numbers, where 1 and 8 are also considered adjacent.
1
2
3
4
5
6
7
8
630
651
500
216+72√3
(16√3)/13
7/72
860/63
315
9
10
11
12
13
14
15
16/45
23/177
149
2/3
-9+√165
351
1/84
Outline solutions
1. Obv 11, 22, ... , 99 sum 11·45 = 495 are solutions. ab must be divisible by a, so only
other possibilities are a(2a), a(3a) etc. Hence none for second digit ≥ 5. Easily find 48
works, 36 works, 39 does not. 24 works, 26, 28 do not. 12, 15 work. So other solns sum
12+15+24+36+48 = 135.
2. Suppose S has n els, total N. Then (N+1)/(n+1) = N/n - 13, or N = 13n2+14n. Also
(N+2001)/(n+1) = N/n + 27, or 1974n-27n2 = N. Hence n = 49, N/n = 651.
3. Expanding, coeff x2001 = 0, coeff x2000 = 2001/2, coeff x1999 = -250·2001. Hence sum
= 500.
4. The trick is that ATC is isosceles, because ∠ C = ∠ T = 75o. Now take CS as an
altitude. Then CS = AC sin 60o = 12√3 and AS = 12. Also ∠ B = ∠ SCB = 45o, so SB =
12√3. Hence AB = 12+12√3. So area = AB·CS/2 = (6+6√3)12√3 = 216+72√3.
5. The line through (0,1) and slope -√3 is (y-1) = -x√3. This intersects the ellipse at x2 +
4(1-x√3)2 = 4, or 13x2 - 8x√3 = 0, so x = (8√3)/13, y = -11/13. Side length is 2x (because
other vertex is (-x,y) ) = (16√3)/13
6. Given the pattern of values (one 6, two 5s, one 3, for example), there is unique
admissible sequence of throws giving those values (eg 3, 5, 5, 6). The no. of patterns is
the same as the number of ways of placing 4 identical balls in 6 urns. That is the same as
the number of ways of placing 4 balls and 5 dividers in a row, or 9C4 = 126. Hence prob
= 126/64 = 7/72.
7. Let area be A, inradius be r, height from base 20 be h. Then A = ½h20 =
½r(20+21+22), so r/h = 20/63. By similar triangles XY = 20(h-r)/h = 20(1-r/h) =
20·43/63.
8. a0+10a1+102a2 + ... = 2(a0+7a1+72a2+...) or a0 + 4a1 = 2a2 + 314a3 + 7599a4 + ... . But
ai ≤ 6, so lhs ≤ 30, so a3, a4, ... = 0 and a0 + 4a1 = 2a2. Largest soln is evidently a2 = 6, a1 =
3, a0 = 0, giving 6307 = 31510 (note that ai must be ≤ 6).
9. area ADC = α area ABC. area ADF = (1-γ)area ADC = α(1-γ) area ABC. Similarly
for the other two triangles, so area DEF/area ABC = 1 - (α+β+γ) + (αβ+βγ+γα) = 1/3 +
½( (α+β+γ)2 - (α2+β2+γ2) ) = 1/3 + 1/45 = 16/45.
10. The midpoint of (x,y,z) and (x',y',z') lies in S iff x,x' have the same parity, y,y' have
the same parity, and z,z' have the same parity. There are 22+12=5 ways of choosing x,x' to
have the same parity, incl 3 with x=x'. Similarly, 8 incl 4 for y,y' and 13 incl 5 for z,z'.
Hence 5·8·13 - 3·4·5 = 460 of choosing distinct points with same parity. There are 60·59
ways of choosing distinct points, so prob 460/3540 = 23/177.
11. Suppose the row 1 point is in col a, the row 2 point in col b and so on. Then a = 5b3, b+N = 5a-4, c+2N = 5d-1, d+3N = 5e, e+4N = 5c-2. First two equs give a = 5b-3, N =
24b-19. Last three give 124e = 89N+7. Hence 124e = 2136b-1684. Now 2136b-1684 =
28b+52 mod 124, and by trial smallest possible solution (for b and hence for N) is b = 7,
giving N = 149, e = 107. Continuing, we get a = 32, d = 88, c = 141. So it is indeed a
solution.
12. We use the formula r x total face area = 3 vol. Triangle (0,0,0), (6,0,0), (0,4,0) has
area 12. Hence vol = 2·12/3 = 8. Triangle (6,0,0), (0,4,0), (0,0,2) has sides √20, √40, √52.
Slogging out Heron gives area 14. Hence total face area = 14 + 12 + 6 + 4 = 36. Hence r
= 24/36 = 2/3
13. Let radius be r, k = d/2, then 2r sin k = 22, 2r sin 2k = x + 20, 2r sin 3k = x. So (sin
2k)/sin k = (x+20)/22, (sin 3k)/sin k = x/22. But (sin 2k)/sin k = 2 cos k, (sin 3k)/sin k = 4
cos2k - 1. Hence x/22 = ( (x+20)/22)2 - 1, or x2 + 18x - 84 = 0. Must take positive root, so
x = √165 - 9.
14. Let an, bn, cn be no. of sequences length n beginning with 1, 01, 001 respectively.
Then an+1 = bn + cn, bn+1 = an, cn+1 = bn. So an+1 = an-1 + bn-1, bn+1 = bn-1 + cn-1, cn+1 = an-1.
Put dn = an + bn + cn. So dn+1 = dn-1 + an-1 + bn-1 = dn-1 + dn-2. Note that d1 = 2 (1, 0), d2 = 3
(10, 01, 00), d3 = 4 (101, 100, 010, 001). So we now calculate successively 4, 5, 7, 9, 12,
16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351.
15.
1. Find the largest positive integer such that each pair of consecutive digits forms a
perfect square (eg 364).
2. A school has 2001 students. Between 80% and 85% study Spanish, between 30% and
40% study French, and no one studies neither. Find m be the smallest number who could
study both, and M the largest number.
3. The sequence a1, a2, a3, ... is defined by a1 = 211, a2 = 375, a3 = 420, a4 = 523, an = an-1
- an-2 + an-3 - an-4. Find a531 + a753 + a975.
4. P lies on 8y = 15x, Q lies on 10y = 3x and the midpoint of PQ is (8,6). Find the
distance PQ.
5. A set of positive numbers has the triangle property if it has three elements which are
the side lengths of a non-degenerate triangle. Find the largest n such that every 10element subset of {4, 5, 6, ... , n} has the triangle property.
6. Find the area of the large square divided by the area of the small square.
7. The triangle is right-angled with sides 90, 120, 150. The common tangents inside the
triangle are parallel to the two sides Find the length of the dashed line joining the centers
of the two small circles.
8. The function f(x) satisfies f(3x) = 3f(x) for all real x, and f(x) = 1 - |x-2| for 1 ≤ x ≤ 3.
Find the smallest positive x for which f(x) = f(2001).
9. Each square of a 3 x 3 board is colored either red or blue at random (each with
probability ½). Find the probability that there is no 2 x 2 red square.
10. How many integers 10i - 10j where 0 ≤ j < i ≤ 99 are multiples of 1001?
11. In a tournament club X plays each of the 6 other sides once. For each match the
probabilities of a win, draw and loss are equal. Find the probability that X finishes with
more wins than losses.
12. The midpoint triangle of a triangle is that obtained by joining the midpoints of its
sides. A regular tetrahedron has volume 1. On the outside of each face a small regular
tetrahedron is placed with the midpoint triangle as its base, thus forming a new
polyhedron. This process is carried out twice more (three times in all). Find the volume
of the resulting polyhedron.
13. ABCD is a quadrilateral with AB = 8, BC = 6, BD = 10, ∠ A = ∠ D and ∠ ABD =
∠ C. Find CD.
14. Find all the values 0 ≤ θ < 360o for which the complex number z = cos θ + i sin θ
satisfies z28 - z8 - 1 = 0.
15. A cube has side 8. A hole with triangular cross-section is bored along a long
diagonal. At one vertex it removes the last 2 units of each of the three edges at that
vertex. The three sides of the hole are parallel to the long diagonal. Find the surface area
of the part of the cube that is left (including the area of the inside of the hole).
1
2
3
4
5
6
7
8
81649
201, 499
898
60/7
253
25
√7250
429
Outline solutions
9
10
11
12
13
14
15
417/512
784
98/243
69/32
64/5
15,75,105,165 and 360-these
372+39√6
1. The two digit squares are 16, 25, 36, 49, 64, 81. So if 16 occurs in the number, then it
must be followed by 4, which must be followed by 9 and nothing can follow 9. Similarly,
16 can only be preceded by 8, which cannot have predecessor. 64 can also be preceded by
36, which cannot have a predecessor. 25 cannot have successors or predecessors. So the
only numbers that cannot be extended at either end are 25, 3649, 81649. The longest is
obviously 81649.
2. 601 to 800 study French, 1601 to 1700 study Spanish. So smallest for both is 201
with 400 just French, 1400 just Spanish. Largest is 499, with 301 just French, 1201 just
Spanish.
3. We find a5 = 267, a6 = -a1, a7 = -a2, a8 = -a3, a9 = -a4, a10 = -a5, a11 = a1, a12 = a2, a13 =
a3, a14 = a4 etc. So a531 = a1, a753 = a3, a975 = a5 and a1 + a3 + a5 = 211 + 420 + 267 = 898
4. The line through (8,6) parallel to OQ is 10y = 3x + 36, and this meets OP at (16/7,
30/7). But that must be the midpoint of OP. So P is (32/7, 60/7). The distance to (8,6) is
√((24/7)2 + (18/7)2) = 30/7. Hence PQ = 60/7.
5. Clearly (4, 5, 9, 14, 23, 37, 60, 97, 157, 254) does not have the triangle property (each
element from 9 on is the sum of the previous two). But if a1 < a2 < ... < a10 does not have
the property, then a10 ≥ a9 + a8 ≥ 2a8 + a7 ≥ 3a7 + 2a6 ≥ ... ≥ 34a2 + 21a1 ≥ 34·5 + 21·4 =
254.
6. Let the large square have side 2, the small square side x. Then the radius is √2. The
lines containing vertical sides of the small square pass a distance x from the center. So by
Pythagoras, 2 = (1 + 2x)2 + x2, so (5x-1)(x+1) = 0, so x = 1/5. So ratio = 1/x2 = 25.
7. Let B be the vertex with angle 90o. Suppose it is a distance x from the points of
contact of the incircle. Then chasing around the triangle using the fact that the tangents
from a point have equal length, we get x = (90 + 120 - 150)/2 = 30. Evidently this is also
the radius of the incircle. The other two triangles are similar. The top one has sides 30,
40, 50 and inradius 10, and the right-hand one has sides 45, 60, 75 and inradius 15. We
can regard the dotted line as the hypoteneuse of a right-angled triangle with vertical side
60+10-15 = 55, horizontal side 60+15-10 = 65. Hence length √(552 + 652) = √7250 =
5√290.
8. The graph of f for x ∈ [1,3] is piecewise linear with f(1) = f(3) = 0 and a peak at f(2)
= 1. The graph for x ∈ [3,9] is a 3x larger copy with peak at 3, the graph for x ∈ [9,27] is
3x larger again, with peak at 9 and so on. f(x) is linear from f(2·729) = 729 to f(2001) =
729-543 = 186. f first reaches this value between x = 243 and x = 486. f(243) = 0 and
f(486) = 243, so f(x) = 186 at x = 243 + 186 = 429.
9. We use the inclusion/exclusion principle. There are 4 possible red squares as shown.
Let pij...k be the probability of getting i and j and ... and k. Obviously p1 = p2 = p3 = p4 =
1/16, so ∑ pi = 1/4. Similarly p12 = p13 = p24 = p34 = 1/64, p14 = p23 = 1/128, so ∑ pij =
5/64. We have pijk = 1/256, so ∑ pijk = 1/64. Finally p1234 = 1/512. Hence p(none) = 1 1/4 + 5/64 - 1/64 _ 1/512 = 417/512.
10. (106 - 1) = (103 + 1)(103 - 1), so 106 - 1 and hence also 106n - 1 is certainly divisible
by 1001. But 7 divides 1001 and 10m - 1 is not divisible by 7 for m not a multiple of 6.
Obviously 10m is never divisible by 7. Thus 10m(10n - 1) is divisible by 1001 iff n is a
multiple of 6. So for j = 0, there are 16 values of i (6, 12, ... , 96). Similarly for j = 1, 2, 3.
For the next 6 values of j there are 15 values of i, total 6·15. For the next 6, there are 14
each, total 6·14, and so on up to the 6 values 88 to 93 for which there is one each, total
6·1. For j > 93, there are no values of i. Hence total = 64 + 6(15 + 14 + ... + 1) = 64 + 720
= 784.
11. By symmetry the prob of more wins than losses equals the prob of more losses than
wins. We calculate the prob of the same number of wins and losses. Prob of no wins, no
losses = 1/36. Prob of 1 win, 1 loss = 6·5/36. Prob of 2 wins, 2 losses = (6C2)(4C2)/36 =
15·6/36. Prob of 3 wins, 3 losses = (6C3)/36 = 20/36. So prob same no. = (1+ 30 + 90 +
20)/729 = 141/729 = 47/243. Hence prob more wins than losses = prob more losses than
wins = 98/243.
12. The side length of each small tetrahedron on the face of the original tetrahedron is
1/2 the side length of the original, so its volume is 1/8. There are 4 of them so total
volume 1/2. The new figure has 24 faces. The second time, each small tetrahedron has
volume 1/64, so their total volume is 24/64 = 3/8. The new figure has 3·24 = 72 faces
which are subsets of faces of the previous figure and another 72 outside, total 144. The
tiny tetrahedra have vol 1/512, so total 144/512 = 9/32. Hence total vol = 1 + 1/2 + 3/8 +
9/32 = (32 + 16 + 12 + 9)/32 = 69/32.
13. Extend AB, CD to meet at E. Then triangles ECB, EBD are similar. Hence BD/ED =
CB/EB, so 10/ED = 6/(ED-8). Hence ED = 20. Also EC/CB = EB/BD, so EC = 7.2.
Hence CD = 12.8 = 64/5.
14. We have sin 28θ - sin 8θ = 0. Hence 28θ = 8θ + 360n or 28θ + 8θ = 180 + 360n. In
the first case we have cos 28θ = cos 8θ. But cos 28θ - cos 8θ = 1 (*). So we must have
36θ = 180 + 360n. Hence θ = 5, 15, 25, ... , or 355 (or 5 + 10n).
(*) gives - 2 sin((28θ+8θ)/2) sin((28θ-8θ)/2) = 1 or sin 18θ sin 10θ = -1/2. Putting θ = 5 +
10n, leads to sin(100n+50) = (-1)n+1/2. Now n odd gives θ = 15 or 75 mod 180 and n even
gives 105 or 165 mod 180.
15. This is messy. Take coordinates with vertices of the cube at (0,0,0), (8,8,8), (8,0,0),
(0,8,0), (0,0,8) etc. Bore the hole from (0,0,0) to (8,8,8). The long edges of the hole are
from A (6,8,8) to B (0,2,2), from C (8,6,8) to D (2,0,2), and from (8,8,6) to (2,2,0). The
plane ACDB has equation 2z = 2+x+y. So it meets an edge of the cube at E (0,0,1).
ACDEB is one face of the tunnel. We find AC = 2√2, AB = 6√3, BE = √5. After some
calculation, area ACDEB = 13√6. So the tunnel faces have total area 39√6.
There are 3 faces which are just original cube faces less a small triangle. These each have
area 62. Another three are original cube faces less a small kite shape removed from the
corner. These also have area 62. So total excluding tunnel faces = 372.
1. A licence plate is 3 letters followed by 3 digits. If all possible licence plates are
equally likely, what is the probability that a plate has either a letter palindrome or a digit
palindrome (or both)?
2. 20 equal circles are packed in honeycomb fashion in a rectangle. The outer rows have
7 circles, and the middle row has 6. The outer circles touch the sides of the rectangle.
Find the long side of the rectangle divided by the short side.
3. Jane is 25. Dick's age is d > 25. In n years both will have two-digit ages which are
obtained by transposing digits (so if Jane will be 36, Dick will be 63). How many
possible pairs (d, n) are there?
4. The sequence x1, x2, x3, ... is defined by xk = 1/(k2 + k). A sum of consecutive terms
xm + xm+1 + ... + xn = 1/29. Find m and n.
5. D is a regular 12-gon. How many squares (in the plane of D) have two or more of their
vertices as vertices of D?
6. The solutions to log225x + log64y = 4, logx225 - logy64 = 1 are (x, y) = (x1, y1) and (x2,
y2). Find log30(x1y1x2y2).
7. What are the first three digits after the decimal point in (102002 + 1)10/7? You may use
the extended binomial theorem: (x + y)r = xr(1 + 4 (y/x) + r(r-1)/2! (y/x)2 + r(r-1)(r-2)/3!
(y/x)3 + ...) for r real and |x/y| < 1
8. Find the smallest integer k for which there is more than one non-decreasing sequence
of positive integers a1, a2, a3, ... such that a9 = k and an+2 = an+1 + an.
9. A, B, C paint a long line of fence-posts. A paints the first, then every ath, B paints the
second then every bth, C paints the third, then every cth. Every post gets painted just
once. Find all possible triples (a, b, c).
10. ABC is a triangle with angle B = 90o. AD is an angle bisector. E lies on the side AB
with AE = 3, EB = 9, and F lies on the side AC with AF = 10, FC = 27. EF meets AD at
G. Find the nearest integer to area GDCF.
11. A cube with two faces ABCD, BCEF, has side 12. The point P is on the face BCEF a
perpendicular distance 5 from the edge BC and from the edge CE. A beam of light leaves
A and travels along AP, at P it is reflected inside the cube. Each time it strikes a face it is
reflected. How far does it travel before it hits a vertex?
12. The complex sequence z0, z1, z2, ... is defined by z0 = i + 1/137 and zn+1 = (zn + i)/(zn
- i). Find z2002.
13. The triangle ABC has AB = 24. The median CE is extended to meet the circumcircle
at F. CE = 27, and the median AD = 18. Find area ABF.
14. S is a set of positive integers containing 1 and 2002. No elements are larger than
2002. For every n in S, the arithmetic mean of the other elements of S is an integer. What
is the largest possible number of elements of S?
15. ABCDEFGH is a polyhedron. Face ABCD is a square side 12. Face ABFG is a
trapezoid with GF parallel to AB and GF = 6, AG = BF = 8. Face CDE is an isosceles
triangle with ED = EC = 14. E is a distance 12 from the plane ABCD. The other faces are
EFG, ADEG and BCEF. Find EG2.
1
2
3
4
5
6
7
8
7/52
7(√3-1)/2
25
28,811
183
12
428
748
9
10
11
12
13
14
15
(3,3,3), (4,2,4)
148
12 √218
1 + 274i
8√55
30
128-16√19
Outline solutions
1. Three letters form a palindrome iff the last matches the first, so the prob that the
letters are not a palindrome is 25/26. Similarly, the prob that the digits are not is 9/10. So
the prob of a palindrome is 1 - (25/26)(9/10) = 7/52.
2. Let the circles have radius 1. Then the long side is 14. The two outer rows of centers
are a distance 2√3 apart, so the short side is 2 + 2√3. Hence ratio 7/(1+ √3) = 7(√3 - 1)/2.
3. Dick's age must be one of: 43; 53, 54; 62, 63, 64, 65; 72, ... , 76; 82, ... , 87; 92, ... ,
98, giving 1 + 2 + 4 + 5 + 6 + 7 possibilities, each corresponding to a unique n. Hence 25.
4. xk = 1/k - 1/(k+1), so the sum is 1/m - 1/(n+1) = 1/29. Put N = n+1. Then mN = 29(Nm), so (m-29)(N+29) + 292 = 0. But N+29 is positive, so m-29 must be negative. But m is
positive, so m-29 is not divisible by 29. Hence N+29 = 292 and m = 28. So n = 28·29 - 1
= 811.
5. Let S be the set of vertices of D. There are 66 pairs of points from S. Each pair gives
2 squares with the pair as adjacent vertices. But the 3 squares with all their vertices in S,
so they are each overcounted 3 times, giving 66·2 - 9 = 123. In addition each pair has one
square with that pair as opposite vertices, but each of the 3 squares with all vertices in S
is again counted twice, an overcount of 6, so 123 + 60 = 183.
6. Put a = log225x, b = log64y. Then logx225 = 1/a, logy64 = 1/b, so equations become 1/a
- 1/b = 1, a + b = 4. Solving (a, b) = (3+√5, 1-√5) or (3-√5, 1+√5). Hence if a1
corresponds to x1 etc, we have log225x1x2 = a1+a2 = 6, so x1x2 = 1512, and log64y1y2 =
b1+b2 = 2, so y1y2 = 212. Hence answer 12.
7. So (102002 + 1)10/7 = 102860 (1 + 10/7 1/102002 + 3/49 1/104004 + ... ) = 102860 + 10859/7 +
(3/49) 1/101143 + ... . Now 1/7 = 0.142857 142857 ... , period 6 and 859 = 1 mod 6, so
10859/7 = ... 1.428... and answer = 428.
8. We have a9 = 13a1 + 21a2, so if the two sequences have (a1, a2) = (a, b) and (A, B),
then k = 13a + 21b = 13A + 21B. So 13(A - a) = 21(b - B). So A - a = 21n and b - B =
13n for some positive integer n. The obvious a = 1, A = 22, b = 14, B = 1 does not work
because A > B. The sequences are non-decreasing iff A <= B and a <= b. So we take a =
1, A = 22, B = 22, b = 35, giving k = 748.
9. Label the posts 1, 2, 3, ... . C cannot paint 4 (or C paints all the posts thereafter).
Suppose A paints 4. C cannot paint 5, or A and C both paint 7, so B paints 5. Hence C
paints 6 and (a, b, c) = (3, 3, 3). Suppose B paints 4. C cannot paint 5 (or there is nothing
left for A to paint). So A paints 5. Hence C paints 7 and (a, b, c) = (4, 2, 4).
10. We have AB = 12, AC = 37, so BC = 35 (Pythagoras). So area ABC = 210. Area
GDCF = area ABC - area ABD - area AGF. We now use the angle bisector theorem
twice. Area ABD = (12/49) area ABC = 51 3/7. Area AGF = (3/13) area AEF = (3/13)
(1/4) area ABF = (10/13) (1/4) (10/37) area ABC = 10 440/481, so area GDCF = 210 - 51
3/7 - 10 440/481 = 148 - (3/7 - 41/481).
11. This needs a trick. Call the cube c. Reflect c in the face BCEF to get cube d. Then
the continuation of the ray reflects into a continuation of the segment AP. If the
continuation strikes face X of d at Q, then reflect d in the face X to get cube e and so on.
Take A as the origin, x-axis along AB, y-axis parallel to BC, z-axis parallel to BF. Then
P is the point (12, 7, 5) and we get the line (12t, 7t, 5t). The question is when this first hits
a vertex (12q, 12r, 12s). Evidently at t = 12, a distance 12 (122 + 72 + 52)1/2 = 12 √218.
12. We find z --> (z+i)/(z-i) --> i(z+1)/(z-1) --> z. So z0 = z3 = ... = z2001. Hence z2002 =
(1/137 + 2i)/(1/137) = 1 + 274i.
13. Let the medians meet at G. Then GE = CE/3 = 9, GC = 2GE = 18, and AG = 2AD/3
= 12. Also AE = BE = 12. Since AB, CF are chords of a circle, we have FE·27 = 12·12,
so FE = 16/3. Now area AFB/area ABC = EF/EC = 16/81. The height of AEG is √(122 (9/2)2) = 3(√55)/2, so area AEG = 27(√55)/4. Hence area AEC = 3 area AEG =
81(√55)/4, and area ABC = 81(√55)/2. So, finally, area AFB = 8√55.
14. Let S have k+1 elements with sum N. Then k divides N-1 and N-2002 and hence
also 2001 = 3·23·29. If m is any element of S, then k divides N-m and hence m-1. In
other words m = 1 mod k. So the largest element is at least 1+k2. Since 2002 is the largest
element, k < 45. Thus the largest possible k is 29. For example: 1, 30, 59, 88, 117, 146,
175, 204, 233, ... , 813, 2002.
15. The trapezoid ABFG is clearly determined. We can regard it as hinged about AB.
Similarly, the triangle CDE is determined and hinges about CD. Moreover, since the
distance of E from the plane ABCD is fixed, the angle at the hinge is fixed. Why then is
the angle at the hinge AB fixed? Because ADEG must be planar.
1. n is an integer between 100 and 999 inclusive, and so is n' the integer formed by
reversing its digits. How many possible values are there for |n-n'|?
2. P (7,12,10), Q (8,8,1) and R (11,3,9) are three vertices of a cube. What is its surface
area?
3. a, b, c are positive integers forming an increasing geometric sequence, b-a is a square,
and log6a + log6b + log6c = 6. Find a + b + c.
4. Hexagons with side 1 are used to form a large hexagon. The diagram illustrates the
case n = 3 with three unit hexagons on each side of the large hexagon. Find the area
enclosed by the unit hexagons in the case n = 202.
5. Find the sum of all positive integers n = 2a3b (a, b ≥ 0) such that n6 does not divide 6n.
6. Find the integer closest to 1000 ∑310000 1/(n2-4).
7. Find the smallest n such that ∑1n k2 is a multiple of 200. You may assume ∑1n k2 =
n(n+1)(2n+1)/6.
8. Find the smallest positive integer n for which there are no integer solutions to [2002/x]
= n.
9. Let S = {1, 2, ... , 10}. Find the number of unordered pairs A, B, where A and B are
disjoint non-empty subsets of S.
10. Find the two smallest positive values of x for which sin(xo) = sin(x rad).
11. Two different geometric progressions both have sum 1 and the same second term.
One has third term 1/8. Find its second term.
12. An unfair coin is tossed 10 times. The probability of heads on each toss is 0.4. Let an
be the number of heads in the first n tosses. Find the probability that an/n ≤ 0.4 for n = 1,
2, ... , 9 and a10/10 = 0.4.
13. ABC is a triangle, D lies on the side BC and E lies on the side AC. AE = 3, EC = 1,
CD = 2, DB = 5, AB = 8. AD and BE meet at P. The line parallel to AC through P meets
AB at Q, and the line parallel to BC through P meets AB at R. Find area PQR/area ABC.
14. Triangle APM has ∠ A = 90o and perimeter 152. A circle center O (on AP) has radius
19 and touches AM at A and PM at T. Find OP.
15. Two circles touch the x-axis and the line y = mx (m > 0). They meet at (9,6) and
another point and the product of their radii is 68. Find m.
1
2
3
4
5
6
7
8
9
294
111
(361803√3)/2
42
521
112
49
Outline solutions
9
10
11
12
13
14
15
28501
360π/(180-π),180π/(180+π)
(√5 - 1)/8
243623/510
225/676
95/3
(12√221)/49
1. n = abc, with 1 ≤ a,c ≤ 9. So |n-n'| = 99|a-c|. There are 9 possible values for |a-c|,
namely 0, 1, ... , 8.
2. We find PQ = QR = PR = √98. If a cube has side a and three vertices form an
equilateral triangle, then the triangle must have side a√2, so a = 7. So area = 6a2 = 294.
3. log6a + log6b + log6c = 6 is equivalent to abc = 66. a, b, c GP is equivalent to b2 = ac,
so b3 = 66, and b = 36. b-a is square, so a = 11, 20, 27, 32, 35. But a divides 64, so a = 27.
Hence c = 48.
4. A large hexagon with n unit hexagons on a side can be divided into 1 + 6(1 + 2 + ...
n-1) = 3n2-3n+1 unit hexagons. So for n=202 the hole contains 3·2012 - 3·201 + 1 =
120601 unit hexagons, each area (3√3)/2. Total area (361803√3)/2
5. We must have either 6a > 2a3b, or 6b > 2a3b. If b = 0, we must have 6a > 2a, so a = 1,
2, 3, 4. If a = 0, we must have 6b > 3b, so b = 1, 2. Suppose b > 0 and a > 0. If 6a > 2a3b,
then 6a > 2a3, so 2a > 2a, which is not possible. Similarly, if 6b > 2a3b, then 3b > 3b,
which is not possible. So only solutions 2, 4, 8, 16, 3, 9.
6. Expr = 250 ∑(1/n-2 - 1/n+2) = 250(1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 1/10002) = 520 + 5/6 - 250(1/9999 + 1/10000 + 1/10001 + 1/10002). The last term is
about -1/10 and certainly > -2/6, so nearest integer 521.
7. We want n(n+1)(2n+1) a multiple of 1200 = 243·52. It is always divisible by 3, so we
need it divisible by 16 and 25. n, n+1, 2n+1 are relatively prime. So (A) 16|n, 25|n+1, (B)
16|n, 25|2n+1, (C) 16|n+1, 25|2n+1, (D) 25|n, 16|n+1. If (A), put n = 16k, so 16k+1 is
multiple of 25. Hence also 16k-24 = 8(2k-3), so smallest k = 14, n = 224. If (B), put n =
16k, so 32k+1 is multiple of 25, so 7k+1 is multiple of 25, smallest k = 7, n = 112. If (C),
put n=16k-1, so 32k-1 is multiple of 25, hence also 7k-1. Smallest k = 18, n = 287. If (D),
put n = 25k, so 25k+1 is multiple of 16, hence also 7k-1. Smallest k = 7, n = 175.
8. Suppose we gradually reduce x from 2002. As we go from k+1 to k, 2002/x is
increased by 2002/(k(k+1)). Since 44·45 = 1980 < 2002, this increase is < 1 whilst x
remains above 44. If we reduce h by < 1, then we cannot reduce [h] by more than 1. So
we cannot miss values whilst x ≥ 44. [2002/44] = 45, so we get 1, 2, ... , 45. [2002/43] =
46, [2002/42] = 47, [2002/41] = 48, [2002/40] = 50. So first missed is 49.
9. To get A, B disjoint we must place each element in just A, just B or neither, 3
choices, or 3n in all (for subsets of {1, 2, ... , n}). That counts all except the empty set
twice, so there are (3n-1)/2 possibilities = 29524. But that includes the 2n-1 = 1023
possibilities where one subset is empty (and the other not). Thus there are 29524 - 1023 =
28501.
10. The smallest value x satisfies 180x/π = 180 - x or x = 180π/(180+π) degrees The
next smallest satisfies 180x/π = 360 + x or x = 360π/(180-π).
11. a + ak + ak2 + ... = a/(1-k). So a = 1-k and |k| < 1. If we have another sequence (1-h),
(1-h)h, (1-h)h2 ... with same second term, then (1-k)k2 = (1-h)h2 implies h-k=0 or h+k=1.
h=k gives same sequence, so must have h = 1-k. Must have k ≠ ½ or that also gives same.
Third term = 1/8 gives 8k3 - 8k + 1 = 0, or (2k-1)(4k2-2k-1) = 0. Hence 4k2-2k-1=0. So k
= (1+√5)/4 or (1-√5)/4. Corresponding h=1-k are (3-√5)/4 and (3+√5)/4. Latter is > 1. So
k = (1+√5)/4 and second term (√5-1)/8.
12. Total Hs must be 0 after tosses 1, 2; ≤1 after 3, 4; ≤2 after 5, 6, 7; ≤3 after 8; 3 after
9; 4 after 10. Thus the sequence of throws must be one of these 23:
1
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
2
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
3
H
H
H
H
H
H
H
T
T
T
T
T
T
T
T
T
T
4
T
T
T
T
T
T
T
H
T
T
T
T
T
T
T
T
T
5
H
H
T
T
T
T
T
6
T
T
H
H
T
T
T
7 8
T H
T T
T H
T T
H H
H T
T H
7 poss
H H T H
H H T T
H T H H
H T H T
H T T H
T H H H
T H H T
T H T H
T T H H
9 10
T H
H H
T H
H H
T H
H H
H H
as above
T H
H H
T H
H H
H H
T H
H H
H H
H H
So prob. 23(0.4)4(0.6)6 = 23·2436/510.
13. Take line through E parallel to AD. Suppose it meets CD at F. Then CF/CD =
CE/CA = 1/4, so CF = 1/2. Hence BP/BE = BD/BF = 5/(6.5) = 10/13. So PQ/EA = 10/13
also (triangles BPQ, BEA). Hence PQ = 30/13. PQR and CAB are similar, so area
PQR/area ABC = (PQ/CA)2 = (30/52)2 = (15/26)2
14. Put OP = x, OPT has sides x, 19, √(x2-361). It is similar to MPA which therefore has
perimeter (19 + x + √(x2-361)) (19+x)/√(x2-361) = 152. Put x = 19y. Dividing through by
19: (1 + y + √(y2-1) ) (1+y)/√(y2-1) = 8. Rearranging: (7-y)√(y2-1) = (1+y)2. Squaring etc,
9y2 - 30y + 25 = 0, or (3y-5) = 0, so y = 5/3, x = 95/3.
15. Suppose centers lie on y = nx. Then using tan 2k = 2 tan k/(1-tan2k), we have m =
2n/(1-n2). If the center is (a,na), then radius must be na. So (a-9)2 + (na-6)2 = (na)2, or a2 (18+12n)a + 117 = 0. So the product of the two possible values of a is 117. Hence the
product of the two radii is 117n2 = 68. Hence m = 2n/(1-n2) = (12√221)/49.
1. Find positive integers k, n such that k·n! = (((3!)!)!/3! and n is as large as possible.
2. Concentric circles radii 1, 2, 3, ... , 100 are drawn. The interior of the smallest circle is
colored red and the annular regions are colored alternately green and red, so that no two
adjacent regions are the same color. Find the total area of the green regions divided by
the area of the largest circle.
3. S = {1, 2, 3, 5, 8, 13, 21, 34}. Find ∑ max(A) where the sum is taken over all 28 twoelement subsets A of S.
4. Find n such that log10sin x + log10cos x = -1, log10(sin x + cos x) = (log10n - 1)/2.
5. Find the volume of the set of points that are inside or within one unit of a rectangular
3 x 4 x 5 box.
6. Let S be the set of vertices of a unit cube. Find the sum of the areas of all triangles
whose vertices are in S.
7. The points A, B, C lie on a line in that order with AB = 9, BC = 21. Let D be a point
not on AC such that AD = CD and the distances AD and BD are integral. Find the sum of
all possible n, where n is the perimeter of triangle ACD.
8. 0 < a < b < c < d are integers such that a, b, c is an arithmetic progression, b, c, d is a
geometric progression, and d - a = 30. Find a + b + c + d.
9. How many four-digit integers have the sum of their two leftmost digits equals the sum
of their two rightmost digits?
10. Triangle ABC has AC = BC and ∠ ACB = 106o. M is a point inside the triangle such
that ∠ MAC = 7o and ∠ MCA = 23o. Find ∠ CMB.
11. The angle x is chosen at random from the interval 0o < x < 90o. Find the probability
that there is no triangle with side lengths sin2x, cos2x and sin x cos x.
12. ABCD is a convex quadrilateral with AB = CD = 180, perimeter 640, AD ≠ BC, and
∠ A = ∠ C. Find cos A.
13. Find the number of 1, 2, ... , 2003 which have more 1s than 0s when written in base
2.
14. When written as a decimal, the fraction m/n (with m < n) contains the consecutive
digits 2, 5, 1 (in that order). Find the smallest possible n.
15. AB = 360, BC = 507, CA = 780. M is the midpoint of AC, D is the point on AC such
that BD bisects ∠ ABC. F is the point on BC such that BD and DF are perpendicular. The
lines FD and BM meet at E. Find DE/EF.
1
2
3
4
5
6
7
8
120, 719
101/200
484
12
154+40π/3
12+12√2+4√3
380
129
9
10
11
12
13
14
15
615
83
(2 tan-12)/π
7/9
1155
127 (32/127)
49/240
Outline solutions
1. ((3!)!)! = 720!, so n = 719.
2. green/π = (22-12) + (42-32) + ... + (1002-992) = (2·1+1) + (2·3+1) + ... + (2·99+1) = 50
+ 50·100. So ratio = 101/200.
3. sum = 7·34 + 6·21 + 5·13 + 4·8 + 3·5 + 2·3 + 1·2 = 484
4. sin x cos x = 1/10, so (sin x + cos x)2 = sin2x + 2 sin x cos x + cos2x = 1 + 1/5 = 6/5.
Hence log(sin x + cos x) = ½ log 6/5 = ½ log 12/10 = (log 12 - 1)/2.
5. The box is 60. There are 4 quarter-cylinders length 3 and similarly for the other
edges, total 3π + 4π + 5 π = 12π. There are 8 one-eighth spheres at the vertices, total 4π/3,
and there are boxes height 1 on each face, 12+12+15+15+20+20 = 94. Total 154 + 40π/3
6. There are 24 half-faces, 24 with one edge a long-diagonal, and 8 equilateral (side √2).
Hence 12 + 12√2 + 4√3.
7. Let M be midpoint of AC. D must lie on perpendicular at M. Suppose DM = x. Note
that x does not have to be an integer. Put k = x2. k+36 = a2 and k+225 = b2 are squares. ba = 1 gives 952-942. b-a = 3 gives 332-302. We must have b-a divides 189 = 337. b-a = 7
gives 172-102. b-a = 9 gives 152-62. b-a = 21 or more does not work. b-a = 9 also does not
work because it gives D on AC. So perimeters 2·95+30, 2·33+30, 2·17+30 total 380.
8. a, a+d, a+2d, (a+2d)2/(a+d). So (a+2d)2-a(a+d) = 30(a+d) or 3ad + 4d2 = 30a + 30d.
Hence d is multiple of 3. Also d(4d-30) = 3a(10-d). So 7.5 < d < 10. Hence d = 9. Hence
a = 18. So 18, 27, 36, 48. Sum 129.
9. Suppose the 2-digit sum is d. We must have 1 ≤ d ≤ 18. For d ≤ 9, ther e are d+1
choices for the 2nd pair and one less for the 1st pair (no. cannot have leading 0). For d >
9, there are 19-d choices for each pair. Hence 1·2 + 2·3 + ... + 9·10 + 92 + 82 + ... + 12 =
615.
10. Take X on the line AM so that angle XBC = 7o. Since CA = CB, X must lie on the
angular bisector of ∠ C. So ∠ BCX = 53o. Hence ∠ XCM = 53o-∠ ACM = 30o. Also
∠ XMC = 7o+23o = 30o. Now XB is perpendicular to MC. But XM = XC, so it must be
the perpendicular bisector of MC. Hence ∠ CMB = ∠ MCB = 83o.
11. If x is too small then cos2x > sin2x + sin x cos x. We have equality when x = k,
where cos 2k = ½ sin 2k, or k = ½ tan-12. There is symmetry about π/4, so prob. of no
triangle is k/(π/4).
12. The idea is shown in the diagram. Take BC' = BD and triangles ABC', CDB
congruent. We have AB = 180 and AC' + AD = 280. So if M is the midpoint of C'D, then
M = 140. Hence cos A = 140/180 = 7/9.
13. There are 2nCn binary strings length 2n with equal numbers of 0s and 1s. Half the
others have more 1s. So (22n-2nCn)/2 with at least as many 1s. Hence (22n-2nCn)/2
strings length 2n+1 with more 1s. Half the strings length 2n-1 have more 1s, so 22n-2
strings length 2n have leading 1 and more 1s. Note that 2C1 = 2, 4C2 = 6, 6C3 = 20, 8C4
= 70, 10C5 = 252. So for strings with leading 1 we have the following numbers with
more 1s:
length 1, 1
length 2, 1
length 3, 2+1=3
length 4, 4
length 5, 8+3=11
length 6, 16
length 7, 32+10=42
length 8, 64
length 9, 128+35=163
length 10, 256
length 11, 512+126=638
Total 1199.
Now 1984 = 11111000000, so all 44 numbers in the range 2004 to 2047 have more 1s
than 0s. They must be excluded, giving 1155.
14. If n x 0. ... 251 ... = integer, then n x 0.251 ... = integer (where we simply delete the
digits before 251. So it is sufficient to consider 0.251... . Obviously if n x 0.251 = integer,
then n must be a multiple of 1000. So assume n < 1000. We require that [n x 0.251] = m,
[n x 0.252] = m+1 for some m. Let {x} denote the fractional part of x. Then since n x
0.252 - n x 0.251 = n/1000, we must have {n x 0.251} > 1 - n/1000. If n is a multiple of
4, then {n x 0.251} = {n/1000}, so we need n > 500. If n = 1 mod 4, then {n x 0.251} =
{0.25 + n/1000}, so n > 375. Similarly, if n = 2 mod 4, then n > 250, and if n = 3 mod 4,
then n > 125. Thus the smallest candidate is 127, and we expect that 32/127 = 0.251... .
That is easily checked.
15. Extend FD to meet the line BA at X. Then BX = BF. Now considering BXF, we
have (AX/AB)(DF/DX)(CB/CF) = 1, so we get BF/BC = 720/867. Since BD is the angle
bisector we have DC/DA = BC/BA. So (390+MD)/(390-MD) = 507/360. Hence MD/MC
= 147/867. Finally, considering DFC, we have (ED/EF)(BF/BC)(MC/MD) = 1, so ED/EF
= (867/720)(147/867) = 147/720 = 49/240.
1. The product N of three positive integers is 6 times their sum. One of the integers is the
sum of the other two. Find the sum of all possible values of N.
2. N is the largest multiple of 8 which has no two digits the same. What is N mod 1000?
3. How many 7-letter sequences are there which use only A, B, C (and not necessarily all
of those), with A never immediately followed by B, B never immediately followed by C,
and C never immediately followed by A?
4. T is a regular tetrahedron. T' is the tetrahedron whose vertices are the midpoints of the
faces of T. Find vol T'/vol T.
5. A log is in the shape of a right circular cylinder diameter 12. Two plane cuts are made,
the first perpendicular to the axis of the log and the second at a 45o angle to the first, so
that the line of intersection of the two planes touches the log at a single point. The two
cuts remove a wedge from the log. Find its volume.
6. A triangle has sides 13, 14, 15. It is rotated through 180o about its centroid to form an
overlapping triangle. Find the area of the union of the two triangles.
7. ABCD is a rhombus. The circumradii of ABD, ACD are 12.5, 25. Find the area of the
rhombus.
8. Corresponding terms of two arithmetic progressions are multiplied to give the
sequence 1440, 1716, 1848, ... . Find the eighth term.
9. The roots of x4 - x3 - x2 - 1 = 0 are a, b, c, d. Find p(a) + p(b) + p(c) + p(d), where p(x)
= x6 - x5 - x3 - x2 - x.
10. Find the largest possible integer n such that √n + √(n+60) = √m for some non-square
integer m.
11. ABC has AC = 7, BC = 24, angle C = 90o. M is the midpoint of AB, D lies on the
same side of AB as C and had DA = DB = 15. Find area CDM.
12. n people vote for one of 27 candidates. Each candidate's percentage of the vote is at
least 1 less than his number of votes. What is the smallest possible value of n? (So if a
candidate gets m votes, then 100m/n ≤ m-1.)
13. A bug moves around a wire triangle. At each vertex it has 1/2 chance of moving
towards each of the other two vertices. What is the probability that after crawling along
10 edges it reaches its starting point?
14. ABCDEF is a convex hexagon with all sides equal and opposite sides parallel. Angle
FAB = 120o. The y-coordinates of A, B are 0, 2 respectively, and the y-coordinates of the
other vertices are 4, 6, 8, 10 in some order. Find its area.
15. The distinct roots of the polynomial x47 + 2x46 + 3x45 + ... + 24x24 + 23x23 + 22x22 +
... + 2x2 + x are z1, z2, ... , zn. Let zk2 have imaginary part bki. Find |b1| + |b2| + ... + |bn|.
1 336
9 6
2
3
4
5
6
7
8
120
192
1/27
216π
112
400
348
10
11
12
13
14
15
48
(527/40)√11
134
171/512
48√3
8 + 4√3
Outline solutions
1. Let the numbers be a, b, a+b. So ab(a+b) = 12(a+b). Hence ab = 12. So {a, b} = {1,
12}, {2, 6} or {3, 4}. Hence N = 156, 96 or 84, sum 336.
2. The best we can hope for is to use each digit just once, and for the larger digits to
occur earlier. Any multiple of 1000 is divisible by 8, so 9876543000 is a multiple of 8.
Now 210 and 201 are not multiples of 8. The next best is 120. So the number is
9876543120, with residue 120.
3. We have 3 choices for the first letter, and 2 for each subsequent letter, so 3·26.
4. The centroid of each face is 1/3 of the way up the median, so the top face of T' is 1/3
of the way up the height of T. T' is obviously regular, so it is 1/3 the scale of T, and its
vol is 1/27 of T.
5. Take another plane cut parallel to the first and with its intersection to the second
parallel to the intersection of the first two and also meeting the log at a single point. Then
the two wedges are congruent, and together they form a cylinder diameter 12, length 12
and hence vol 12·π62.
6. The overlap is a hexagon, and there are six congruent triangles outside the hexagon,
each similar to the original triangle and 1/3 the (linear) size (because the centroid divides
the median in the ratio 2:1). So the area of the union is 4/3 times the area of the triangle.
We have the semiperimeter s = 21, so by Heron's formula the triangle has area
√(21·8·7·6) = 84.
7. AD = 2 AO cos x and = 2 AO' cos y = 2 AO' sin x = 4 AO sin x, so cos x = 2 sin x, so
tan x = 1/2, cos x = 2/√5. Area rhombus = 2 AD2 sin x cos x = 25 cos4x = 400.
8. (A-D)(a-d) = 1440, Aa = 1716, (A+D)(a+d) = 1848, so 2Dd = 1440 + 1848 - 2·1716
= -144. Hence (Ad + Da) = 204, so 8th term (A+6D)(a+6d) = Aa + 36Dd + 6(Ad+Da) =
1716 - 36·72 + 6·204 = 348.
9. x4-x3-x2-1 = (x+1)(x3-2x2+x-1), so a = -1, b+c+d = 2, bc+cd+db = 1. Hence b2+c2+d2
= 22-2 = 2. We have p(x) = (x3-2x2+x-1)(x3+x2+x+1)+x2-x+1. Hence p(a) = 3, p(b) = b2b+1, p(c) = c2-c+1, p(d) = d2-d+1, so p(a)+p(b)+p(c)+p(d) = 6 + (b2+c2+d2)-(b+c+d) = 6.
10. Squaring, we see that n(n+60) must be a square. If either of n, n+60 is a square, then
the other is and hence m is a square. Contradiction. So neither n nor n+60 is a square.
Thus we must have n = a2d, n+60 = b2d for some non-square d. So 60 = d(b2-a2). b+a and
b-a have the same parity, so b2-a2 must be odd or a multiple of 4. Hence d = 3, 5 or 15.
This leads to n = 20 or 48 as the only solutions.
11. Extend AC and MD to meet at X. Let Y be the foot of the perpendicular from C to
DM. XMA and BCA are similar, so XA = 625/14. Hence CY/AM = XC/XA = 527/625,
and CY = 527/50. DM = (5/2)√11, so area CDM = (527/40)√11.
12. n = 135 works with 5 votes per candidate. Then each has 5/135 = 3.7%. No
candidate can have only 1 vote. If a candidate has just 2 votes, then 2/n <= 1/100, so n >=
200. If a candidate has 3 votes, then n >= 150. So in a minimal solution each candidate
must have at least 4 votes. If all have at least 5, then n >= 135. If a candidate has 4, then
4/n <= 3/100, so n >= 134. This can be achieved: 1 candidate has 4 votes, the other 26
have 5 each. Then 5/134 = 3.7%, 4/134 = 2.99%.
13. Label vertices 0, 1, 2. Suppose A moves +1 mod 3 and B moves +2 mod 3. Then A
+ 2B = 0 mod 3 and A + B = 10. Hence A = B = 2 mod 3. So (A, B) = (2, 8), (5, 5) or (8,
2). There are 10C2, 5C5, 10C8 possibilities of each type. Hence 45 + 252 + 45 = 342 in
all. So prob = 171/512.
14. wlog B is lower than F, so the y-coords of B, C, D, E, F must be 2, 6, 10, 8, 4
respectively. Angle A = 120o, so y = 60o - x. B is half the height of F, so sin(60o-x) = 2
sin x, giving tan x = (√3)/5 and side AB = 4√(7/3) = k. E is twice the height of F and FE
= FA, so E must be vertically above A. Similarly, D is vertically above B. Area AFE =
area BCD = 4k cos y. Area ABDE = 8k cos x. So area hexagon = 8k(cos x + cos y) =
4k(3 cos x + √3 sin x) = 48√3.
15. The polynomial factorises as x(x23 + x22 + ... + 1)2, with distinct roots x = 0, and
e2πki/24 for k = 1, 2, .. , 23. We can ignore x = 0, whose square is real. The square of e2πki/24
is e2πki/12 with imaginary part sin(πk/6). So the sum is 4 + 8 sin(π/6) + 8 sin(&pi/3) (and 3
zero terms) = 8 + 4√3.
1. n has 4 digits, which are consecutive integers in decreasing order (from left to right).
Find the sum of the possible remainders when n is divided by 37.
2. The set A consists of m consecutive integers with sum 2m. The set B consists of 2m
consecutive integers with sum m. The difference between the largest elements of A and B
is 99. Find m.
3. P is a convex polyhedron with 26 vertices, 60 edges and 36 faces. 24 of the faces are
triangular and 12 are quadrilaterals. A space diagonal is a line segment connecting two
vertices which do not belong to the same face. How many space diagonals does P have?
4. A square X has side 2. S is the set of all segments length 2 with endpoints on adjacent
sides of X. The midpoints of the segments in S enclose a region with area A. Find 100A
to the nearest whole number.
5. A and B took part in a two-day maths contest. At the end both had attempted questions
worth 500 points. A scored 160 out of 300 attempted on the first day and 140 out of 200
attempted on the second day, so his two-day success ratio was 300/500 = 3/5. B's
attempted figures were different from A's (but with the same two-day total). B had a
positive integer score on each day. For each day B's success ratio was less than A's. What
is the largest possible two-day success ratio that B could have achieved?
6. An integer is snakelike if its decimal digits d1d2...dk satisfy di < di+1 for i odd and di >
di+1 for i even. How many snakelike integers between 1000 and 9999 have four distinct
digits?
7. Find the coefficient of x2 in the polynomial (1-x)(1+2x)(1-3x)...(1+14x)(1-15x).
8. A regular n-star is the union of n equal line segments P1P2, P2P3, ... , PnP1 in the plane
such that the angles at Pi are all equal and the path P1P2...PnP1 turns counterclockwise
through an angle less than 180o at each vertex. There are no regular 3-stars, 4-stars or 6stars, but there are two non-similar regular 7-stars. How many non-similar regular 1000stars are there?
9. ABC is a triangle with sides 3, 4, 5 and DEFG is a 6 x 7 rectangle. A line divides ABC
into a triangle T1 and a trapezoid R1. Another line divides the rectangle into a triangle T2
and a trapezoid R2, so that T1 and T2 are similar, and R1 and R2 are similar. Find the
smallest possible value of area T1.
10. A circle radius 1 is randomly placed so that it lies entirely inside a 15 x 36 rectangle
ABCD. Find the probability that it does not meet the diagonal AC.
11. The surface of a right circular cone is painted black. The cone has height 4 and its
base has radius 3. It is cut into two parts by a plane parallel to the base, so that the
volume of the top part (the small cone) divided by the volume of the bottom part (the
frustrum) equals k and painted area of the top part divided by the painted are of the
bottom part also equals k. Find k.
12. Let S be the set of all points (x,y) such that x, y ∈ (0,1], and [log2(1/x)] and
[log5(1/y)] are both even. Find area S.
13. The roots of the polynomial (1 + x + x2 + ... + x17)2 - x17 are rk ei2πak, for k = 1, 2, ... ,
34 where 0 < a1 ≤ a2 ≤ ... ≤ a34 < 1 and rk are positive. Find a1 + a2 + a3 + a4 + a5.
14. A unicorn is tethered by a rope length 20 to the base of a cylindrical tower. The rope
is attached to the tower at ground level and to the unicorn at height 4 and pulled tight.
The unicorn's end of the rope is a distance 4 from the nearest point of the tower. Find the
length of the rope which is in contact with the tower.
15. Define f(1) = 1, f(n) = n/10 if n is a multiple of 10 and f(n) = n+1 otherwise. For each
positive integer m define the sequence a1, a2, a3, ... by a1 = m, an+1 = f(an). Let g(m) be the
smallest n such that an = 1. For example, g(100) = 3, g(87) = 7. Let N be the number of
positive integers m such that g(m) = 20. How many distinct prime factors does N have?
1
2
3
4
5
6
7
8
217
201
241
858
349/500
882
-588
199
9
10
11
12
13
14
15
3/32
375/442
125/387
5/9
159/323
(60 - √750)/3
29509
Outline solutions
1. n = 9876 = 34 mod 37, or 8765 = 33 mod 37, or 7654 = 32 mod 37, or 6543 = 31 mod
37, or 5432 = 30 mod 37, or 4321 = 29 mod 37, or 3210 = 28 mod 37. So sum = 34 + 33
+ 32 + 31 + 30 + 29 + 28 = 217.
2. (a+1) + ... + (a+m) = m(2a+m+1)/2 = 2m, and (b+1) + ... + (b+2m) = m(2b+2m+1) =
m. Also b+m-a = 99. Solving a = -99, b = -201, m = 201.
3. Label vertices i = 1 to 26. Let i belong to ti triangles and qi quadrilaterals. Then i
belongs to (25-ti-2qi) space diagonals. So total = ½∑(25-ti-2qi) = 325 - ½∑ ti - ∑ qi = 325
- 72/2 - 48 = 241.
4. If ER = 2, and O is the midpoint of ER, then O is the center of the circle through
E,Q,R, so OQ = 1. Hence O lies on the quarter circle center Q radius 1. Thus the
midpoints enclose the region shown in the second diagram.
So A = 4 - π, and 100A is about 858.
5. A's ratio was 8/15 on day 1, 7/10 on day 2, and 8/15 < 7/10. We want B's ratio to be
weighted towards day 2, so we take 1/2 on day 1 and 348/498 on day 2 (we must have a
total of 500 attempted), giving 349/500 overall.
6. Given digits a < b < c < d, the possible snakelike numbers are acbd, adbc, bcad, bdac,
cdab, unless a = 0, in which case only the last three are possible. There are 9C4 = 126
ways of choosing 4 non-zero digits, and 9C3 = 84 ways of choosing 3 non-zero digits, so
5·126 + 3·84 = 882 snakelike 4-digit nos.
7. -1+2-3+4-...+14-15 = -8. So twice coeff is -1(-8+1) + 2(-8-2) - 3(-8+3) + ... -15(8+15) = (-1+2-3+...-15)(-8) - (12+22+...+152) = 64 - 15·16·31/6 = -1176, and coeff is 588.
8. Evidently Pi must be the vertices of a regular n-gon (but in a different order). So we
want number of m st 2 ≤ m ≤ 499 with m coprime to 1000, in other words odd and a not a
multiple of 5. There 249 odd nos. 3, 5, 7, ... , 499, of which 50 are multiples of 5: 5, 15, ...
, 495. Hence 199 non-similar stars.
9. The line dividing the rectangle must pass through a vertex, so take it to be DZ.
If D is not the smallest angle of DZG, then GZ ≥ 4/3 DG. Contradiction. So D is the
smallest angle, so either DG = 6 and GZ = 4½, or DG = 7 and GZ = 5¼. We want area
AXY to be as small as possible, so BX/BC to be as big as possible and hence EF/DE to
be as big as possible. So we take EF = 7. Hence AX = 1/2, XY = 3/8, area AXY = 3/32.
10. If the circle lies entirely inside the rectangle, then its center must lie inside a 13x34
rectangle, area 442. If it does not intersect the diagonal, then its center must lie inside one
of the two triangles shown. Note that a 5, 12, 13 triangle has inradius 2 (because if the
inradius is r, then calculating the area two ways we get 5·12/2 = r(5+12+13)/2). So the
triangles are 12½,30, 32½. Thus their area is 30·12½ = 375. So prob = 375/442
11. Let small cone have height h. Then vol cone/vol frustrum = h3/(64-h3). If we unroll
large cone, it forms part of circle with radius 5, arc length 6π, so area (6π/10π)25π = 15π.
Base has area 9π. So painted area of small cone is h215π/16, painted area of frustrum = 9π
+ (16-h2)15π/16. Thus k = h3/(64-h3) = 5h2/(128-5h2). So h = 5/2, k = 125/387.
12. We must have x ∈ (1/2, 1] ∪ (1/8, 1/4] ∪ (1/32, 1/16] ∪ ... and y ∈ (1/5, 1] ∪
(1/125, 1/25] ∪ (1/3125, 1/625] ∪ ... So we have a set of rectangles with total area (1/2 +
1/8 + 1/32 + ... )(4/5 + 4/125 + 4/3125 + ... ) = 1/2(1 + 1/4 + 1/42 + ... )4/5(1 + 1/25 +
1/252 + ... ) = (2/5)(1/(1 - 1/4))(1/(1 - 1/25)) = (2/5)(4/3)(25/24) = 5/9.
13. We have x36 - 2x18 + 1 = x19 - 2x18 + x17, so (x19 - 1)(x17 - 1) = 0. We exclude the two
roots x = 1 (which we introduced by multiplying by (x-1)2). So the ai are 1/19, 2/19, ... ,
18/19 and 1/17, 2/17, ... , 16/17. Thus sum = 1/19 + 2/19 + 3/19 + 1/17 + 2/17 = 6/19 +
3/17 = 159/323.
14. The first diagram shows a horizontal plane through the point where the rope is
attached to the unicorn. Evidently the tangent is length 4√5.
Unroll the surface of the tower. We have AC = 20, CD = 4√5, DE = 4, so AB = (8√6 4√5)20/(8√6) = (60 - √750)/3.
15. Most integers have two possible predecessors. Let M be the operation of moving
back from n to n-1 and D the operation of moving back from n to 10n. There are two
related difficulties. We cannot take M10 on a multiple of 10 (because M does not apply to
multiples of 10 plus 1); we cannot even take M9 on 10 (because having reached 1 we
must stop). The predecessor of 1 must be 10. Then we need a further 18 moves back. So,
leaving aside the difficulties above, there are 218 possibilities. We must exclude the 29
M9X...X, where X denotes M or D. Then we must exclude the 27 DM10X...X. Similarly,
we must exclude the 27 XDM10X...X, the 27 XXDM10X...X, ... , and the 27 X...XDM10.
Thus we are left with 218 - 29 - 8·27 = 218 - 3·29 = 29509.