SW-ARML Practice 9-16-12
Individual Problems
1.
Compute the largest prime divisor of 15! – 13!.
[Hint: Use factoring.]
Factor 15! – 13! to obtain 13!(1514 – 1) = 13!209. The largest prime divisor of 13! is 13, so
continue by factoring 209 = 1119. Thus the largest prime divisor of 15! – 13! is 19.
2.
3. A circle with center O and radius 1 contains chord AB of length 1, and point M is the midpoint
of AB. If the perpendicular to AO through M intersects AO at P, compute the area of MAP.
[Hints: ABO is an equilateral triangle. Also, in a 30–60–90 triangle, the side facing the 30
angle is half of the hypotenuse in length.]
Draw auxiliary segment OB, as shown in the diagram below.
Triangle OAB is equilateral, so mOAB = 60.
Then MAP is a 30–60–90 triangle with hypotenuse
AM = 1/2.
Thus AP = 1/4 and MP =
=
3.
3 / 4 , so the area of MAP
1 1 3
3
 

.
2 4 4
32
Suppose that p and q are two-digit prime numbers such that p2 – q2 = 2p + 6q + 8. Compute the
largest possible value of p + q.
[Hints: Complete squares. X2 – Y2 = (X – Y)( X + Y). The largest two-digit prime is 97.]
Subtract from both sides and regroup to obtain p2 – 2p – q2 – 6q = 8. Completing both squares
yields (p – 1)2 – (q + 3)2 = 0. The left side is a difference of two squares; factor to obtain (p – 1 +
q + 3)(p – 1 – q – 3) = 0, whence (p + q +2)(p – q – 4) = 0. For positive primes p and q, the first
factor p + q +2 must also be positive. Therefore the second factor p – q – 4 must be zero, hence p
– 4 = q. Now look for primes starting with 97 and working downward. If p = 97, then q = 93,
which is not prime; if p = 89, then q = 85, which is also not prime. But if p = 83, then q = 79,
which is prime. Thus the largest possible value of p + q is 83 + 79 = 162.
4.
The four zeros of the polynomial x4 + jx2 + kx + 225 are distinct real numbers in arithmetic
progression. Compute the value of j.
[Hints: For any polynomial xn + axn-1 +  + c (with leading coefficient 1), the sum of all of its
zeros is equal to –a, and the product of all of its zeros is equal to (–1)nc. An arithmetic
progression means there is a common difference between any two consecutive terms or,
graphically, they are evenly spaced on the number line.]
1
Let the four zeros be p  q  r  s. The coefficient of x3 is 0, so p + q + r + s = 0. The mean of four
numbers in arithmetic progression is the mean of the middle two numbers, so q = –r. Then the
common difference is r – q = r – (–r) = 2r, so s = r + 2r = 3r and p = q – 2r = –3r. Therefore the
four zeros are –3r, –r, r, 3r. The product of the zeros is 9r4; referring to the original polynomial
and using the product of roots formula gives 9r4 = 225: Thus r = 5 the zeros are

3 5,  5, 5,3 5 , and the polynomial can be factored as x  5
 x  5  x  3 5  x  3 5  .
Expanding this product yields (x – 5)(x – 45)
= x4 – 50x2 + 225, so j = –50.
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5.
2
If n is a positive integer, then n!! is defined to be n(n – 2)(n – 4)2 if n is even and
n(n – 2)(n – 4) 1if n is odd. For example, 8!! = 864 2 = 384 and 9!! = 97531 = 945.
Compute the number of positive integers n such that n!! divides 2012!!.
[Hints: The case when n is even should be obvious. Almost all odd solutions can be obtained by
cleverly rearranging the terms of 2012!! and factoring out 2’s. 1007 = 19  53; 1009 is prime.]
If n is even and n  2012, then n!!  2012!! trivially, while if n > 2012, 2012!! < n!!, so n!! cannot
divide 2012!!. Thus there are a total of 1006 even values of n such that n!!  2012!!.
If n is odd and n < 1006, then n!!  2012!!. To show this, rearrange the terms of 2012!! and factor:
2012!! = 2  4  6    2010  2012 = (2  6  10    2010)(4  8  12    2012)
= 2503(1  3  5    1005)(4  8  12    2012).
However, the condition n < 1006 and odd is not necessary, only sufficient, because n!! also
divides 2012!! if 10071009   n  (4  8  12    2012). (The factor of 2503 is irrelevant because all
the factors on the left side are odd.) The expression (4  8  12    2012) can be factored as 4503(1 
2  3    503) = 4503  503!. Examining the numbers 1007, 1009, … in sequence shows that 1007 is
satisfactory, because 1007 = 19  53. On the other hand, 1009 is prime, so it cannot be a factor of
4503  503!. Thus the largest possible odd value of n is 1007, and there are 504 odd values of n
altogether. The total is 1006 + 504 = 1510.
6.
One face of a 222 cube is painted (not the entire cube), and the cube is cut into eight 111
cubes. The small cubes are reassembled randomly into a 222 cube. Compute the probability
that no paint is showing.
[Hint: Focus on the 111 cubes that have painted faces.]
Call each 111 cube a cubelet. Then four cubelets are each painted on one face, and the other
four cubelets are completely unpainted and can be ignored. For each painted cubelet, the painted
face can occur in six positions, of which three are hidden from the outside, so the probability that
a particular painted cubelet has no paint showing is 3/6 = 1/2. Thus the probability that all four
painted cubelets have no paint showing is (1/2)4 = 1/16.
Team Problems
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7.
In ABC, mA = mB = 45 and AB = 16. Mutually tangent circular arcs are drawn centered at
all three vertices; the arcs centered at A and B intersect at the midpoint of AB. Compute the area of
the region inside the triangle and outside of the three arcs.
[Hints: Calculate the length of AC and BC. Proportionally compute the area of the three circular
sectors.]
16
 8 2 . Then each of the large arcs has radius 8, and the small
2
arc has radius 8 2  8 . Each large arc has measure 45 and the small arc has measure 90.
45
 π  82  8π , and the area enclosed by the
Therefore the area enclosed by each large arc is
360
90
 π  (8 2  8) 2  48π  32 2π . Thus the sum of the areas enclosed by the three
small arc is
360
2
1
8 2  64 . So the area of
arcs is 64π  32 2π . On the other hand, the area of the triangle is
2
the desired region is 64  64π  32 2π .
Because AB = 16, AC = BC =

8.

Compute the number of ordered pairs of integers (a, b) such that 1 < a  50, 1 < b  50, and logb a
is rational.
[Hints: logb a is rational if and only if a and b are integral powers of some integer d. Partition {2,
3, 4…., 50} accordingly.]
Begin by partitioning {2, 3, …, 50} into the subsets
A = {2, 4, 8, 16, 32}
B = {3, 9, 27}
C = {5, 25}
D = {6; 36}
E = {7; 49}
F = all other integers between 2 and 50, inclusive:
logb a is rational, then either a and b are both members of one of the sets A, B, C, D, or E, or a = b
in F. Then the number of possible ordered pairs is
2
2
2
2
2
A  B  C  D  E  F = 25 + 9 + 4 + 4 + 4 + 35 = 81.
9.
Suppose that 5-letter “words" are formed using only the letters A, R, M, and L. Each letter need
not be used in a word, but each word must contain at least two distinct letters. Compute the
number of such words that use the letter A more than any other letter.
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[Hints: Depending on how many times the letter A is used, calculate the number of possibilities
for each case. The number of distinct permutations of distinct objects, where the first object is
used n1 times, the second object n2 times, …., and the (last) kth object nk times, is equal to
(n1  n2   nk )!
. For example, if X is used 3 times, Y is used 2 times, and Z is used 4 times,
n1 !n2 ! nk !
9!
then there are
possible ways to form a nine-letter “word” using X, Y, and Z according to
3!2!4!
the specified multiplicities.]
Let the number n of A's that appear in the word; n is at least two, because of the requirement that
A occur more often than any other letter, and n is at most 4, because of the requirement that there
be at least two distinct letters.
In the case n = 4, there are 3 choices for the other letter, and 5 choices for where to place it, for a
total of 15 possibilities.
In the case n = 3, there are two possibilities to consider: either a second letter occurs twice, or
there are two distinct letters besides A. If a second letter occurs twice, there are 3 choices for the
5! 1  2  3  4  5

 10 ways to arrange the three A's and two non-A's for their
other letter, and
3!2! 1  2  3 1  2
locations, for a total of 30 choices. If there are two distinct letters besides A, then there are
 3
5!
1 2  3  4  5

 20 ways to arrange them, for a total
   3 ways to pick the two letters, and
3!1!1!
1 2  3
 2
of 60 words. Thus there are a combined total of 90 words when n = 3.
In the case n = 2, no other letter can occur twice, so all the letters R,M, L, must appear in the
5!
1 2  3  4  5

 60 ways.
word; they can be arranged in
2!1!1!1!
1 2
The total number of words satisfying the conditions is therefore 15 + 90 + 60 = 165.
10. Positive integers a1; a2; a3, …. form an arithmetic sequence. If a1 = 10 and aa2 = 100, compute
aaa .
3
[Hints: aaa means a
3
 a  
a3
. For example, if a3 = 5, then aaa = aa5 . Then, if a5 = 17, then aaa =
3
3
a17 . The main goal is to figure out the common difference d. In an arithmetic sequence, an = a1 +
(n – 1)d. Write out aa2 in terms of a2 and re-apply the formula to a2.]
Let d be the common difference of the sequence. Then aa2 = a1 + (a2 – 1)d = 100
 (a2 – 1)d = 90. But a2 = a1 + d = 10 + d, so (9 + d)d = 90. Solving the quadratic yields
d = –15 or d = 6, but the requirement that ai be positive for all i rules out the negative value, so d =
6 and an = 10 + (n – 1)6. Thus a3 = 10 + 2(6) = 22, and aa3 = a22 = 10 + 21(6) = 136. Finally,
aaa = a136 = 10 + 135(6) = 820.
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