Leap Frog Relay II Solutions–2006
No calculators allowed
Correct Answer = 4, Incorrect Answer = −1, Blank = 0
1. The sum of the prime divisors (each distinct prime divisor counted once
regardless of exponent) of 1776 is
.
(a) 19
(b) 21
(c) 113
(d) 42
(e) None of these
Solution. (d) 1776 = 24 × 3 × 37, which can easily be ascertained
from a factor tree. So the sum of the prime divisors is
2 + 3 + 37 = 42.
2.
2005
X
k=1
log2006
k+1
k
!
=
.
(a) − log2006 2005
(b) log2006 2005
(c) log2006 2 − log2006 2005
(d) log2006 2 + log2006 2005
(e) None of these
Solution. (e) Use the rule log(A/B) = log(A) − log(B) to expand
log2006
k+1
= log2006 (k + 1) − log2006 (k).
k
1
Then expand the sum, canceling terms. You will be left with
− log2006 1 + log2006 2006 = 1.
3. In the figure below, E is the center of the circle and bisects of the side
DC of the inscribed square ABCD. Then, the ratio of square ABCD
.
area to circle area is
A
D
B
E
C
5
6π
2
(d)
3π
3
4π
4
(c)
5π
(b)
(a)
(e) None of these
Solution. (c) Draw the radius of the circle as EB as pictured below.
A
D
B
E
C
Now scale the picture so EC = 1 and BC = 2. By the pythagorean
√
theorem, EB = 5. So the area of the circle is 5π. The area of square
ABCD is 4. So the ratio is
4
.
5π
4. Let (0.xyxyxy . . .)b and (0.yxyxyx . . .)b be the base b representations
of the two numbers A and B respectively, where x and y represent base
2
b digits, not both of which are zero. Then,
A
=
B
y+b
x+b
xb + y
(c)
yb + x
.
x+b
y+b
yb + x
(d)
xb + y
(a)
(b)
(e) None of these
Solution. (c) Let’s first convert A to base 10.
x
y
x
y
A =
+ 2 + 3 + 4 + ···
b b
b b x
y
y
x
=
+ 3 + ··· + 2 + 4 + ···
b b
b
b
x
1
1
y
1
1
=
1 + 2 + 4 + ··· + 2 1 + 2 + 4 + ···
b
b
b
b
b
b
y
1
1
x
=
+
1 + 2 + 4 + ···
b b2
b
b
1
bx + y
·
=
b2
1 − b12
bx + y
= 2
b −1
By reversing the roles of x and y, we get B =
by + x
and so
b2 − 1
A
bx + y
=
.
B
by + x
Here is an alternate solution. From the given base-b representations
we get two equations (1) bA = B + x and (2) bB = A + y. Equation
(1) gives B = bA − x, and substituting this into Equation (2) gives
b(bA − x) = A + y =⇒ b2 A − bx = A + y
=⇒ (b2 − 1)A = bx + y
bx + y
=⇒ A = 2
.
b −1
3
Then proceed as in solution 1.
5. In the figure below, the large circle has radius equal to 2 inches and
the small circle has radius equal to 1 inch. What is the area of the
circumscribing square?
√
1
(9 + 27 2) inches2
2
√
1
(c) (27 + 38 2) inches2
3
√
1
(27 + 18 2) inches2
2
√
1
(d) (40 + 27 2) inches2
3
(a)
(b)
(e) None of these
Solution. (b) Label the respective centers of the two circles A and B
as pictured.
Y
B
A
X
The diagonal length XY can be computed using the Pythagorean as
follows. BY is the hypotenuse
√ of an equilateral right triangle whose leg
lengths are 1 inch. So BY = 2. Similarly, AX is the hypotenuse of√
the
equilateral right triangle whose leg lengths are 2 inches, so AX = 2 2.
Finally, AB = 1 + 2 = 3, so the diagonal of the square is
√
XY = 3 + 3 2.
4
If we let s denote the square side length,
again by the Pythagorean
√ then
2
2
2
theorem, we have 2s = XY = (3+3 2) . From this, we can compute
the square area s2 ,
√
1
(3 + 3 2)2
2
√
1
=
(27 + 18 2).
2
s2 =
6. Let r1 and r2 be the roots of the polynomial x2 + 2006x + 1776. Then
1
1
+
=
.
r1 r 2
1003
888
1
(c) −
1776
(a) −
(b) −2006
(d) −
1
3782
(e) None of these
Solution. (a) If r1 and r2 are the roots of a polynomial x2 + ax + b,
then
x2 + ax + b = (x − r1 )(x − r2 )
= x2 + (−r1 − r2 )x + r1 r2 ,
and so r1 + r2 = −a and r1 r2 = b. Thus
1
r 1 + r2
1
+
=
r1 r2
r 1 r2
a
= − .
b
Substituting the values a = 2006 and b = 1776 gives
1
1
2006
+
= −
r1 r 2
1776
1003
= −
.
888
5
7. Suppose N is the 3-digit number N = abc where a, b and c are the
respective hundreds, tens and units digits. Assuming that when N 2 is
divided by 100 the remainder is 1, then the possible values for b are
what?
(a) b = 0, 4, 5 or 9
(b) b = 0, 4, 5, 6 or 9
(c) b = 0, 4, 5 or 8
(d) b = 0, 4, 6, 8 or 9
(e) None of these
Solution. (a) We first observe that the answer is independent of a.
This is because N = 100a+(bc) and so N 2 = 10000a2 +200a(bc)+(bc)2 .
Since 10000a2 + 200a(bc) is divisible by 100, the remainder of N is the
same as the remainder of (bc). Thus, in our analysis we may replace N
by N 0 = bc = 10b + c. Square N 0 to get
N 02 = (10b + c)2
= 100b2 + 20bc + c2 .
The 100b2 term will not contribute to the remainder, and so may be
ignored. We are then left with dividing 20bc + c2 by 100 to get a
remainder of 1. Clearly this means c = 1 or 9.
If c = 1, then 20bc + c2 = 20b + 1, which by checking the possibilities
exhaustively, will have a remainder of 1 only if b = 0 or 5. And this
can happen if say N = 101 or 151.
Similarly, if c = 9, then 20bc + c2 = 180b + 81, which will have a
remainder of 1 only if b = 4 or 9. This can happen if say N = 149 or
N = 199.
So the possible values for b are 0, 4, 5 or 9.
8. A formula for the height h in terms of the base length b and the two
.
base angles 6 A and 6 B for the triangle pictured below is
6
C
h
A
B
b
b
cot 6 A + cot 6 B
b
(c)
6
1 + cot A cot 6 B
b
A + cot2 6 B
b
(d) √
26
1 + cot A cot2 6 B
(b) √
(a)
cot2 6
(e) None of these
Solution. (a) Add the labels x and b − x as the obvious lengths in
the picture below.
C
h
x
A
b-x
b
B
From the picture we can see that tan 6 A =
These two formulas can be rewritten as
h
h
and tan 6 B =
.
x
b−x
x = h cot 6 A and b − x = h cot 6 B.
Adding these two formulas together gives
b = h cot 6 A + h cot 6 B.
Solve for h to get
h=
b
.
cot 6 A + cot 6 B
7
9. Suppose that the average of 10 numbers is 90. One of the numbers in
the list is deleted, and the resulting average of nine numbers is equal
to 91. What is the value of the deleted number?
(a) 80
(b) 81
(c) 82
(d) 83
(e) None of these
Solution. (b) Let the 10 numbers be denoted by N1 , N2 , . . . , N10 ,
where N10 will be the deleted number. We are given that the average
of the original 10 numbers is 90. Thus
N1 + N2 + · · · + N10
= 90,
10
which impies N1 + N2 + · · · + N10 = 900. We are also given that
N1 + N2 + · · · + N9
= 91,
9
and so N1 + N2 + · · · + N9 = 819. Putting this together gives
819 + N10 = 900
and so N10 = 81.
10. You are playing the game Yatze. You roll five dice, which results in
four 1s and one 2. You hope to roll the die that displays the number
2 to get the fifth 1, and you have two tries to do this. What is the
probability that you will get a 1 in one or two rolls of a single die?
(Note, if you get a 1 on the first roll, you stop.)
1
4
11
(c)
36
5
18
1
(d)
3
(a)
(b)
(e) None of these
8
Solution. (c) We’ll break the problem into two parts. First, the
1
probability that a single roll of the die will result in a 1 is . On the
6
5
other hand, if the first roll is not a 1 (with probability ), a second roll
6
1
will display a 1 also with probability . So the probability of obtaining
6
5 1
5
a 1 with two rolls is · = .
6 6
36
The two events described above (one roll or two rolls) are mutually
exclusive, so the probability of either occurring is the sum of the probabilities of each,
5
11
1
+
= .
6 36
36
9