Download Solution

2006 MathCounts Training, Team Round 4
1. For how many positive integers n is n2 − 3n + 2 a prime number?
If n ≥ 4, then
Solution:
n2 − 3n + 2 = (n − 1)(n − 2)
is the product of two integers greater than 1, and thus is not prime.
For n = 1, 2, and 3 we have, respectively,
(1 − 1)(1 − 2) = 0,
(2 − 1)(2 − 2) = 0,
and (3 − 1)(3 − 2) = 2.
Therefore, n2 − 3n + 2 is prime only when n = 3.
2. Find the value of x that satisfies the equation
25−2 =
548/x
.
526/x · 2517/x
Solution:
Write all the terms with the common base 5. Then
−4
5
−2
= 25
548/x
548/x
= 26/x
= 26/x 34/x = 5(48−26−34)/x = 5−12/x .
17/x
5
· 25
5
·5
It follows that − 12
= −4, so x = 3.
x
OR
First write 25 as 52 . Raising both sides to the x power gives
5−4x =
So −4x = −12 and x = 3.
548
= 548−26−34 = 5−12 .
526 534
3. Nebraska, the home of the AMC, changed its license plate scheme. Each
old license plate consisted of a letter followed by four digits. Each new
license plate consists of three letters followed by three digits. By how
many more license plates are possible under the new system?
Solution: In the old scheme 26 × 104 different plates could be constructed. In the new scheme 263 × 103 different plates can be constructed. There are 263 × 103 − 26 × 104 = 103 · 26(262 − 10) =
26000 · 666 = 17316000 more licences available under the new scheme.
4. A line with slope 3 intersects a line with slope 5 at the point (10, 15).
What is the distance between the x-intercepts of these two lines?
Solution: The two lines have equations
y − 15 = 3(x − 10) and y − 15 = 5(x − 10).
The x-intercepts, obtained by setting y = 0 in the respective equations,
are 5 and 7. The distance between the points (5, 0) and (7, 0) is 2.
5. On a trip from the United States to Canada, Isabella took d U.S.
dollars. At the border she exchanged them all, receiving 10 Canadian
dollars for every 7 U.S. dollars. After spending 60 Canadian dollars,
she had d Canadian dollars left. What is the d?
Solution: Isabella received 10d/7 Canadian dollars at the border and
spent 60 of them. Thus 10d/7 − 60 = d, from which it follows that
d = 140.
6. A square has sides of length 10, and a circle centered at one of its
vertices has radius 10. What is the area of the union of the regions
enclosed by the square and the circle? Express your answer in terms
of π.
Solution: The areas of the regions enclosed by the square and the
circle are 102 = 100 and π(10)2 = 100π, respectively. One quarter of
the second region is also included in the first, so the area of the union
is
100 + 100π − 25π = 100 + 75π.
7. For what values of k does the equation
for x?
x−1
x−k
=
have no solution
x−2
x−6
Solution: From the given equation we have (x − 1)(x − 6) = (x −
2)(x − k). This implies that
x2 − 7x + 6 = x2 − (2 + k)x + 2k,
so
2k − 6
.
k−5
Hence a value of x satisfying the equation occurs unless k = 5. Note
that when k = 6, there is also no solution for x.
(k − 5)x = 2k − 6 and x =
8. Find the value(s) of x such that 8xy − 12y + 2x − 3 = 0 is true for all
values of y.
Solution: The given equation can be factored as
0 = 8xy − 12y + 2x − 3 = 4y(2x − 3) + (2x − 3) = (4y + 1)(2x − 3).
For this equation to be true for all values of y we must have 2x − 3 = 0,
that is, x = 3/2.
9. The number 2564 · 6425 is the square of a positive integer N . In decimal
representation, the sum of the digits of N is
Solution: We have
p
· · · 0} .
N = (52 )64 · (26 )25 = 564 ·23·25 = (5·2)64 ·211 = 1064 ·2048 = 2048 000
| {z
64 digits
The zeros do not contribute to the sum, so the sum of the digits of N
is 2 + 4 + 8 = 14.
10. The positive integers A, B, A − B, and A + B are all prime numbers.
What is the sum of these four primes?
Solution: The numbers A − B and A + B are both odd or both
even. However, they are also both prime, so they must both be odd.
Therefore, one of A and B is odd and the other even. Because A is a
prime between A − B and A + B, A must be the odd prime. Therefore,
B = 2, the only even prime. So A − 2, A, and A + 2 are consecutive
odd primes and thus must be 3, 5, and 7. The sum of the four primes
2, 3, 5, and 7 is 17.