1994 AJHSME Problems/Problem 10

Problem

For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?

$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

Solution

We should list all the positive divisors of $36$ and count them. By trial and error, the divisors of $36$ are found to be $1,2,3,4,6,9,12,18,36$ , for a total of $9$ . However, $1$ and $2$ can't be expressed as $N+2$ for a POSITIVE integer N, so the number of possibilities is $\boxed{\text{(A)}\ 7}$ .

Solution 2

To find the number of positive divisors of $36$ , first prime factorize $36$ to get $2^2*3^2$ . Then add $1$ to the power of both $2$ and $3$ to get $3$ . Multiply $3*3$ to get $9$ . Since the problem is asking only for positive integer values of N, subtract $2$ from $9$ (since $2-2$ and $1-2$ result in integers that are not positive) to get $\boxed{\text{(A)}\ 7}$ .

~ spoamath321

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1994 AJHSME Problems/Problem 10

Contents

Problem

Solution

Solution 2

See Also