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 equal to $N+2$ 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

See Also

1994 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AJHSME/AMC 8 Problems and Solutions

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