1956 AHSME Problems/Problem 48

Problem 48

If $p$ is a positive integer, then $\frac {3p + 25}{2p - 5}$ can be a positive integer, if and only if $p$ is:

$\textbf{(A)}\ \text{at least }3\qquad \textbf{(B)}\ \text{at least }3\text{ and no more than }35\qquad \\ \textbf{(C)}\ \text{no more than }35 \qquad \textbf{(D)}\ \text{equal to }35 \qquad \textbf{(E)}\ \text{equal to }3\text{ or }35$


Lets begin by noticing that: \[\frac{3p+25}{2p-5} = 1 + \frac{p+30}{2p-5}\]

Therefore, in order for $\frac{3p+25}{2p-5}$ to be a positive integer, $\frac{p+30}{2p-5}$ must be a non-negative integer. Since the bottom the the fraction is an odd number, we can multiply the top of $\frac{p+30}{2p-5}$ by 2 without changing whether it is an integer or not. Therefore, in order for $\frac{3p+25}{2p-5}$ to be an integer, $\frac{2p+60}{2p-5} = 1 + \frac{65}{2p-5}$ must also be an integer. As a result, $2p-5$ must be a factor of 65, or $p = 3, 5, 9, 35$. Therefore $p$ must be at least 3, and less than or equal to 35. So the answer which best fits these constraints is $\boxed{\textbf{(B)}}$.