# Difference between revisions of "1967 AHSME Problems/Problem 24"

## Problem

The number of solution-pairs in the positive integers of the equation $3x+5y=501$ is:

$\textbf{(A)}\ 33\qquad \textbf{(B)}\ 34\qquad \textbf{(C)}\ 35\qquad \textbf{(D)}\ 100\qquad \textbf{(E)}\ \text{none of these}$

## Solution

We have $y = \frac{501 - 3x}{5}$. Thus, $501 - 3x$ must be a positive multiple of $5$. If $x = 2$, we find our first positive multiple of $5$. From there, we note that $x = 2 + 5k$ will always return a multiple of $5$ for $501 - 3x$. Our first solution happens at $k=0$.

We now want to find the smallest multiple of $5$ that will work. If $x = 2 + 5k$, then we have $501 - 3x = 501 - 3(2 + 5k)$, or $495 - 15k$. When $k = 32$, the expression is equal to $15$, and when $k = 33$, the expression is equal to $0$, which will no longer work.

Thus, all integers from $k = 0$ to $k = 32$ will generate an $x = 2 + 5k$ that will be a positive integer, and which will in turn generate a $y$ that is also a positive integer. So, the answer is $\fbox{A}$.