# 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$ (Error making remote request. No response to HTTP request)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}\$.

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