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 $\fbox{A}$ .

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1967 AHSME Problems/Problem 24

Problem

Solution

See also