1963 AMC 12 Problem 31

Attempt Problem 31 of the 1963 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1963 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

31.

The number of solutions in positive integers of 2x+3y=7632x+3y=763 is:
255255
254254
128128
127127
00
Answer: D
Concepts:Diophantine Equationparitycounting integers in a range
Small hint:
Solve for x=763−3y2x=\frac{763-3y}{2}
Big hint:
Positive integral xx requires yy to be an odd positive integer below 7633\frac{763}{3}
Solution:
We need yy odd so that 763−3y763-3y is even, and positivity gives 1≤y≤253.1\leq y\leq253. The odd values 1,1, 3,3, …,\ldots, and 253253 number 253+12=127.\frac{253+1}{2}=127.
Therefore, the correct answer is D.
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