2004 AMC 12A Problem 3

Attempt Problem 3 of the 2004 AMC 12A 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 2004 AMC 12A solutions, or check the answer key.

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

3.

For how many ordered pairs of positive integers (x,y)(x, y) is x+2y=100?x + 2y = 100?
3333
4949
5050
9999
100100
Answer: B
Concepts:Diophantine Equationcounting integers in a range
Small hint:
Solve for xx in terms of yy
Big hint:
x=100−2yx = 100 - 2y must be positive, which bounds how large the positive integer yy can be
Solution:
Writing x=100−2y,x = 100 - 2y, the value of xx is a positive integer precisely when yy is a positive integer with 1≤y≤49.1 \le y \le 49.
This gives 4949 valid ordered pairs.
Thus, the correct answer is B.
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