2012 AIME II Problems/Problem 1

Revision as of 17:28, 6 March 2016 by Budu (talk | contribs) (Solution)

Problem 1

Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$.

Solution

Solving for $m$ gives us $m = \frac{503-3n}{5},$ so in order for $m$ to be an integer, we must have $3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.$ The smallest possible value of $n$ is obviously $1,$ and the greatest is $\frac{503 - 5}{3} = 166,$ so the total number of solutions is $\frac{166-1}{5}+1 = \boxed{034}$

See Also

2012 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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