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2012 AIME II Problems/Problem 1: Difference between revisions

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== See also ==
== See also ==
{{AIME box|year=2012|n=II|before=First Problem|num-a=2}}
{{AIME box|year=2012|n=II|before=First Problem|num-a=2}}
[[Category:Introductory Number Theory Problems]]

Revision as of 20:22, 1 April 2012

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 (Problems • Answer Key • Resources)
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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