2012 AIME II Problems/Problem 1

Problem 1

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

Solution

Solution 1

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 $\frac{166-1}{5}+1 = \boxed{034}$

Solution 2

Dividing by $4$ gives us $5m + 3n = 503$ . Solving for $n$ gives $n \equiv 1 \pmod 5$ . The solutions are the numbers $n = 1, 6, 11, ... , 166$ . There are $\boxed{034}$ solutions.

2012 AIME II (Problems • Answer Key • Resources)
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2012 AIME II Problems/Problem 1

Contents

Problem 1

Solution

Solution 1

Solution 2

See Also