2012 AIME II Problems/Problem 1

Revision as of 12:22, 26 May 2018 by Bad math (talk | contribs) (Added one more solution.)

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.

Solution 3

Because the x-intercept of the equation is $\frac{2012}{20}$, and the y-intercept is $\frac{2012}{12}$, the slope is $\frac{\frac{-2012}{12}}{\frac{2012}{20}} = \frac{-5}{3}$. Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: $(100,1), (97,6), (94,11)...$ Because the solutions are only positive, we can generate only 33 more solutions, so in total we have $33+1=\boxed{034}$ solutions.

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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