Difference between revisions of "2012 AIME II Problems/Problem 1"

Revision as of 14:43, 9 August 2018

Problem

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.

@@ Line 1: / Line 1: @@
+==Problem==
 == Problem 1 ==
 Find the number of ordered pairs of positive integer solutions <math>(m, n)</math> to the equation <math>20m + 12n = 2012</math>.

2012 AIME II (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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