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

Revision as of 13:15, 17 July 2017

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 $253437 = 1, 6, 11, ... , 166$ . There are $\boxed{34}$ solutions.

@@ Line 3: / Line 3: @@
 == Solution ==
+==Solution 1==
 Solving for <math>m</math> gives us <math>m = \frac{503-3n}{5},</math> so in order for <math>m</math> to be an integer, we must have <math>3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.</math> The smallest possible value of <math>n</math> is obviously <math>1,</math> and the greatest is <math>\frac{503 - 5}{3} = 166,</math> so the total number of solutions is <math>\frac{166-1}{5}+1 = \boxed{034}</math>
+==Solution 2==
+Dividing by <math>4</math> gives us <math>5m + 3n = 503</math>. Solving for <math>n</math> gives <math>n \equiv 1 \pmod 5</math>. The solutions are the numbers <math> 253437
+ = 1, 6, 11, ... , 166</math>. There are <math>\boxed{34}</math> solutions.
 == 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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