Difference between revisions of "2012 AIME II Problems/Problem 1"
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− | 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{ | + | 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> |
== See Also == | == See Also == |
Revision as of 17:28, 6 March 2016
Problem 1
Find the number of ordered pairs of positive integer solutions to the equation .
Solution
Solving for gives us so in order for to be an integer, we must have The smallest possible value of is obviously and the greatest is so the total number of solutions is
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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All AIME Problems and Solutions |
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