Difference between revisions of "2012 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
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+ | ==Solution 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{034}</math> | 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> | ||
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+ | ==Solution 2== | ||
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+ | 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 == | == See Also == |
Revision as of 13:15, 17 July 2017
Problem 1
Find the number of ordered pairs of positive integer solutions to the equation .
Solution
Solution 1
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
Solution 2
Dividing by gives us . Solving for gives . The solutions are the numbers . There are 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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