Difference between revisions of "2012 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
+ | 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 == | ||
{{AIME box|year=2012|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2012|n=II|before=First Problem|num-a=2}} |
Revision as of 17:32, 31 March 2012
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) | ||
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 |