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{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> | ||
| − | == See | + | == 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}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 14:42, 3 April 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 | ||
