2012 AIME II Problems/Problem 1: Difference between revisions
No edit summary |
No edit summary |
||
| Line 3: | Line 3: | ||
== 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 16:32, 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 | ||
