Difference between revisions of "2012 AIME II Problems/Problem 1"
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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> | + | 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>n |
= 1, 6, 11, ... , 166</math>. There are <math>\boxed{34}</math> solutions. | = 1, 6, 11, ... , 166</math>. There are <math>\boxed{34}</math> solutions. | ||
Revision as of 13:16, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
