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
Contents
[hide]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.