2012 AMC 10A Problems/Problem 24
- The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.
- The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.
Problem
Let , , and be positive integers with such that and .
What is ?
Solution
Add the two equations.
.
Now, this can be rearranged:
and factored:
, , and are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that .
, since is the biggest difference. It is impossible to determine by inspection whether or , or whether or .
We want to solve for , so take the two cases and solve them each for an expression in terms of . Our two cases are or . Plug these values into one of the original equations to see if we can get an integer for .
, after some algebra, simplifies to . is not divisible by , so is not an integer.
The other case gives , which simplifies to . Thus, and the answer is .
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.