Difference between revisions of "1999 AIME Problems/Problem 3"
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Find the sum of all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]]. | Find the sum of all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]]. | ||
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We conclude that the answer is <math>1 + 9 + 10 + 18 = \boxed{38}.</math> | We conclude that the answer is <math>1 + 9 + 10 + 18 = \boxed{38}.</math> | ||
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== See also == | == See also == |
Latest revision as of 01:18, 29 January 2021
Problem
Find the sum of all positive integers for which is a perfect square.
Solution 1
If for some positive integer , then rearranging we get . Now from the quadratic formula,
Because is an integer, this means for some nonnegative integer . Rearranging gives . Thus or , giving or . This gives or , and the sum is .
Solution 2
Suppose there is some such that . Completing the square, we have that , that is, . Multiplying both sides by 4 and rearranging, we see that . Thus, . We then proceed as we did in the previous solution.
Solution 3
When , we have
So if and is a perfect square, then
or .
For , it is easy to check that is a perfect square when and ( using the identity
We conclude that the answer is
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.