Difference between revisions of "1999 AIME Problems/Problem 3"
(→Solution) |
|||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Find the sum of all [[positive integer]]s <math> | + | Find the sum of all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]]. |
== Solution == | == Solution == | ||
Line 17: | Line 17: | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:39, 4 July 2013
Problem
Find the sum of all positive integers for which
is a perfect square.
Solution
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
.
Alternate Solution
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.
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.