Difference between revisions of "1999 AIME Problems/Problem 3"
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:<math>(2x + q)(2x - q) = 35</math> | :<math>(2x + q)(2x - q) = 35</math> | ||
− | <math>35</math> has two pairs of positive [[divisor | factors]]: <math>\{1,\ 35\}</math> and <math>\{5,\ 7\}</math>. Respectively, these yield <math>9</math> and <math>3</math> for <math>x</math>, which results in <math>n = 1,\ 9,\ 10,\ 18</math>. The sum is therefore <math>038</math>. | + | <math>35</math> has two pairs of positive [[divisor | factors]]: <math>\{1,\ 35\}</math> and <math>\{5,\ 7\}</math>. Respectively, these yield <math>9</math> and <math>3</math> for <math>x</math>, which results in <math>n = 1,\ 9,\ 10,\ 18</math>. The sum is therefore <math>\boxed{038}</math>. |
==Alternate Solution== | ==Alternate Solution== |
Revision as of 22:25, 24 March 2011
Contents
[hide]Problem
Find the sum of all positive integers for which
is a perfect square.
Solution
If the perfect square is represented by , then the equation is
. The quadratic formula yields
In order for this to be an integer, the discriminant must also be a perfect square, so for some nonnegative integer
. This factors to
has two pairs of positive factors:
and
. Respectively, these yield
and
for
, which results in
. The sum is therefore
.
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 |