Difference between revisions of "1999 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | {{ | + | If the [[perfect square]] is represented by <math>x^2</math>, then the equation is <math>n^2 - 19n + 99 - x^2 = 0</math>. The [[quadratic formula]] yields: |
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| + | :<math>\frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}</math> | ||
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| + | The [[discriminant]] must also be a perfect square (<math>y^2</math>), so <math>x^2 - 35 = y^2</math>. This factors to: | ||
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| + | :<math>(x + y)(x - y) = 35</math> | ||
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| + | <math>35</math> has two pairs of factors: <math>\{1,\ 35\}</math> and <math>\{5,\ 7\}</math>. Respectively, these yield 18 and 6 for <math>x</math>, which results in <math>n = 1,\ 9,\ 10,\ 18</math>. The sum is therefore <math>038</math>. | ||
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== See also == | == See also == | ||
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* [[1999 AIME Problems]] | * [[1999 AIME Problems]] | ||
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| + | {{AIME box|year=1999|num-b=2|num-a=4}} | ||
Revision as of 22:22, 9 February 2007
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:
The discriminant must also be a perfect square (
), so 
. This factors to:
has two pairs of factors: 
and 
. Respectively, these yield 18 and 6 for 
, which results in 
. The sum is therefore 
.
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 | ||
