Difference between revisions of "1999 AIME Problems/Problem 3"

Revision as of 11:04, 10 February 2007

Problem

Find the sum of all positive integers $\displaystyle n$ for which $\displaystyle n^2-19n+99$ is a perfect square.

Solution

If the perfect square is represented by $x^2$ , then the equation is $n^2 - 19n + 99 - x^2 = 0$ . The quadratic formula yields

$n = \frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}$

In order for this to be an integer, the discriminant must also be a perfect square, so $x^2 - 35 = y^2$ for some nonnegative integer $y$ . This factors to

$(x + y)(x - y) = 35$

$35$ has two pairs of positive factors: $\{1,\ 35\}$ and $\{5,\ 7\}$ . Respectively, these yield 18 and 6 for $x$ , which results in $n = 1,\ 9,\ 10,\ 18$ . The sum is therefore $038$ .

@@ Line 15: / Line 15: @@
 == See also ==
 {{AIME box|year=1999|num-b=2|num-a=4}}
+[[Category:Intermediate Number Theory Problems]]
+[[Category:Intermediate Algebra Problems]]

1999 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1999 AIME Problems/Problem 3"

Revision as of 11:04, 10 February 2007

Problem

Solution

See also