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

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== Problem ==
 
== Problem ==
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Find the sum of all positive integers <math>\displaystyle n</math> for which <math>\displaystyle n^2-19n+99</math> is a perfect square.
+
Find the sum of all [[positive integer]]s <math>\displaystyle n</math> for which <math>\displaystyle n^2-19n+99</math> is a [[perfect square]].
  
 
== 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:
+
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
  
−
:<math>\frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}</math>
+
:<math> n = \frac{19 \pm \sqrt{361 - 4(99 - x^2)}}{2}</math>
  
−
The [[discriminant]] must also be a perfect square (<math>y^2</math>), so <math>x^2 - 35 = y^2</math>. This factors to:
+
In order for this to be an [[integer]], the [[discriminant]] must also be a perfect square, so <math>x^2 - 35 = y^2</math> for some [[nonnegative]] integer <math>y</math>. This [[factoring | factors]] to
  
 
:<math>(x + y)(x - y) = 35</math>  
 
:<math>(x + y)(x - y) = 35</math>  
  
−
<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>.
+
<math>35</math> has two pairs of positive [[divisor | 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>.
  
 
== See also ==
 
== See also ==
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* [[1999 AIME Problems]]
 
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{{AIME box|year=1999|num-b=2|num-a=4}}
 
{{AIME box|year=1999|num-b=2|num-a=4}}

Revision as of 11:02, 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 $038$.

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
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