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

Latest revision as of 00:18, 29 January 2021

Problem

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

Solution 1

If $n^2-19n+99=x^2$ for some positive integer $x$ , then rearranging we get $n^2-19n+99-x^2=0$ . Now from the quadratic formula,

$n=\frac{19\pm \sqrt{4x^2-35}}{2}$

Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$ . Rearranging gives $(2x+q)(2x-q)=35$ . Thus $(2x+q, 2x-q)=(35, 1)$ or $(7,5)$ , giving $x=3$ or $9$ . This gives $n=1, 9, 10,$ or $18$ , and the sum is $1+9+10+18=\boxed{38}$ .

Solution 2

Suppose there is some $k$ such that $x^2 - 19x + 99 = k^2$ . Completing the square, we have that $(x - 19/2)^2 + 99 - (19/2)^2 = k^2$ , that is, $(x - 19/2)^2 + 35/4 = k^2$ . Multiplying both sides by 4 and rearranging, we see that $(2k)^2 - (2x - 19)^2 = 35$ . Thus, $(2k - 2x + 19)(2k + 2x - 19) = 35$ . We then proceed as we did in the previous solution.

Solution 3

When $n \geq 12$ , we have $(n-10)^2 < n^2 -19n + 99 < (n-8)^2.$

So if $n \geq 12$ and $n^2 -19n + 99$ is a perfect square, then $n^2 -19n + 99 = (n-9)^2$

or $n = 18$ .

For $1 \leq n < 12$ , it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-10)^2 + n - 1.)$

We conclude that the answer is $1 + 9 + 10 + 18 = \boxed{38}.$

@@ Line 1: / Line 1: @@
 == Problem ==
-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>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]].
+== Solution 1==
+If <math>n^2-19n+99=x^2</math> for some positive integer <math>x</math>, then rearranging we get <math>n^2-19n+99-x^2=0</math>. Now from the quadratic formula,
+:<math>n=\frac{19\pm \sqrt{4x^2-35}}{2}</math>
+Because <math>n</math> is an integer, this means <math>4x^2-35=q^2</math> for some nonnegative integer <math>q</math>. Rearranging gives <math>(2x+q)(2x-q)=35</math>. Thus <math>(2x+q, 2x-q)=(35, 1)</math> or <math>(7,5)</math>, giving <math>x=3</math> or <math>9</math>. This gives <math>n=1, 9, 10,</math> or <math>18</math>, and the sum is <math>1+9+10+18=\boxed{38}</math>.
+==Solution 2==
+Suppose there is some <math>k</math> such that <math>x^2 - 19x + 99 = k^2</math>. Completing the square, we have that <math>(x - 19/2)^2 + 99 - (19/2)^2 = k^2</math>, that is, <math>(x - 19/2)^2 + 35/4 = k^2</math>. Multiplying both sides by 4 and rearranging, we see that <math>(2k)^2 - (2x - 19)^2 = 35</math>. Thus, <math>(2k - 2x + 19)(2k + 2x - 19) = 35</math>. We then proceed as we did in the previous solution.
+==Solution 3==
+When <math> n \geq 12 </math>, we have
+<cmath> (n-10)^2 < n^2 -19n + 99 < (n-8)^2. </cmath>
+So if <math> n \geq 12</math> and <math> n^2 -19n + 99 </math> is a perfect square, then
+<cmath> n^2 -19n + 99 = (n-9)^2 </cmath>
+or <math> n = 18 </math>.
+For <math> 1 \leq n < 12 </math>, it is easy to check that <math> n^2 -19n + 99 </math> is a perfect square when <math> n = 1, 9 </math> and <math> 10 </math> ( using the identity <math> n^2 -19n + 99 = (n-10)^2 + n - 1.) </math>
+We conclude that the answer is <math>1 + 9 + 10 + 18 = \boxed{38}.</math>
-== Solution ==
-{{solution}}
 == See also ==
-* [[1999_AIME_Problems/Problem_2|Previous Problem]]
+{{AIME box|year=1999|num-b=2|num-a=4}}
-* [[1999_AIME_Problems/Problem_4|Next Problem]]
+[[Category:Intermediate Number Theory Problems]]
-* [[1999 AIME Problems]]
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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