Difference between revisions of "1999 AIME Problems/Problem 3"
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| − | == Problem == | + | == Problem == |
Find the sum of all [[positive integer]]s <math>n</math> for which <math>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]]. | ||
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:<math>n=\frac{19\pm \sqrt{4x^2-35}}{2}</math> | :<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{ | + | 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== | ==Solution 2== | ||
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==Solution 3== | ==Solution 3== | ||
| − | When <math> n \geq 12 </math>, | + | When <math> n \geq 12 </math>, we have |
<cmath> (n-10)^2 < n^2 -19n + 99 < (n-8)^2. </cmath> | <cmath> (n-10)^2 < n^2 -19n + 99 < (n-8)^2. </cmath> | ||
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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> | 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 = 38.</math> | + | We conclude that the answer is <math>1 + 9 + 10 + 18 = \boxed{38}.</math> |
== See also == | == See also == | ||
Latest revision as of 00:18, 29 January 2021
Problem
Find the sum of all positive integers 
for which 
is a perfect square.
Solution 1
If 
for some positive integer 
, then rearranging we get 
. Now from the quadratic formula,
Because is an integer, this means 
for some nonnegative integer 
. Rearranging gives 
. Thus 
or 
, giving 
or 
. This gives 
or 
, and the sum is 
.
Solution 2
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.
Solution 3
When 
, we have
![\[(n-10)^2 < n^2 -19n + 99 < (n-8)^2.\]](http://latex.artofproblemsolving.com/1/d/b/1dbd841be4e04181821cb5b138cefcb0329c3a5c.png)
So if and 
is a perfect square, then
![\[n^2 -19n + 99 = (n-9)^2\]](http://latex.artofproblemsolving.com/6/2/b/62bbea318696aacc4e1a63ef35356b064342d473.png)
or 
.
For 
, it is easy to check that is a perfect square when 
and 
( using the identity 
We conclude that the answer is 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
