Difference between revisions of "1999 AIME Problems/Problem 3"
Line 1: | Line 1: | ||
== 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]]. | ||
+ | |||
+ | |||
+ | == nosaj's Solution == | ||
+ | We have <math> n^2-19n+99 =m^2</math>, where <math>m</math> is a positive integer. Rearranging gives us | ||
+ | <cmath>n^2-19n+(-m^2+99)=0.</cmath> | ||
+ | Applying the quadratic formula yields | ||
+ | <cmath>n=\frac{19 \pm \sqrt{361-4(1)(-m^2+99)}}{2}.</cmath> Now, we simplyfy: | ||
+ | <cmath>n=\frac{19 \pm \sqrt{4m^2-35}}{2}.</cmath> | ||
+ | In order for <math>n</math> to be an integer, the discriminant <math>4m^2-35</math> must be a perfect square. In other words, | ||
+ | <cmath>4m^2-35=t^2,</cmath> | ||
+ | where <math>t</math> is a positive integer. Rearranging gives us | ||
+ | <cmath>4m^2-t^2=35.</cmath> Aha! Difference of squares. | ||
+ | <cmath>(2m+t)(2m-t)=35.</cmath> | ||
+ | Remember that both of these factors are integers, so we have a very limited amount of choices. Either | ||
+ | <cmath>2m+t=35 \ 2m-t=1</cmath> or <cmath>2m+t=7 \ 2m-t=5</cmath> | ||
+ | These two systems yield <math>m=9</math> and <math>m=3</math>. Plugging <math>m</math> back in gives us <math>n=9</math>, <math>n=10</math>, <math>n=18</math>, <math>n=1</math>. Therefore, our answer is <math>9+10+18+1=\boxed{38}</math> | ||
== Solution == | == Solution == |
Revision as of 21:31, 14 December 2014
Problem
Find the sum of all positive integers for which
is a perfect square.
nosaj's Solution
We have , where
is a positive integer. Rearranging gives us
Applying the quadratic formula yields
Now, we simplyfy:
In order for
to be an integer, the discriminant
must be a perfect square. In other words,
where
is a positive integer. Rearranging gives us
Aha! Difference of squares.
Remember that both of these factors are integers, so we have a very limited amount of choices. Either
or
These two systems yield
and
. Plugging
back in gives us
,
,
,
. Therefore, our answer is
Solution
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
.
Alternate Solution
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.
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.