Difference between revisions of "1999 AIME Problems/Problem 3"
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<math>35</math> has two pairs of positive [[divisor | factors]]: <math>\{1,\ 35\}</math> and <math>\{5,\ 7\}</math>. Respectively, these yield <math>9</math> and <math>3</math> for <math>x</math>, which results in <math>n = 1,\ 9,\ 10,\ 18</math>. The sum is therefore <math>\boxed{038}</math>. | <math>35</math> has two pairs of positive [[divisor | factors]]: <math>\{1,\ 35\}</math> and <math>\{5,\ 7\}</math>. Respectively, these yield <math>9</math> and <math>3</math> for <math>x</math>, which results in <math>n = 1,\ 9,\ 10,\ 18</math>. The sum is therefore <math>\boxed{038}</math>. | ||
| + | |||
| + | We have | ||
| + | |||
| + | :<math>n^2-19n+99=x^2\longleftrightarrow</math> | ||
==Alternate Solution== | ==Alternate Solution== | ||
Revision as of 15:21, 13 May 2012
Problem
Find the sum of all positive integers 
for which 
is a perfect square.
Solution
If the perfect square is represented by 
, then the equation is 
. The quadratic formula yields
In order for this to be an integer, the discriminant must also be a perfect square, so 
for some nonnegative integer 
. This factors to
has two pairs of positive factors: 
and 
. Respectively, these yield 
and 
for 
, which results in 
. The sum is therefore 
.
We have
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 | ||
