Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution: Added an alternate solution.)
Line 12: Line 12:
  
 
<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>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>038</math>.
 +
 +
==Alternate Solution==
 +
 +
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.
  
 
== See also ==
 
== See also ==

Revision as of 22:43, 7 March 2009

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 $4x^2 - 35 = q^2$ for some nonnegative integer $q$. This factors to

$(2x + q)(2x - q) = 35$

$35$ has two pairs of positive factors: $\{1,\ 35\}$ and $\{5,\ 7\}$. Respectively, these yield $9$ and $3$ for $x$, which results in $n = 1,\ 9,\ 10,\ 18$. The sum is therefore $038$.

Alternate Solution

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.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AIME_Problems/Problem_3&oldid=30626"