Difference between revisions of "2024 SSMO Relay Round 3 Problems/Problem 2"
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| + | We have <math>n^2+99n = s^2,</math> for some integer <math>s.</math> Factoring, we have <cmath>\left(\left(n+\frac{99}{2}\right)+s\right)\left(\left(n+\frac{99}{2}\right)-s\right)=\frac{99^2}{4}\implies</cmath><cmath>(2n+99-2s)(2n+99+2s) = 99^2.</cmath> To maximize <math>n,</math> we let <math>2n+99-2s = 1</math> and <math>2n+99+2s = 99^2 = 9801.</math> This gives <math>n = \frac{\frac{9801+1}{2}-99}{2} = \boxed{2401}.</math> | ||
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| + | ~SMO_Team | ||
Latest revision as of 14:46, 10 September 2025
Problem
Let 
Find the greatest odd integer 
for which 
is a perfect square.
Solution
We have 
for some integer 
Factoring, we have ![\[\left(\left(n+\frac{99}{2}\right)+s\right)\left(\left(n+\frac{99}{2}\right)-s\right)=\frac{99^2}{4}\implies\]](http://latex.artofproblemsolving.com/3/9/c/39ce11a9c4e3d9745a111046804840247897bd5a.png)
![\[(2n+99-2s)(2n+99+2s) = 99^2.\]](http://latex.artofproblemsolving.com/6/0/e/60e8a0ceaf4da8621eebac82ef8583907d44bcf9.png)
To maximize 
we let 
and 
This gives 
~SMO_Team
