# Math Help - A weird problem.. Please take a look and help!

1. ## A weird problem.. Please take a look and help!

Hello, I got an "application" from my teacher and this is what the problem says. Please help I kind of let this go to the last minute and need some help!

Find all integers n>3 such that n-3 divides evenly into n^2-n.

(find all integers n is greater than 3, such that n minus 3 divides evenly into n squared minus n.) That was the phonetics of it, i hope i wrote it right!

2. Hello, Airjunkie!

Find all integers $n>3$ such that $n-3$ divides evenly into $n^2-n$

Divide: . $\frac{n^2-n}{n-3} \;=\;n + 2 + \frac{6}{n-3}$

Since that fraction, $\frac{6}{x-3}$, must be an integer, $(n-3)$ must be a factor of 6.

This happens when: . $n \;=\;4, 5, 6, 9$

3. ty so much