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

  1. #1
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    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!
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  2. #2
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    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

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  3. #3
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    ty so much
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