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Division by 6

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Division by 6

by kbikov » Wed Jun 10, 2009 11:14 am
What is the proof that the multiple of three consequtive integers is divisible by 6 i.e. k(k-1)(k+1) mod 6 = 0?

Thank you,
Kal
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by kbikov » Wed Jun 10, 2009 11:41 am
Actually, I figured it out.

To prove division by 6 we need to prove division by 2 and by 3.

Obviuosly, there will be at least one intiger among the three consequtive intigers which will be even and so their product will also be an even number (divisible by 2).

Since every third integer starting from 1 is divisible by 3, we know that exactly one of the three consequtive intigers will be divisible by 3 and therefore their product will also be divisible by 3.
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by gmatplayer » Sun Jun 14, 2009 6:44 pm
Multiple of any 3 consicative number is divisible by 6.
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by Ian Stewart » Mon Jun 15, 2009 7:17 am
Yes, exactly. You can extend your proof to any number of consecutive integers:

The product of n consecutive integers is always divisible by n!

So, if you multiply any five consecutive integers, you can be sure that the result will be divisible by 5! = 120, for example.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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