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August 29, 2014

August 29, 2014

Posted by **charlie** on Monday, May 3, 2010 at 9:49am.

The product of any two (whole) numbers each of which leave a remainder of 1 on dividing by 7, also leaves a remainder of 1 on dividing by 7. Why?

I THINK that I can see a quadratic in there ( (n+1)(2n+1) ); and when I multiply any variation out, there's always a remainder 1.

Can anyone confirm the link; and point me where to go next? Could i use a diagram to explain it? Thanks.

Charlie

[math - MathMate, Sunday, May 2, 2010 at 5:49pm

An integer that leaves a remainder of 1 when divided by 7 can be represented by

7m+1, or 7n+1, where m, n are integers.

The product is thus:

(7m+1)(7n+1)

Expand the product and complete the proof.]

The expansion seems to be:

49mn+7m+7n+1

I'm not seeing where's next in explaining WHY?

Thanks

Charlie

- Math algebra -
**Reiny**, Monday, May 3, 2010 at 12:17pmyou are looking at it ...

49mn+7m+7n+1

= 7(mn+m+n) +1

isn't mn+m+n an integer?

so the last expression has the form of 7k+1 which would leave a remainder of 1 when divided by 7

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