Ask a New Question
Menu

I'm reposting this, because Mathmate's answer hasn't directed me to the WHY? part of the question. Any prompts appreciated.

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

1 answer

you 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