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

Math algebra  Reiny, Monday, May 3, 2010 at 12:17pm
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