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
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
To explain why the product of two whole numbers that each leave a remainder of 1 when divided by 7 also leaves a remainder of 1 when divided by 7, let's break it down step by step.
1. Start with the expression you have expanded: 49mn + 7m + 7n + 1.
2. Remember that we are looking for a remainder when dividing by 7. When we divide any number by 7, the remainder can only be in the range of 0 to 6.
3. Focus on the terms involving multiples of 7: 49mn and 7m.
4. Notice that when we divide 49 by 7, the remainder is 0 since 49 is a multiple of 7. So, the term 49mn is divisible by 7 without any remainder.
5. Now consider the term 7m. Since m is an integer, multiplying it by 7 gives us a multiple of 7.
6. Combining the terms 49mn and 7m, we have a product that is divisible by 7 without any remainder.
7. Now, let's add the other terms: 7n + 1.
8. The term 7n is also a multiple of 7, meaning it is divisible evenly by 7.
9. Finally, we have the term 1. Any number divided by 7 will have a nonzero remainder unless it is a multiple of 7. Therefore, the remainder when dividing 1 by 7 is 1.
10. Since there is a term (1) that leaves a remainder of 1 and no other term that leaves a remainder other than 0 when divided by 7, the product 49mn + 7m + 7n + 1 will also leave a remainder of 1 when divided by 7.
In summary, the WHY behind the product of two whole numbers, each leaving a remainder of 1 when divided by 7, also leaving a remainder of 1 when divided by 7 is that we have a combination of terms where some are divisible by 7 without any remainder, while the term 1 ensures that there is a nonzero remainder of 1 when dividing the entire expression by 7.