Homework Help: A proof of divisibility problem

1. Feb 5, 2013

junglebobo

If a divides b, and a divides b+c then a divides 3c.

How do I go forward with this?

This is what I've done so far:

suppose a|b and a|b+c

then b = an for some integer n
and c = am for some integer m

∴ b+c = an+am
= a(n+m)
= ak for some integer k

but I don't feel like this is getting me anywhere

2. Feb 5, 2013

junglebobo

Alright, I got it

If a|3c then a is a multiple of 3

then b is also a multiple of 3, as a divides it

therefore a|b+c = b/a + c/a

and thus a|c

therefore a|3c

3. Feb 7, 2013

ramsey2879

why doesn't 'a' divide c and not 3. Honestly the teacher put 3c in the problem to confuse you. He could have made it 13c. That wouldn't mean 'a' divides 13 since it can be shown that 'a' divides c. P.S you have shown that c = am so what does 3c equal?

4. Feb 7, 2013

junglebobo

we start out by assuming that a|3c is correct,
therefore a has to be a multiple of 3, if it divides 3c

5. Feb 7, 2013

ramsey2879

No 'a' can be 7, 'c' can be 14 and 'b' can be 35. 3c would then be 42, a divides 3c because a divides c which is am. 3c would be 3am. a divides 3am because the factor a is in the product 3am. there is nothing in the problem that would suggest that 'a' divides 3.

Last edited: Feb 7, 2013
6. Feb 7, 2013

junglebobo

but it divides it into a multiple of 3. That's my point

7. Feb 7, 2013

ramsey2879

That doesn't mean that 3 divides 7,14 and 21. Thus neither a, b or c need be multiples of 3which is contrary to what you said.

8. Feb 7, 2013

junglebobo

hmm.. alright, point taken..

9. Feb 7, 2013

ramsey2879

All you need to do is show that c = am, so that 3c = 3am = a multiple of a.

Last edited: Feb 7, 2013
10. Feb 8, 2013

HallsofIvy

3c= 3(b+ c)- 3b.

Really what you are proving is that "if a divides b+ c and a divides b, then a divides c." The "3" is irrelevant.