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## #1 2008-12-06 09:25:57

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### My brain isn't working

Prove that if three numbers are in arithmetic progression, at least one of them is divisible by 3.

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## #2 2008-12-06 09:37:19

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### Re: My brain isn't working

Maybe the reason I can't prove it is because it's not true.

4, 7, 10.

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## #3 2008-12-06 09:41:43

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### Re: My brain isn't working

In fact, I've realised that if three numbers are in arithmetic progression, at least one of them is divisible by 3 unless the common difference is divisible by 3.

Easy to prove (by considering possible values of a and d (mod 3).

Last edited by Daniel123 (2008-12-06 09:44:49)

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## #4 2008-12-06 10:08:46

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: My brain isn't working

The question should be: Prove that if three numbers are in arithmetic progression whose common difference is not a multiple of 3, at least one of them is divisible by 3. (If fact exactly one of them is.)

http://www.mathisfunforum.com/viewtopic.php?id=8520

There I proved a more general result, namely that if k integers form an arithmetic progression whose common difference is coprime with k, then (exactly) one of the terms in the AP is divisible by k.

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## #5 2008-12-06 10:15:53

Daniel123
Member
Registered: 2007-05-23
Posts: 663

### Re: My brain isn't working

Ahh, thanks for sharing that

Also, "if three numbers are in arithmetic progression whose common difference is not a multiple of 3, at least one of them is divisible by 3" is pretty much what I stated in post #3

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