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

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

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

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

4, 7, 10.

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

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

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.)

See my exercises thread:

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

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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