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 Lasts a year (Posted on 2011-10-24)
Every 8-digit number structured ABCXABCX, where A,B and C are not necessarily distinct, is divisible by 365 iff X equals either 5 or 0.

Why?

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 Finishing the proof | Comment 2 of 5 |
iff = if and only if.

The first post proves that ABCXABCX is divisible by 365 if it is divisible by 5.

The "only if" part is not proven, however.  Fortunately, it is obvious. If ABCXABCX is not divisible by 5, then it cannot be divisible by 365, because 5 is a factor of 365.  But it has to be stated to finish the proof in both directions.  So, this paragraph completes the proof started in the first post.

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Here's a simple proof that goes in both directions:

365 = 5 * 73, so ABCXABCX is divisible by 365 iff (if and only if) it is divisible by each of the factors of 365: 5 and 73.

But it is always divisible by 73, because ABCXABCX = ABCX * 10001 = ABCX * 73* 137, so this condition is always met.

Therefore, it is divisible by 365 iff it is divisible by 5, and this is the case iff X is 0 or 5.

q.e.d.

 Posted by Steve Herman on 2011-10-24 10:48:52

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