Consider six consecutive positive integers. Show that there is a prime number that divides exactly one of them.

(In reply to n = 8 by Steve Herman)

Steve,
since you proved the statement for n=8..12, I felt I had to go a little further and verified it for all n=2..14. The case n=9 is the most interesting one, don't you think? I got a bit stuck on n=15, probably even more interesting...

I am sharing your feeling that the statement is true for all n. But despite having verified a number of cases now, I don't see a promising path to prove the general case. Here is why: In all the cases n<=14 I did actual prove the following (I suppose you did the same???):

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Consider n consecutive numbers. Then it is impossible that each prime number p<=n divides at least two of the n numbers!
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The above is a much stronger statement than that of the original puzzle. I have however a feeling that this stronger statement does not hold for all n. I guess for n=15 it still holds, but you probably need to distinguish quite a few cases. The way I have proven the above for up to 14, was by trying to distribute all primes below n in a manner to cover n consecutive numbers, and somehow leading this to a contradiction. For every n only a finite number of cases must be considered, so this can be done in finite time for all n. Assume I am right with my suspicion that the strong statement does not hold, how would one prove the original puzzle for all n? Currently I see no promising path.

This leads to the following challenges:
- Can you prove or disprove the above, stronger statment?
- In case of disproval, what is a counter example with minimal n?
- For this minimal n, can you still prove the original puzzle?


Edited, cause my previous "strong statement" was rubbish... Corrected to what I actually meant.

Edited on September 6, 2006, 1:56 pm

Posted by JLo on 2006-09-05 17:54:50