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

(In reply to re(3): n = 8... and more questions by JLo)

JLo:

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JLo's conjecture:  Consider n consecutive numbers. Then it is impossible that each prime number p<=n divides at least two of the n numbers and that each of the n numbers is divisible by a p<=n.
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25 is the first n for which I found a counterexample, although it is clear to me now that I could have found a counterexamples for n = 20 or 21 or 22 or 23 or 24.  At any rate, It is possible to have 25 consecutive numbers beginning with an odd number such that:

  23 divides the 1st and  24th
  19 divides the 6th and 25th
  17 divides the 6th and 23rd
  13 divides the 11th and 24th

Of the odd numbers,
  number   is divisible by
  --------    ---------------
       1       23
       3       3
       5       7
       7       5
       9       3
       11     13
       13     11
       15     3
       17     5
       19     7
       21     3
       23    17
       25    19

It is interesting (but not significant, I think)  that each of these is divisible by exactly one prime less than 25.

So, what is the first of the infinite number of 25-sequences like this?   This is a lot like a diophantine equation.

The smallest positive counterexample I found (with a little help from excel) was the 25 numbers starting with 185,065,429.  
The rest are given by 185,065,429 + k*223,092,870 where k is integral.

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Do any of these sequences disprove Steve Herman's conjecture?

"Consider k consecutive positive integers, K > 1. Show that there is a prime number that divides exactly one of them."

No, while these sequences disprove JLo's conjecture, they do not disprove Steve Herman's.  Every one of the odd numbers in every sequence is divisible by at most one prime under 25.  So every one of the odd numbers is divisible by at least one prime which divides none of the others, unless it happens to be an exact power of a prime under n!   And that is also true of the even numbers, unless one of the even numbers happens to be an an exact power of two. 

And no, I can't prove Steve Herman's conjecture, at least not without some new tools.

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By the way, this has turned into a  great puzzle.  We've sure come a long way from n = 6.  Too bad nobody but JLo and I will likely ever read it.  Perhaps I should turn the counterexample into a perplexus puzzle.  What do you think, JLo?

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Edited 9/8/06 to remove faulty reasoning that ruled out n = 20, 21, 22, 23, 24.  Any of the sequences above, with the first five numbers removed, are also counterexamples.

It is possible to have 20 consecutive numbers beginning with an even number such that:

  19 divides the 1st and 20th
  17 divides the 1st and 18th
  13 divides the 6th and 19th
  11 divides the 8th and 19th

Of the odd numbers,
  number   is divisible by
  --------    ---------------
       2       5
       4       3
       6      13
       8      11
       10     3
       12     5
       14     7
       16     3
       18    17
       20    19

I haven't tried to find the first such sequence, although I could

 

Edited on September 8, 2006, 8:16 am

Edited on September 8, 2006, 8:21 am

Edited on September 8, 2006, 8:23 am

Posted by Steve Herman on 2006-09-08 01:20:41