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

First let's assume there exist six consecutive positive integers such that there is not a prime number that divides exactly one of the set.

First we deduce that the first and last numbers are divisible by five.  Therefore, the least number in the set must be at least 5.

No matter what set is chosen, exactly three of the integers are even (exactly one of which is also divisible by 5).  Also, exactly two of the integers are divisible by 3 (exactly one of which is also even).  This means that 4 or 5* of the numbers are divisible by 2, 3 or 5.

The last number (or one of the last numbers) in the set, not divisible by 2, 3, or 5, must be divisible by at least one higher prime.  Choose one of its prime divisors.  Since this prime divisor is greater than 5, none of the other numbers in the set may share this divisor.  This contradicts our original assumption, therefore any chosen set of six consecutive positive integers has a prime that divides exactly one of the set.

QED

*corrected, thanks to Richard

Edited on August 29, 2006, 9:21 pm

Posted by Tristan on 2006-08-29 17:18:56