Consider six consecutive positive integers. Show that there is a prime number that divides exactly one of them.
n = 8 is almost as easy as n = 6
Assume that there is no
prime number that divides exactly one of the 8. Then the
first and last must both be divisible by 7.
There are 4 odd
numbers out of the eight consecutive, and the one on the end is
divisible by 7. So there are three "consecutive" odd
numbers that are not divisible by 2 or 7. At most one is
divisible by 3 and at most one is divisible by 5. So there
is at least one odd number which is not divisible by 3, 5, or 7.
It must be divisible by a prime > 7, and that prime divides no other
number in the consecutive 8.
Therefore, our initial
assumption is wrong, and at least one prime number divides exactly one
of every 8 consecutive positive integers.
The proof for n = 9, 10 or 12 is trickier, but I assert that they are all provable (since I have done it)
Edited on September 2, 2006, 12:07 am
n = 8
