I draw numbers 1 through k (k≤10) out of a hat ten times at random, replacing the numbers after drawing them. If I disregard the case where I draw "1" all ten times, explain why the number of possible sequences is divisible by 11. (Result by a calculator is insufficient because anyone can do that easily.)
Now if I change the number '10' to another integer n in the above paragraph, can I still have a similar result; i.e., the total possible number of configurations is divisible by n+1? Does this work for all integers n? If so, prove it; if not, find all integers n it works for.
(In reply to
re(2): Replace '10' by n by Bractals)
n+1 must divide k^n  1 for all k=1,2,...,n which includes n^n  1. This is true if and only if n+1 is a prime. The "if" is Little Fermat.
The "only if" is because if n=1 is not prime, there is a prime divisor
p of n+1 in [1,n], and p^n  1 = m(n+1) = mpq is impossible since it
requires 1 to be a multiple of p.

Posted by Richard
on 20060904 20:21:48 