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(6): Replace '10' by n by Bractals)

Difficulty is in the eye of the beholder, but also depends on what you can refer to instead of  proving yourself.  "Little Fermat" might be D4 or D5 to prove, but it is D0 to refer to. What can be somewhat of a stumbling block for this problem is knowing that "Little Fermat" doesn't work backwards.  However, there are no nonprimes that work for this problem.  That is because in this problem, the cases where k and (n+1) have a common factor are NOT filtered out, and they are what makes any nonprime not work. Hence a Carmichael number will not work here. 

Posted by Richard on 2006-09-06 02:55:58