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 Replace '10' by n by Bractals)

I draw numbers 1 through k (k≤n) out of a hat n times at random, replacing the numbers after drawing them. If I disregard the case where I draw "1" all n times, explain why the number of possible sequences is divisible by n+1 (OR NOT).

The above is the original with 10 and ten replaced by n, and with 11 replaced by n+1, and with (OR NOT) added.  This is what Charlie was doing in his program, and seems to be what was intended.

BTW, it is easy to see that a^n cannot be congruent to 1 mod n+1 if a and n+1 have a common factor.  For example, 3^560 is not congruent to 1 mod 561, but 4^560 is congruent to 1 mod 561 (because 561=3*11*17 is a Carmichael number). So this leaves only the primes (no pseudoprimes are allowed in) as (n+1)'s that work.

Posted by Richard on 2006-09-04 16:17:22