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 20060904 16:17:22 