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.

Oops, an imprecise click sent an empty post earlier... apologies.

To continue Richard's research, here are results for larger ranges of numbers to be drawn:

(11^10-1)/11
2357947690.90909090909090909090
(12^10-1)/11
5628851293.00000000000000000000
(13^10-1)/11
12532590168.00000000000000000000
(14^10-1)/11
26295877725.00000000000000000000
(15^10-1)/11
52422762784.00000000000000000000
(16^10-1)/11
99955602525.00000000000000000000
(17^10-1)/11
183272172768.00000000000000000000
(18^10-1)/11
324587929693.00000000000000000000
(19^10-1)/11
557369659800.00000000000000000000
(20^10-1)/11
930909090909.00000000000000000000
(21^10-1)/11
1516352816200.00000000000000000000
(22^10-1)/11
2414538435583.90909090909090909090
(23^10-1)/11
3766046473968.00000000000000000000

It seems k^10-1 is divisible by 11 whenever k is not. (That it doesn't work for k=multiples of 11 is of course trivial).

It does't seem to work when the divisor is not prime:
(5^7-1)/8
9765.50000000000000000000
(even though base, exponent and divisor have no common factors.)

Posted by vswitchs on 2006-09-04 12:32:06