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:
(11^101)/11
2357947690.90909090909090909090
(12^101)/11
5628851293.00000000000000000000
(13^101)/11
12532590168.00000000000000000000
(14^101)/11
26295877725.00000000000000000000
(15^101)/11
52422762784.00000000000000000000
(16^101)/11
99955602525.00000000000000000000
(17^101)/11
183272172768.00000000000000000000
(18^101)/11
324587929693.00000000000000000000
(19^101)/11
557369659800.00000000000000000000
(20^101)/11
930909090909.00000000000000000000
(21^101)/11
1516352816200.00000000000000000000
(22^101)/11
2414538435583.90909090909090909090
(23^101)/11
3766046473968.00000000000000000000
It seems k^101 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:
(5^71)/8
9765.50000000000000000000
(even though base, exponent and divisor have no common factors.)

Posted by vswitchs
on 20060904 12:32:06 