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.
Using a calculator (and thereby getting an insufficient result according to the proposer) gives me the following results
2^10-1=1023, 3^10-1=59048, 4^10-1=1048575, 5^10-1=9765624,
6^10-1=60466175, 7^10-1=282475248, 8^10-1=1073741823,
and these are all divisible by 11. For example, 282475248=11*25679568.
The relation between this and the "Hint" of the previous post is not clear to me as yet.
Edited on September 4, 2006, 12:32 pm
Posted by Richard
on 2006-09-04 11:14:29