Imagine a row of 100 closed doors. Now, make 100 passes along the row, and at each pass "toggle" the doors whose number is divisible by the number of the pass. (By "toggle", we mean to open a closed door, or close an opened one.
For example, on the first pass, we will toggle all the doors, on the second, we will toggle only the even-numbered doors, on the third - only doors whose number is divisible by three, and so on.
At the end of the 100 passes, how many doors will be left open?
I made a mistake on 9, it comes out to open. It seems that perfect squares of primes > 2 will always be open. So there will be at least 4 open. :/
Posted by Dominic
on 2002-05-01 10:54:00