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?

(from techInterview.org)

we can see that a door is open if and only if it had been toggled an odd number of times, which happens if and only if the door number have odd number of factor (since each factor contribute one toggle). Now for each number n which have factor d, then n/d is also a divisor. so for 'almost' all number we can pair the factors as (d,n/d), yielding odd number of factors. the only exception, of course if when d=n/d so that there'll be no pair, and this is possible if and only if n=d^2, i.e n is a perfect square. so, the answer is the number of perfect squares <= 100, namely 10.

Posted by theBal on 2002-05-01 18:02:46