There coins placed in a line, and each one is facing up. A man flips every other coin, going down the line from left to right. He then starts on the left and flips every third coin, then flipping every fourth, and so on until he has gone down the line times. How many coins are facing up at the end?
(This problem is not original)
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The numbers not flipped are those with an even number of factors including one. So, since square numbers are the only numbers with an odd number of divisors (including one), this means all the coins that are a square number from the left face up. ⌊ 1 0 0 0 ⌋ = 3 1 , so 3 1 face up.