A Christmas tree was decorated with 100 bulbs in a row. Then the bulbs started switching according to the following algorithm: all bulbs lit up, after a second every second bulb turned off, after another second every third bulb switched: if it was on, it turned off and vice versa. After a second, every fourth bulb switched, then every fifth bulb after another second, and so on. After 100 seconds, everything was over. Find the probability that a randomly chosen bulb is on after this (the bulbs do not burn out or break).
Obviously, the bulb with number $n$ will remain on only if it has been switched an odd number of times, that is, if the number $n$ has an odd number of natural divisors. It is clear that this condition is only satisfied by squares: $n=1,4,9, \ldots, 100$ (see problem $\underline{30365}$). Thus, 10 bulbs out of 100 will remain on. Therefore, the probability of randomly selecting a bulb that is on is $0.1$.

## Answer

$0.1$.

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