11. There are 10,000 light bulbs numbered starting from 1, each of which can be turned on and off with a regular switch. Initially, all the light bulbs are off; then the switches of all the light bulbs marked by multiples of 1 (thus all the light bulbs are turned on) are pressed once, followed by pressing the switches of all those in even positions (i.e., multiples of 2), then those marked with multiples of 3, then those corresponding to multiples of 4, and so on, up to the multiples of 10,000. Which of the following light bulbs remains on at the end of the operations?
(A) The number 9405
(B) the number 9406
(C) the number 9407
(D) the number 9408
(E) the number 9409.
11. The answer is $\mathbf{( E )}$. The switch at position $n$ is touched once for each positive divisor of $n$. Therefore, the $n$-th light bulb remains on at the end if and only if $n$ has an odd number of divisors. This happens only for perfect squares: if indeed $n$ is not a perfect square, we can pair its divisors into pairs of the form $(d, n / d)$, and this tells us that they are in even number. If $n=m^{2}$ is a perfect square, we can pair all its divisors except $m$ by again pairing $\left(d, m^{2} / d\right)$: thus $m^{2}$ has an odd number of divisors. At this point, it is easy to verify that the only perfect square among the possible answers is $97^{2}=9409$.