10. Caboyara, a famous Australian circus performer, is also performing a great trick this year. He prepares a spectacular ladder with $N=p_{1} \cdot p_{2} \cdot \ldots \cdot p_{2015}$ steps, where $p_{1}, p_{2}, \ldots, p_{2015}$ are distinct prime numbers; the steps that correspond to divisors of $N$ (including the first and the $N$-th step) are special and are initially lit in green.

During the performance, 2015 trained kangaroos climb the ladder one after another; for $i=$ $1,2, \ldots, 2015$, the $i$-th kangaroo jumps $p_{i}$ steps at a time, starting at the bottom of the ladder (jumping to step $p_{i}$, then to $2 p_{i}$, and so on until reaching step $N$). Each time a kangaroo jumps on a special step, it changes color: from green to red, from red to green.

How many special steps will be lit in green at the end of the performance?
(A) $2^{2015}-2^{1008}$
(B) $2^{2014}$
(C) $2^{2014}-2^{1007}$
(D) $2^{2013}$
(E) $2015 \cdot 2^{1008}$
10. The answer is (B). Given the special step corresponding to the divisor $d$ of $N$, the $i$-th kangaroo will jump on it if and only if the prime $p_{i}$ is a factor of $d$; the light of the step will therefore change color as many times as there are prime factors of $d$: it will be green at the end of the performance if and only if $d$ has an even number of prime factors. On the other hand, for every divisor $d$ of $N$ with an odd number of prime factors, the divisor $N / d$ (which is different from $d$, since $N$ is not a square) has an even number, and vice versa: thus the green lights will be exactly half of the divisors of $N$. The total number of special steps (i.e., of positive divisors of $N$) is equal to the number of subsets of the set $\left\{p_{1}, \ldots, p_{2015}\right\}$, i.e., $2^{2015}$. Therefore, the final number of green lights will be $2^{2014}$.