## Divisor Pairs

### Problem 561 Published on Saturday, 21st May 2016, 10:00 pm; Solved by 559;Difficulty rating: 30%

Let $S(n)$ be the number of pairs $(a,b)$ of distinct divisors of $n$ such that $a$ divides $b$.
For $n=6$ we get the following pairs: $(1,2), (1,3), (1,6),( 2,6)$ and $(3,6)$. So $S(6)=5$.
Let $p_m\#$ be the product of the first $m$ prime numbers, so $p_2\# = 2*3 = 6$.
Let $E(m, n)$ be the highest integer $k$ such that $2^k$ divides $S((p_m\#)^n)$.
$E(2,1) = 0$ since $2^0$ is the highest power of 2 that divides S(6)=5.
Let $Q(n)=\sum_{i=1}^{n} E(904961, i)$
$Q(8)=2714886$.

Evaluate $Q(10^{12})$.