Divisibility by $10^6$? Let $p_k$ be the $k^{th}$ prime number. 
Find the least $n$ for which $(p_1^2+1)(p_2^2+1) \cdots (p_n^2+1)$ is divisible by $10^6$.
I have no idea where to start on this problem. Any help would be appreciated.

Let's evaluate each $(p^2 + 1)$ expression, counting up the factors of $2$ and $5$, stopping when they both are at least $6$:
$n = 1$
$p = 2$, and $(p^2 + 1) = 5$.  2s so far = $0$, 5s so far = $1$.
$n = 2$
$p = 3$, and $(p^2 + 1) = 10$.  2s so far = $1$, 5s so far = $2$.
$n = 3$
$p = 5$, and $(p^2 + 1) = 26$.  2s so far = $2$, 5s so far = $2$.
$n = 4$
$p = 7$, and $(p^2 + 1) = 50$.  2s so far = $3$, 5s so far = $4$.
$n = 5$
$p = 11$, and $(p^2 + 1) = 122$.  2s so far = $4$, 5s so far = $4$.
$n = 6$
$p = 13$, and $(p^2 + 1) = 170$.  2s so far = $5$, 5s so far = $5$.
$n = 7$
$p = 17$, and $(p^2 + 1) = 290$.  2s so far = $6$, 5s so far = $6$.
We have 6 $2$s and 6 $5$s, enough for the product to be divisible by $1000000$.  $n = 7$.  (The actual product here is $390,949,000,000$.)
