For a positive integer $n$ and digits $d$, we define $F(n, d)$ as the number of the divisors of $n$ whose last digits equal $d$.
For example, $F(84, 4) = 3$. Among the divisors of $84$ ($1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84$), three of them ($4, 14, 84$) have the last digit $4$.
We can also verify that $F(12!, 12) = 11$ and $F(50!, 123) = 17888$.
Find $F(10^6!, 65432)$ modulo ($10^{16} + 61$).
Solution
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