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Задача 474: Последние цифры делителей
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(106!, 65432) modulo (1016 + 61).
/**
* Your test output will go here.
*/