Problem 474
Last digits of divisors
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).
因数的尾数
对于一个正整数n和数字d，我们记F(n, d)为n的尾数是d的因数数目。
例如F(84, 4) = 3，因为在84的因数(1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84)中，有三个因数(4, 14, 84)的尾数是4。
可以验证，F(12!, 12) = 11，F(50!, 123) = 17888。
求F(106!, 65432) modulo (1016 + 61)。