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Problem 926

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Total Roundness

A round number is a number that ends with one or more zeros in a given base.

Let us define the roundness of a number $n$ in base $b$ as the number of zeros at the end of the base $b$ representation of $n$.

For example, $20$ has roundness $2$ in base $2$, because the base $2$ representation of $20$ is $10100$, which ends with $2$ zeros.

Also define $R(n)$, the total roundness of a number $n$, as the sum of the roundness of $n$ in base $b$ for all $b > 1$.

For example, $20$ has roundness $2$ in base $2$ and roundness $1$ in base $4$, $5$, $10$, $20$, hence we get $R(20)=6$.

You are also given $R(10!) = 312$.

Find $R(10\ 000\ 000!)$. Give your answer modulo $10^9 + 7$.

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总取整度

取整数是指在给定进制下以一个或多个零结尾的数。

定义数$n$在进制$b$下的取整度为$n$在$b$进制表示下末尾零的个数。

例如，$20$在$2$进制下的取整度为$2$，因为$20$的$2$进制表示是$10100$，末尾有$2$个零。

再定义$R(n)$为数$n$的总取整度，即$n$在所有$b > 1$的进制$b$下的取整度之和。

例如，$20$在$2$进制下的取整度为$2$，在$4$、$5$、$10$、$20$进制下的取整度为$1$，因此$R(20)=6$。

已知$R(10!) = 312$。

求$R(10\ 000\ 000!)$，并对$10^9 + 7$取余作为你的答案。

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