## Consider the number $48$.

There are five pairs of integers $a$ and $b$ ($a \leq b$) such that $a \times b=48$: $(1,48)$, $(2,24)$, $(3,16)$, $(4,12)$ and $(6,8)$.

It can be seen that both $6$ and $8$ have $4$ divisors.

So of those five pairs one consists of two integers with the same number of divisors.

In general:

Let $C(n)$ be the number of pairs of positive integers $a \times b=n$, ($a \leq b$) such that $a$ and $b$ have the same number of divisors; so $C(48)=1$.

You are given $C(10!)=3$: $(1680, 2160)$, $(1800, 2016)$ and $(1890,1920)$.

Find $C(100!)$.

### This problem is unrealistic. The solution should not be calculated directly by hand because the number $100!$ – which denotes $100$ factorial, or the product of all numbers from $1$ to $100$ – is too large even to write out, let alone find the pairs of factors that have the same number of divisors.

However, we can make some general conclusions about the types of integer pairs for any factorial. To have the same number of divisors, two numbers likely share many factors in common – and since factorials multiply together all numbers up to a certain integer, the pairs of integers for a factorial will likely be very close to each other.

The method to solve the $C(n)$ where $n$ is a factorial requires advanced programming or mathematical software. The algorithm would involve factoring $n$ down to its prime factors, generating all pair combinations of those factors where $a \leq b$, and counting the number of divisors in each pair. The pairs with equal number of divisors are the one we’re interested in. Without such tools, this problem is essentially impossible to solve.

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