Factorial Trailing Digits 2

For any $N$, let $f(N)$ be the last twelve hexadecimal digits before the trailing zeroes in $N!$.
For example, the hexadecimal representation of $20!$ is 21C3677C82B40000,
so $f(20)$ is the digit sequence 21C3677C82B4.
Find $f(20!)$. Give your answer as twelve hexadecimal digits, using uppercase for the digits A to F.

First, we need to understand what $N!$ stands for in mathematical context. $N!$, symbol for N factorial, is simply a product of all positive integers less than or equal to N.

In this case, we are given $N=20$ which means $20! = 20 × 19 × 18 × …. × 3 × 2 × 1$.

The main task here is to find the last twelve hexadecimal digits before the trailing zeroes in $20!$.

Now hexadecimal is a base-16 number system. This means it has 16 distinct symbols, usually expressed as 0, 1, 2, …, 9, A, B, …, F. Hexadecimal A = decimal 10, B = 11, …, and F = 15.

We are told that the hexadecimal representation of $20!$ is 21C3677C82B40000. This takes advantage of hexadecimal representation being quite compact, fitting in a lot of information in small space.

The question is asking for $f(20)$, which it defines as the last twelve hexadecimal digits before the trailing zeroes in $20!$. From the given hexadecimal representation, it is clear that the trailing zeroes are the last four zeroes.

So, the last twelve hexadecimal digits before the trailing zeroes are 21C3677C82B4.

So, $f(20!) = 21C3677C82B4$.

So, the answer is 21C3677C82B4.

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