To determine the remainder when $2009!$ is divided by 131, we use Wilson's theorem and modular arithmetic.

Wilson's Theorem:

For a prime number $p$, $(p-1)! \equiv -1 \pmod{p}$.

Here, $p = 131$, which is prime. Thus: $130! \equiv -1 \pmod{131}$

Extend to $2009!$:

We observe that $2009! = (131k + r)!$, where $k$ is an integer and $r = 2009 \mod 131$. First, calculate $2009 \mod 131$: $2009 \div 131 = 15 \text{ remainder } 44$ Thus: $2009 \equiv 44 \pmod{131}$

This reduces the factorial $2009!$ to $44! \pmod{131}$ because all multiples of $131$ beyond $44$ contribute nothing under modulo arithmetic.

Simplifying $44!$:

To simplify $44! \pmod{131}$, note that directly computing $44!$ is infeasible due to its size. However, $44!$ can be written as: $44! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot 44$

Using modular arithmetic, we multiply the terms iteratively modulo 131. Using computational tools or a systematic reduction approach, we compute: $44! \equiv 0 \pmod{131}$ because $131$ divides one of the terms in $2009!$ (any term greater than 131 contributes a factor of 0).

Thus, the remainder when $2009!$ is divided by $131$ is: $\boxed{0}$

Let me know if you'd like a detailed step-by-step walkthrough of the modular arithmetic or any further clarifications!

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Related Questions:

1. How does Wilson's theorem apply to other primes?
2. What is the factorial remainder of $2009!$ modulo $13$?
3. Can we generalize $n!$ modulo $p$ for composite $p$?
4. How can modular arithmetic simplify large factorials?
5. What are other applications of Wilson's theorem in number theory?

Tip: Modular arithmetic often simplifies problems involving large numbers by reducing terms to smaller equivalences under the modulus.