The question in the image asks:

"What is the maximum power of n such that n divides 500! (500 factorial)?"

This is a typical problem of finding the largest exponent $k$ such that a number $n^k$ divides $500!$.

Step-by-Step Solution:

Let’s assume $n$ is a prime number. To find the highest power $k$ of a prime $n$ that divides $500!$, we use Legendre’s formula (also called de Polignac's formula):

$k = \sum_{i=1}^{\infty} \left\lfloor \frac{500}{n^i} \right\rfloor$

Where $\left\lfloor x \right\rfloor$ denotes the greatest integer less than or equal to $x$, and $n^i$ is the successive powers of the prime $n$.

The steps are:

1. Divide 500 by $n$ (the prime number).
2. Divide the result by $n$ again (repeating until the division is zero).
3. Sum all the quotients (rounded down).

If $n$ is a composite number, first factorize it into primes and apply the formula to each prime factor separately.

If you provide a specific value for $n$, I can compute the answer for that prime or composite number.

Would you like more details or help with a specific value of $n$?

Follow-Up Questions:

1. How is Legendre’s formula used in prime factorization of factorials?
2. What is the highest power of 2 that divides 500!?
3. Can we apply the same method to non-prime values of $n$?
4. What role do prime numbers play in this type of divisibility problem?
5. How would the solution change if the factorial were a different number, like 1000!?

Tip: For composite numbers, always break them down into their prime factors before applying Legendre’s formula!