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The problem in the image is asking for the **index of the highest power of 24 that is a factor of 124!** (124 factorial).

The process to find the highest power of a number \( n \) in a factorial \( N! \) is based on the formula:
\[
\text{Index of } p \text{ in } N! = \left\lfloor \frac{N}{p} \right\rfloor + \left\lfloor \frac{N}{p^2} \right\rfloor + \left\lfloor \frac{N}{p^3} \right\rfloor + \dots
\]
where \( p \) is a prime factor of \( n \).

### Step 1: Prime factorization of 24
First, we find the prime factorization of 24:
\[
24 = 2^3 \times 3
\]
Thus, we need to find the highest power of \( 2^3 \times 3 \) in \( 124! \).

### Step 2: Finding the highest power of 2 in \( 124! \)
Using the formula, calculate the highest power of 2 in \( 124! \):
\[
\left\lfloor \frac{124}{2} \right\rfloor = 62, \quad \left\lfloor \frac{124}{4} \right\rfloor = 31, \quad \left\lfloor \frac{124}{8} \right\rfloor = 15
\]
\[
\left\lfloor \frac{124}{16} \right\rfloor = 7, \quad \left\lfloor \frac{124}{32} \right\rfloor = 3, \quad \left\lfloor \frac{124}{64} \right\rfloor = 1
\]
Adding them up:
\[
62 + 31 + 15 + 7 + 3 + 1 = 119
\]
So, the highest power of 2 in \( 124! \) is 119.

### Step 3: Finding the highest power of 3 in \( 124! \)
Next, we find the highest power of 3 in \( 124! \):
\[
\left\lfloor \frac{124}{3} \right\rfloor = 41, \quad \left\lfloor \frac{124}{9} \right\rfloor = 13, \quad \left\lfloor \frac{124}{27} \right\rfloor = 4, \quad \left\lfloor \frac{124}{81} \right\rfloor = 1
\]
Adding them up:
\[
41 + 13 + 4 + 1 = 59
\]
So, the highest power of 3 in \( 124! \) is 59.

### Step 4: Finding the highest power of 24 in \( 124! \)
Now, we combine the results for the prime factors \( 2^3 \) and \( 3 \). The highest power of 24 is determined by how many complete sets of \( 2^3 \times 3 \) we can form.

- Since 24 contains three factors of 2, divide the power of 2 by 3:
\[
\left\lfloor \frac{119}{3} \right\rfloor = 39
\]
- The limiting factor is 39 (since 59 powers of 3 are available, but only 39 complete sets of \( 2^3 \times 3 \)).

### Final Answer:
The highest power of 24 that divides \( 124! \) is **39**.

Would you like further details, or do you have any questions? Here are five related questions:

1. How do you calculate the highest power of a prime factor in \( N! \)?
2. What happens if the number has more than two distinct prime factors?
3. How does the factorial grow in terms of prime factors?
4. Can this method be generalized for numbers other than factorials?
5. What if the number has a prime factor with a higher exponent than the others (e.g., \( 2^4 \times 3 \))?

**Tip:** When solving factorial-based questions, always start with the prime factorization of the number and break it down step by step.

The index of the highest power of 24 which is a factor of 124!

This problem explores how to find the highest power of 24 that divides 124! using prime factorization and Legendre's theorem. Learn to break down factorials and calculate the highest powers of 2 and 3.

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