123! Is exactly divisible by `(25!)^(x)` Find the max value of x.

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the maximum value of \( x \) such that \( 123! \) is exactly divisible by \( (25!)^x \), we need to determine how many times \( 25! \) can fit into \( 123! \). This involves finding the prime factorization of \( 25! \) and then determining how many times each prime factor appears in \( 123! \).

### Step-by-Step Solution:

1. **Determine the Prime Factorization of \( 25! \)**:
We need to find the prime factorization of \( 25! \). The prime factors of \( 25! \) are all the prime numbers less than or equal to 25.

2. **Count the Occurrences of Each Prime Factor in \( 25! \)**:
For each prime \( p \), the exponent of \( p \) in \( 25! \) can be calculated using the formula:
\[ e_p(25!) = \left\lfloor \frac{25}{p} \right\rfloor + \left\lfloor \frac{25}{p^2} \right\rfloor + \left\lfloor \frac{25}{p^3} \right\rfloor + \ldots \]

3. **Count the Occurrences of Each Prime Factor in \( 123! \)**:
Similarly, for \( 123! \), we calculate the exponent of each prime \( p \):
\[ e_p(123!) = \left\lfloor \frac{123}{p} \right\rfloor + \left\lfloor \frac{123}{p^2} \right\rfloor + \left\lfloor \frac{123}{p^3} \right\rfloor + \ldots \]

4. **Calculate the Maximum \( x \)**:
For each prime \( p \), we need to find:
\[ x_p = \left\lfloor \frac{e_p(123!)}{e_p(25!)} \right\rfloor \]
The maximum value of \( x \) will be the minimum of all \( x_p \) values for each prime factor.

5. **Perform the Calculations**:
Let's calculate for the prime factors of \( 25! \) which are \( 2, 3, 5, 7, 11, 13, 17, 19, 23 \).

- For \( p = 2 \):
\[ e_2(25!) = \left\lfloor \frac{25}{2} \right\rfloor + \left\lfloor \frac{25}{4} \right\rfloor + \left\lfloor \frac{25}{8} \right\rfloor + \left\lfloor \frac{25}{16} \right\rfloor = 12 + 6 + 3 + 1 = 22 \]
\[ e_2(123!) = \left\lfloor \frac{123}{2} \right\rfloor + \left\lfloor \frac{123}{4} \right\rfloor + \left\lfloor \frac{123}{8} \right\rfloor + \left\lfloor \frac{123}{16} \right\rfloor + \left\lfloor \frac{123}{32} \right\rfloor + \left\lfloor \frac{123}{64} \right\rfloor = 61 + 30 + 15 + 7 + 3 + 1 = 117 \]
\[ x_2 = \left\lfloor \frac{117}{22} \right\rfloor = 5 \]

- For \( p = 3 \):
\[ e_3(25!) = \left\lfloor \frac{25}{3} \right\rfloor + \left\lfloor \frac{25}{9} \right\rfloor + \left\lfloor \frac{25}{27} \right\rfloor = 8 + 2 + 0 = 10 \]
\[ e_3(123!) = \left\lfloor \frac{123}{3} \right\rfloor + \left\lfloor \frac{123}{9} \right\rfloor + \left\lfloor \frac{123}{27} \right\rfloor + \left\lfloor \frac{123}{81} \right\rfloor = 41 + 13 + 4 + 1 = 59 \]
\[ x_3 = \left\lfloor \frac{59}{10} \right\rfloor = 5 \]

- For \( p = 5 \):
\[ e_5(25!) = \left\lfloor \frac{25}{5} \right\rfloor + \left\lfloor \frac{25}{25} \right\rfloor = 5 + 1 = 6 \]
\[ e_5(123!) = \left\lfloor \frac{123}{5} \right\rfloor + \left\lfloor \frac{123}{25} \right\rfloor = 24 + 4 = 28 \]
\[ x_5 = \left\lfloor \frac{28}{6} \right\rfloor = 4 \]

- For \( p = 7 \):
\[ e_7(25!) = \left\lfloor \frac{25}{7} \right\rfloor + \left\lfloor \frac{25}{49} \right\rfloor = 3 + 0 = 3 \]
\[ e_7(123!) = \left\lfloor \frac{123}{7} \right\rfloor + \left\lfloor \frac{123}{49} \right\rfloor = 17 + 2 = 19 \]
\[ x_7 = \left\lfloor \frac{19}{3} \right\rfloor = 6 \]

- For \( p = 11 \):
\[ e_{11}(25!) = \left\lfloor \frac{25}{11} \right\rfloor = 2 \]
\[ e_{11}(123!) = \left\lfloor \frac{123}{11} \right\rfloor = 11 \]
\[ x_{11} = \left\lfloor \frac{11}{2} \right\rfloor = 5 \]

- For \( p = 13 \):
\[ e_{13}(25!) = \left\lfloor \frac{25}{13} \right\rfloor = 1 \]
\[ e_{13}(123!) = \left\lfloor \frac{123}{13} \right\rfloor = 9 \]
\[ x_{13} = \left\lfloor \frac{9}{1} \right\rfloor = 9 \]

- For \( p = 17 \):
\[ e_{17}(25!) = \left\lfloor \frac{25}{17} \right\rfloor = 1 \]
\[ e_{17}(123!) = \left\lfloor \frac{123}{17} \right\rfloor = 7 \]
\[ x_{17} = \left\lfloor \frac{7}{1} \right\rfloor = 7 \]

- For \( p = 19 \):
\[ e_{19}(25!) = \left\lfloor \frac{25}{19} \right\rfloor = 1 \]
\[ e_{19}(123!) = \left\lfloor \frac{123}{19} \right\rfloor = 6 \]
\[ x_{19} = \left\lfloor \frac{6}{1} \right\rfloor = 6 \]

- For \( p = 23 \):
\[ e_{23}(25!) = \left\lfloor \frac{25}{23} \right\rfloor = 1 \]
\[ e_{23}(123!) = \left\lfloor \frac{123}{23} \right\rfloor = 5 \]
\[ x_{23} = \left\lfloor \frac{5}{1} \right\rfloor = 5 \]

6. **Find the Minimum \( x \)**:
Now we have:
- \( x_2 = 5 \)
- \( x_3 = 5 \)
- \( x_5 = 4 \)
- \( x_7 = 6 \)
- \( x_{11} = 5 \)
- \( x_{13} = 9 \)
- \( x_{17} = 7 \)
- \( x_{19} = 6 \)
- \( x_{23} = 5 \)

The minimum value among these is \( 4 \).

### Final Answer:
The maximum value of \( x \) such that \( 123! \) is exactly divisible by \( (25!)^x \) is \( \boxed{4} \).