If 123! is divisible by `12^(n)` then find the maximum value of n.

Options

A. 58
B. 50
C. 59
D. 60

Answer

To find the maximum value of \( n \) such that \( 123! \) is divisible by \( 12^n \), we first need to express \( 12 \) in terms of its prime factors:

\[ 12 = 2^2 \times 3^1 \]

This means that \( 12^n = (2^2 \times 3^1)^n = 2^{2n} \times 3^n \). Therefore, we need to find the maximum \( n \) such that \( 123! \) has at least \( 2^{2n} \) and \( 3^n \).

### Step 1: Calculate the highest power of 2 in \( 123! \)

To find the highest power of a prime \( p \) in \( n! \), we use the formula:

\[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]

For \( p = 2 \):

\[ \left\lfloor \frac{123}{2} \right\rfloor + \left\lfloor \frac{123}{4} \right\rfloor + \left\lfloor \frac{123}{8} \right\rfloor + \left\lfloor \frac{123}{16} \right\rfloor + \left\lfloor \frac{123}{32} \right\rfloor + \left\lfloor \frac{123}{64} \right\rfloor \]

Calculating each term:

\[ \left\lfloor \frac{123}{2} \right\rfloor = 61 \]
\[ \left\lfloor \frac{123}{4} \right\rfloor = 30 \]
\[ \left\lfloor \frac{123}{8} \right\rfloor = 15 \]
\[ \left\lfloor \frac{123}{16} \right\rfloor = 7 \]
\[ \left\lfloor \frac{123}{32} \right\rfloor = 3 \]
\[ \left\lfloor \frac{123}{64} \right\rfloor = 1 \]

Now, summing these values:

\[ 61 + 30 + 15 + 7 + 3 + 1 = 117 \]

So, the highest power of \( 2 \) in \( 123! \) is \( 117 \).

### Step 2: Calculate the highest power of 3 in \( 123! \)

Now, for \( p = 3 \):

\[ \left\lfloor \frac{123}{3} \right\rfloor + \left\lfloor \frac{123}{9} \right\rfloor + \left\lfloor \frac{123}{27} \right\rfloor + \left\lfloor \frac{123}{81} \right\rfloor \]

Calculating each term:

\[ \left\lfloor \frac{123}{3} \right\rfloor = 41 \]
\[ \left\lfloor \frac{123}{9} \right\rfloor = 13 \]
\[ \left\lfloor \frac{123}{27} \right\rfloor = 4 \]
\[ \left\lfloor \frac{123}{81} \right\rfloor = 1 \]

Now, summing these values:

\[ 41 + 13 + 4 + 1 = 59 \]

So, the highest power of \( 3 \) in \( 123! \) is \( 59 \).

### Step 3: Determine the maximum \( n \)

We have:
- The highest power of \( 2 \) is \( 117 \), which means \( 2^{2n} \) can be satisfied for \( n \leq \frac{117}{2} = 58.5 \) (so \( n \leq 58 \)).
- The highest power of \( 3 \) is \( 59 \), which means \( n \) can be at most \( 59 \).

Since \( n \) must satisfy both conditions, the limiting factor is the power of \( 2 \):

\[ n \leq 58 \]

Thus, the maximum value of \( n \) such that \( 123! \) is divisible by \( 12^n \) is:

\[ \boxed{58} \]