If 122! is divisible by 6^n then find the maximum value of n.

Options

A. 58
B. 62
C. 40
D. 48

Answer

To find the maximum value of \( n \) such that \( 122! \) is divisible by \( 6^n \), we need to consider the prime factorization of \( 6 \). The number \( 6 \) can be expressed as:

\[ 6 = 2 \times 3 \]

Thus, \( 6^n = 2^n \times 3^n \). This means that for \( 122! \) to be divisible by \( 6^n \), it must have at least \( n \) factors of \( 2 \) and \( n \) factors of \( 3 \).

### Step 1: Calculate the number of factors of \( 2 \) in \( 122! \)

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

\[ \text{Number of factors of } p = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]

For \( p = 2 \) and \( n = 122 \):

\[ \left\lfloor \frac{122}{2} \right\rfloor + \left\lfloor \frac{122}{4} \right\rfloor + \left\lfloor \frac{122}{8} \right\rfloor + \left\lfloor \frac{122}{16} \right\rfloor + \left\lfloor \frac{122}{32} \right\rfloor + \left\lfloor \frac{122}{64} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{122}{2} \right\rfloor = 61 \)
- \( \left\lfloor \frac{122}{4} \right\rfloor = 30 \)
- \( \left\lfloor \frac{122}{8} \right\rfloor = 15 \)
- \( \left\lfloor \frac{122}{16} \right\rfloor = 7 \)
- \( \left\lfloor \frac{122}{32} \right\rfloor = 3 \)
- \( \left\lfloor \frac{122}{64} \right\rfloor = 1 \)

Now, summing these values:

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

So, the total number of factors of \( 2 \) in \( 122! \) is \( 117 \).

### Step 2: Calculate the number of factors of \( 3 \) in \( 122! \)

Now, we will calculate the number of factors of \( 3 \) in \( 122! \) using the same formula:

\[ \left\lfloor \frac{122}{3} \right\rfloor + \left\lfloor \frac{122}{9} \right\rfloor + \left\lfloor \frac{122}{27} \right\rfloor + \left\lfloor \frac{122}{81} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{122}{3} \right\rfloor = 40 \)
- \( \left\lfloor \frac{122}{9} \right\rfloor = 13 \)
- \( \left\lfloor \frac{122}{27} \right\rfloor = 4 \)
- \( \left\lfloor \frac{122}{81} \right\rfloor = 1 \)

Now, summing these values:

\[ 40 + 13 + 4 + 1 = 58 \]

So, the total number of factors of \( 3 \) in \( 122! \) is \( 58 \).

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

Since \( 6^n = 2^n \times 3^n \), we need both \( n \) factors of \( 2 \) and \( n \) factors of \( 3 \). The limiting factor will be the smaller of the two counts:

\[ n \leq \min(117, 58) = 58 \]

Thus, the maximum value of \( n \) such that \( 122! \) is divisible by \( 6^n \) is:

\[ \boxed{58} \]