If 133 ! Is exactly divisible by `6^(n)` find the max value of n.

Answer

To find the maximum value of \( n \) such that \( 133! \) is exactly divisible by \( 6^n \), we start by recognizing that \( 6 \) can be factored into its prime components:

\[ 6 = 2 \times 3 \]

Thus, \( 6^n = 2^n \times 3^n \). This means we need to determine how many times \( 2 \) and \( 3 \) appear in the prime factorization of \( 133! \).

### Step 1: Calculate the power of \( 2 \) in \( 133! \)

The power of a prime \( p \) in \( n! \) can be calculated using the formula:

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

For \( p = 2 \) and \( n = 133 \):

\[ \left\lfloor \frac{133}{2} \right\rfloor + \left\lfloor \frac{133}{4} \right\rfloor + \left\lfloor \frac{133}{8} \right\rfloor + \left\lfloor \frac{133}{16} \right\rfloor + \left\lfloor \frac{133}{32} \right\rfloor + \left\lfloor \frac{133}{64} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{133}{2} \right\rfloor = 66 \)
- \( \left\lfloor \frac{133}{4} \right\rfloor = 33 \)
- \( \left\lfloor \frac{133}{8} \right\rfloor = 16 \)
- \( \left\lfloor \frac{133}{16} \right\rfloor = 8 \)
- \( \left\lfloor \frac{133}{32} \right\rfloor = 4 \)
- \( \left\lfloor \frac{133}{64} \right\rfloor = 2 \)

Now, summing these values:

\[ 66 + 33 + 16 + 8 + 4 + 2 = 129 \]

So, the power of \( 2 \) in \( 133! \) is \( 129 \).

### Step 2: Calculate the power of \( 3 \) in \( 133! \)

Using the same formula for \( p = 3 \):

\[ \left\lfloor \frac{133}{3} \right\rfloor + \left\lfloor \frac{133}{9} \right\rfloor + \left\lfloor \frac{133}{27} \right\rfloor + \left\lfloor \frac{133}{81} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{133}{3} \right\rfloor = 44 \)
- \( \left\lfloor \frac{133}{9} \right\rfloor = 14 \)
- \( \left\lfloor \frac{133}{27} \right\rfloor = 4 \)
- \( \left\lfloor \frac{133}{81} \right\rfloor = 1 \)

Now, summing these values:

\[ 44 + 14 + 4 + 1 = 63 \]

So, the power of \( 3 \) in \( 133! \) is \( 63 \).

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

Since \( 6^n = 2^n \times 3^n \), the maximum \( n \) will be determined by the smaller of the two powers calculated:

\[ n = \min(129, 63) = 63 \]

Thus, the maximum value of \( n \) such that \( 133! \) is exactly divisible by \( 6^n \) is:

\[ \boxed{63} \]