If 100! divisible by 3^n then find the maximum value of n.

Options

A. 48
B. 44
C. 40
D. 33

Answer

To find the maximum value of \( n \) such that \( 100! \) is divisible by \( 3^n \), we need to determine how many times the prime number 3 appears in the factorization of \( 100! \). This can be calculated using the formula for finding the highest power of a prime \( p \) that divides \( n! \):

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

In this case, \( n = 100 \) and \( p = 3 \). We will calculate each term until \( p^k \) exceeds \( n \).

### Step 1: Calculate \( \left\lfloor \frac{100}{3} \right\rfloor \)

\[ \left\lfloor \frac{100}{3} \right\rfloor = \left\lfloor 33.33 \right\rfloor = 33 \]

### Step 2: Calculate \( \left\lfloor \frac{100}{3^2} \right\rfloor \)

\[ \left\lfloor \frac{100}{9} \right\rfloor = \left\lfloor 11.11 \right\rfloor = 11 \]

### Step 3: Calculate \( \left\lfloor \frac{100}{3^3} \right\rfloor \)

\[ \left\lfloor \frac{100}{27} \right\rfloor = \left\lfloor 3.70 \right\rfloor = 3 \]

### Step 4: Calculate \( \left\lfloor \frac{100}{3^4} \right\rfloor \)

\[ \left\lfloor \frac{100}{81} \right\rfloor = \left\lfloor 1.23 \right\rfloor = 1 \]

### Step 5: Calculate \( \left\lfloor \frac{100}{3^5} \right\rfloor \)

\[ \left\lfloor \frac{100}{243} \right\rfloor = \left\lfloor 0.41 \right\rfloor = 0 \]

Since \( 3^5 = 243 \) is greater than 100, we stop here.

### Step 6: Sum all the contributions

Now, we add all the values we calculated:

\[ 33 + 11 + 3 + 1 = 48 \]

### Conclusion

Thus, the maximum value of \( n \) such that \( 100! \) is divisible by \( 3^n \) is:

\[ \boxed{48} \]