If 187! is divisible by `15^(n)` then find the maximum value of n.

Options

A. 45
B. 50
C. 46
D. 48

Answer

To find the maximum value of \( n \) such that \( 187! \) is divisible by \( 15^n \), we start by recognizing that \( 15 = 3 \times 5 \). Therefore, we need to determine how many times \( 3 \) and \( 5 \) appear in the prime factorization of \( 187! \).

### Step 1: Calculate the highest power of \( 3 \) in \( 187! \)

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 \( p = 3 \) and \( n = 187 \):

\[ \left\lfloor \frac{187}{3} \right\rfloor + \left\lfloor \frac{187}{3^2} \right\rfloor + \left\lfloor \frac{187}{3^3} \right\rfloor + \left\lfloor \frac{187}{3^4} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{187}{3} \right\rfloor = 62 \)
- \( \left\lfloor \frac{187}{9} \right\rfloor = 20 \)
- \( \left\lfloor \frac{187}{27} \right\rfloor = 6 \)
- \( \left\lfloor \frac{187}{81} \right\rfloor = 2 \)
- \( \left\lfloor \frac{187}{243} \right\rfloor = 0 \) (stop here since further terms will be 0)

Adding these together:

\[ 62 + 20 + 6 + 2 = 90 \]

Thus, the highest power of \( 3 \) in \( 187! \) is \( 90 \).

### Step 2: Calculate the highest power of \( 5 \) in \( 187! \)

Now, we do the same for \( p = 5 \):

\[ \left\lfloor \frac{187}{5} \right\rfloor + \left\lfloor \frac{187}{5^2} \right\rfloor + \left\lfloor \frac{187}{5^3} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{187}{5} \right\rfloor = 37 \)
- \( \left\lfloor \frac{187}{25} \right\rfloor = 7 \)
- \( \left\lfloor \frac{187}{125} \right\rfloor = 1 \)
- \( \left\lfloor \frac{187}{625} \right\rfloor = 0 \) (stop here)

Adding these together:

\[ 37 + 7 + 1 = 45 \]

Thus, the highest power of \( 5 \) in \( 187! \) is \( 45 \).

### Step 3: Determine the limiting factor for \( n \)

Since \( 15^n = 3^n \times 5^n \), the maximum \( n \) is limited by the smaller of the two powers we calculated:

- The highest power of \( 3 \) is \( 90 \).
- The highest power of \( 5 \) is \( 45 \).

Thus, the maximum value of \( n \) such that \( 15^n \) divides \( 187! \) is:

\[ n = \min(90, 45) = 45 \]

### Final Answer

The maximum value of \( n \) such that \( 187! \) is divisible by \( 15^n \) is \( \boxed{45} \).