Find the number of highest power of 3 in 270!

Options

A. 134
B. 130
C. 140
D. 135

Answer

To find the highest power of 3 in \( 270! \), we can use the formula that calculates the highest power of a prime \( p \) in \( n! \):

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

In this case, \( n = 270 \) and \( p = 3 \). We will calculate the sum of the integer divisions of \( 270 \) by increasing powers of \( 3 \) until the result is zero.

### Step-by-Step Solution:

1. **Calculate \( \left\lfloor \frac{270}{3^1} \right\rfloor \)**:
\[ \left\lfloor \frac{270}{3} \right\rfloor = \left\lfloor 90 \right\rfloor = 90 \]

2. **Calculate \( \left\lfloor \frac{270}{3^2} \right\rfloor \)**:
\[ \left\lfloor \frac{270}{9} \right\rfloor = \left\lfloor 30 \right\rfloor = 30 \]

3. **Calculate \( \left\lfloor \frac{270}{3^3} \right\rfloor \)**:
\[ \left\lfloor \frac{270}{27} \right\rfloor = \left\lfloor 10 \right\rfloor = 10 \]

4. **Calculate \( \left\lfloor \frac{270}{3^4} \right\rfloor \)**:
\[ \left\lfloor \frac{270}{81} \right\rfloor = \left\lfloor 3.33 \right\rfloor = 3 \]

5. **Calculate \( \left\lfloor \frac{270}{3^5} \right\rfloor \)**:
\[ \left\lfloor \frac{270}{243} \right\rfloor = \left\lfloor 1.11 \right\rfloor = 1 \]

6. **Calculate \( \left\lfloor \frac{270}{3^6} \right\rfloor \)**:
\[ \left\lfloor \frac{270}{729} \right\rfloor = \left\lfloor 0.37 \right\rfloor = 0 \]

Since the result is now zero, we stop here.

7. **Sum all the results**:
\[ 90 + 30 + 10 + 3 + 1 = 134 \]

Thus, the highest power of \( 3 \) in \( 270! \) is \( 134 \).

### Final Answer:
The highest power of \( 3 \) in \( 270! \) is \( 134 \).