The highest power of 7 which divides 1000 ! is

Options

A. `164`
B. `162`
C. `167`
D. `142`

Answer

To find the highest power of 7 that divides \(1000!\), we can use the formula for finding the highest power of a prime \(p\) that divides \(n!\):

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

Where:
- \(n\) is the number for which we are calculating the factorial (in this case, \(1000\)),
- \(p\) is the prime number (in this case, \(7\)),
- \(\left\lfloor x \right\rfloor\) denotes the greatest integer less than or equal to \(x\).

### Step-by-step Solution:

1. **Identify \(n\) and \(p\)**:
- Here, \(n = 1000\) and \(p = 7\).

2. **Calculate the terms of the series**:
- We will calculate \(\left\lfloor \frac{1000}{7^i} \right\rfloor\) for \(i = 1, 2, 3, \ldots\) until the terms become zero.

- For \(i = 1\):
\[ \left\lfloor \frac{1000}{7^1} \right\rfloor = \left\lfloor \frac{1000}{7} \right\rfloor = \left\lfloor 142.857 \right\rfloor = 142 \]

- For \(i = 2\):
\[ \left\lfloor \frac{1000}{7^2} \right\rfloor = \left\lfloor \frac{1000}{49} \right\rfloor = \left\lfloor 20.408 \right\rfloor = 20 \]

- For \(i = 3\):
\[ \left\lfloor \frac{1000}{7^3} \right\rfloor = \left\lfloor \frac{1000}{343} \right\rfloor = \left\lfloor 2.915 \right\rfloor = 2 \]

- For \(i = 4\):
\[ \left\lfloor \frac{1000}{7^4} \right\rfloor = \left\lfloor \frac{1000}{2401} \right\rfloor = \left\lfloor 0.416 \right\rfloor = 0 \]

- Since the term for \(i = 4\) is zero, we can stop here.

3. **Sum the results**:
- Now, we sum the non-zero terms:
\[ k = 142 + 20 + 2 = 164 \]

### Final Answer:
The highest power of 7 that divides \(1000!\) is \(164\).

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