If the product of first fifty positive consecutive integers be divisible by `7^(n)`, where n is an integer , then the largest possible value of n is

Options

A. 7
B. 8
C. 10
D. 5

Answer

To find the largest integer \( n \) such that the product of the first 50 positive integers is divisible by \( 7^n \), we can use the formula for finding the highest power of a prime \( p \) that divides \( n! \):

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

In this case, \( n = 50 \) and \( p = 7 \).

### Step 1: Calculate \( \left\lfloor \frac{50}{7} \right\rfloor \)

\[ \left\lfloor \frac{50}{7} \right\rfloor = \left\lfloor 7.14 \right\rfloor = 7 \]

This means there are 7 multiples of 7 between 1 and 50 (7, 14, 21, 28, 35, 42, 49).

### Step 2: Calculate \( \left\lfloor \frac{50}{7^2} \right\rfloor \)

\[ \left\lfloor \frac{50}{49} \right\rfloor = \left\lfloor 1.02 \right\rfloor = 1 \]

This means there is 1 multiple of \( 49 \) (which is \( 7^2 \)) between 1 and 50 (only 49).

### Step 3: Calculate \( \left\lfloor \frac{50}{7^3} \right\rfloor \)

\[ \left\lfloor \frac{50}{343} \right\rfloor = \left\lfloor 0.145 \right\rfloor = 0 \]

Since \( 7^3 = 343 \) is greater than 50, there are no multiples of \( 343 \) in this range.

### Step 4: Sum the results

Now, we sum the results from the previous steps to find the total power of 7 in \( 50! \):

\[ \text{Total power of } 7 = 7 + 1 + 0 = 8 \]

Thus, the largest possible value of \( n \) such that the product of the first 50 positive integers is divisible by \( 7^n \) is:

\[ \boxed{8} \]