Question

The largest \( n \in \mathbb{N} \), for which \( 7^n \) divides \( 101! \), is :

• A. 15
• B. 19
• C. 16
• D. 18

Hint:

Always divide the number by powers of the prime until the result is less than 1.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To find the exponent of a prime \( p \) in \( m! \), we use Legendre's Formula.

Step 2: Key Formula or Approach:
\[ E_p(m!) = \lfloor \frac{m}{p} \rfloor + \lfloor \frac{m}{p^2} \rfloor + \dots \]

Step 3: Detailed Explanation:
Here \( m = 101 \) and \( p = 7 \).
\[ n = \lfloor \frac{101}{7} \rfloor + \lfloor \frac{101}{49} \rfloor + \lfloor \frac{101}{343} \rfloor + \dots \]
\[ n = \lfloor 14.42 \dots \rfloor + \lfloor 2.06 \dots \rfloor + 0 \]
\[ n = 14 + 2 = 16 \]

Step 4: Final Answer:
The largest \( n \) is 16.