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

To determine the largest natural number \( n \) for which \( 7^n \) divides \( 101! \), we use the formula to find the highest power of a prime \( p \) that divides \( n! \). The formula is: 

\(\text{Power of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots\)

Here, \( p = 7 \) and \( n = 101 \). We can calculate each term:

1. \(\left\lfloor \frac{101}{7} \right\rfloor = \left\lfloor 14.4286 \right\rfloor = 14\)
2. \(\left\lfloor \frac{101}{7^2} \right\rfloor = \left\lfloor \frac{101}{49} \right\rfloor = \left\lfloor 2.0612 \right\rfloor = 2\)
3. \(\left\lfloor \frac{101}{7^3} \right\rfloor = \left\lfloor \frac{101}{343} \right\rfloor = \left\lfloor 0.2945 \right\rfloor = 0\)

Higher powers of 7 (like \( 7^4 \) or greater) will divide 101! less than once, hence provide negligible contributions to the sum. Adding these up gives:

\(14 + 2 + 0 = 16\)

Thus, the largest \( n \) for which \( 7^n \) divides \( 101! \) is 16.