Question

The largest value of \( n \in \mathbb{N} \) such that \( 7^n \) divides \( 101! \) is ______.

Hint:

Use Legendre’s formula to find the largest power of a prime that divides \( n! \) by summing the terms \( \left\lfloor \frac{n}{p^k} \right\rfloor \), where \( p \) is the prime and \( k \) increases until \( p^k>n \).

Answer 1

Correct Answer: 16

Step 1: Understanding the problem.
We are asked to find the largest value of \( n \) such that \( 7^n \) divides \( 101! \). This is equivalent to finding the highest power of \( 7 \) that divides \( 101! \).
Step 2: Use Legendre's Formula.
Legendre’s formula is used to find the highest power of a prime \( p \) that divides \( n! \). The formula is: \[ \text{Power of } p \text{ in } n! = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \] For \( n = 101 \) and \( p = 7 \), we calculate the sum of the terms for powers of \( 7 \).
Step 3: Apply Legendre's Formula.
We need to compute the following terms: \[ \left\lfloor \frac{101}{7} \right\rfloor = 14 \] \[ \left\lfloor \frac{101}{7^2} \right\rfloor = \left\lfloor \frac{101}{49} \right\rfloor = 2 \] \[ \left\lfloor \frac{101}{7^3} \right\rfloor = \left\lfloor \frac{101}{343} \right\rfloor = 0 \] Thus, the total power of \( 7 \) in \( 101! \) is: \[ 14 + 2 + 0 = 16 \]
Step 4: Conclusion.
Therefore, the largest value of \( n \) such that \( 7^n \) divides \( 101! \) is \( 16 \). Thus, the correct answer is 16.