Question

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

Hint:

Whenever you are asked for the highest power of a prime dividing a factorial, directly apply Legendre’s formula. Stop the calculation once \(p^{k}>n\).

Answer 1

Correct Answer: 16

Concept: To find the highest power of a prime number \(p\) dividing \(n!\), we use Legendre’s formula: \[ \text{Power of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^{2}} \right\rfloor + \left\lfloor \frac{n}{p^{3}} \right\rfloor + \cdots \] Here, \(p = 7\) and \(n = 101\).
Step 1: Apply Legendre’s formula \[ \left\lfloor \frac{101}{7} \right\rfloor = 14 \] \[ \left\lfloor \frac{101}{7^{2}} \right\rfloor = \left\lfloor \frac{101}{49} \right\rfloor = 2 \] \[ \left\lfloor \frac{101}{7^{3}} \right\rfloor = \left\lfloor \frac{101}{343} \right\rfloor = 0 \]
Step 2: Add all contributions \[ n = 14 + 2 + 0 = 16 \] Final Answer: The largest power of \(7\) that divides \(101!\) is \[ \boxed{7^{16}} \] Hence, \[ \boxed{n = 16} \]