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

Step 1: Understanding the problem

We need to find the largest integer \( n \) such that \( 7^n \) divides \( 101! \). This requires finding the highest power of 7 that divides \( 101! \).

Step 2: Using Legendre's formula

To find the largest power of a prime \( p \) that divides \( n! \), we use Legendre's formula, which states: \[ \text{Exponent of } p \text{ in } n! = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \] where \( \left\lfloor x \right\rfloor \) denotes the greatest integer less than or equal to \( x \). For our case, \( p = 7 \) and \( n = 101 \), so we need to compute: \[ \sum_{k=1}^{\infty} \left\lfloor \frac{101}{7^k} \right\rfloor \]

Step 3: Calculate each term in the sum

- For \( k = 1 \): \[ \left\lfloor \frac{101}{7^1} \right\rfloor = \left\lfloor \frac{101}{7} \right\rfloor = \left\lfloor 14.428 \right\rfloor = 14 \] - For \( k = 2 \): \[ \left\lfloor \frac{101}{7^2} \right\rfloor = \left\lfloor \frac{101}{49} \right\rfloor = \left\lfloor 2.061 \right\rfloor = 2 \] - For \( k = 3 \): \[ \left\lfloor \frac{101}{7^3} \right\rfloor = \left\lfloor \frac{101}{343} \right\rfloor = \left\lfloor 0.294 \right\rfloor = 0 \] Since \( 7^k \) for \( k \geq 4 \) will be greater than 101, all further terms will contribute 0.

Step 4: Add the results

Now, we sum the non-zero terms: \[ 14 + 2 + 0 = 16 \]

Step 5: Final Answer

The largest value of \( n \) such that \( 7^n \) divides \( 101! \) is \( \boxed{16} \).