Question

The greatest value of \( n \), where \( n \in \mathbb{N} \), such that \( 3^n \) divides \( 50! \) is:

• A. 20
• B. 21
• C. 22
• D. 23

Hint:

For finding the highest power of a prime dividing \( n! \), use the formula involving the sum of floors of divisions by powers of the prime.

Answer 1

The Correct Option is A

We are asked to find the greatest value of \( n \) such that \( 3^n \) divides \( 50! \). The number of times a prime \( p \) divides \( n! \) is given by the formula: \[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \] where \( \left\lfloor x \right\rfloor \) denotes the greatest integer less than or equal to \( x \). In this case, \( p = 3 \) and \( n = 50 \). So, we need to find: \[ \sum_{k=1}^{\infty} \left\lfloor \frac{50}{3^k} \right\rfloor \] Step 1: Calculate the individual terms. For \( k = 1 \): \[ \left\lfloor \frac{50}{3} \right\rfloor = \left\lfloor 16.67 \right\rfloor = 16 \] For \( k = 2 \): \[ \left\lfloor \frac{50}{9} \right\rfloor = \left\lfloor 5.56 \right\rfloor = 5 \] For \( k = 3 \): \[ \left\lfloor \frac{50}{27} \right\rfloor = \left\lfloor 1.85 \right\rfloor = 1 \] For \( k = 4 \): \[ \left\lfloor \frac{50}{81} \right\rfloor = \left\lfloor 0.62 \right\rfloor = 0 \] Since higher powers of 3 will not divide 50, we stop here. Step 2: Add the results. \[ 16 + 5 + 1 = 22 \] Thus, the greatest value of \( n \) such that \( 3^n \) divides \( 50! \) is \( 22 \).