Question

The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:

• A. 21
• B. 22
• C. 23
• D. 25

Hint:

Use Legendre’s formula to find the exponent of a prime in factorials.

Answer 1

The Correct Option is B

To determine the largest \( n \in \mathbb{N} \) such that \( 3^n \) divides \( 50! \), we need to calculate the number of times the prime number 3 appears in the factorization of \( 50! \). This can be done using Legendre's formula, which states that the exponent of a prime \( p \) in \( n! \) is given by:

\[\text{Exponent of } p = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor\]

In this case, \( n = 50 \) and \( p = 3 \). We calculate:

\[\left\lfloor \frac{50}{3} \right\rfloor = 16\]

\[\left\lfloor \frac{50}{9} \right\rfloor = 5\]

\[\left\lfloor \frac{50}{27} \right\rfloor = 1\]

\[\left\lfloor \frac{50}{81} \right\rfloor = 0\]

For higher powers of 3, the floor function results in zero.

Adding these values gives the largest \( n \) for which \( 3^n \) divides \( 50! \):

\[16 + 5 + 1 + 0 = 22\]

Therefore, the largest \( n \) is 22.