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 find the highest power of 3 that divides 50!, we use Legendre's formula, which states that the exponent of a prime \( p \) in the prime factorization of \( n! \) is given by:

\(\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor\)
For \( n = 50 \) and \( p = 3 \), this becomes:

\(\left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor + \ldots\)

Evaluating each term:

• \(\left\lfloor \frac{50}{3} \right\rfloor = 16\)
• \(\left\lfloor \frac{50}{9} \right\rfloor = 5\)
• \(\left\lfloor \frac{50}{27} \right\rfloor = 1\)
• \(\left\lfloor \frac{50}{81} \right\rfloor = 0\)

Subsequent terms are zero. The sum is:

\(16 + 5 + 1 = 22\)

Therefore, the largest integer \( n \) such that \( 3^n \) divides \( 50! \) is 22.