Question:

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

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Use Legendre’s formula to find the exponent of a prime in factorials.
Updated On: Sep 9, 2025
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The Correct Option is B

Approach Solution - 1

To determine the largest \( n \in \mathbb{N} \) such that \( 3^n \) divides \( 50! \), we will use the method of finding the highest power of a prime \( p \) dividing \( n! \). The formula is:

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

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

Calculating each term step by step:

  • \(\left\lfloor \frac{50}{3} \right\rfloor = \left\lfloor 16.6667 \right\rfloor = 16\)
  • \(\left\lfloor \frac{50}{9} \right\rfloor = \left\lfloor 5.5556 \right\rfloor = 5\)
  • \(\left\lfloor \frac{50}{27} \right\rfloor = \left\lfloor 1.8519 \right\rfloor = 1\)
  • \(\left\lfloor \frac{50}{81} \right\rfloor = \left\lfloor 0.6173 \right\rfloor = 0\)

Since all further terms will also be zero, the sum is:

\(16 + 5 + 1 = 22\)

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

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Approach Solution -2

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.

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