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 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.

Answer 2

We are asked to find the largest natural number \( n \) such that \( 3^n \) divides \( 50! \). This is equivalent to finding the exponent of the highest power of the prime number 3 in the prime factorization of \( 50! \).

Concept Used:

To solve this problem, we use Legendre's Formula. This formula calculates the exponent of a prime number \( p \) in the prime factorization of \( n! \). The formula is given by:

\[ E_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots \]

where \( \lfloor x \rfloor \) represents the greatest integer function, which gives the largest integer less than or equal to \( x \).

Step-by-Step Solution:

• In this problem, we are given \( n = 50 \) and we need to find the highest power of the prime \( p = 3 \).
• We will apply Legendre's formula by substituting these values. The summation continues until the term \( \left\lfloor \frac{n}{p^k} \right\rfloor \) becomes zero.
• Let's calculate each term of the series:
  • For \( k = 1 \): The term is \( \left\lfloor \frac{50}{3^1} \right\rfloor = \left\lfloor \frac{50}{3} \right\rfloor = \lfloor 16.66\dots \rfloor = 16 \). This represents the number of multiples of 3 up to 50.
  • For \( k = 2 \): The term is \( \left\lfloor \frac{50}{3^2} \right\rfloor = \left\lfloor \frac{50}{9} \right\rfloor = \lfloor 5.55\dots \rfloor = 5 \). This counts the multiples of 9, which contribute an additional factor of 3.
  • For \( k = 3 \): The term is \( \left\lfloor \frac{50}{3^3} \right\rfloor = \left\lfloor \frac{50}{27} \right\rfloor = \lfloor 1.85\dots \rfloor = 1 \). This counts the multiples of 27, which contribute yet another factor of 3.
  • For \( k = 4 \): The term is \( \left\lfloor \frac{50}{3^4} \right\rfloor = \left\lfloor \frac{50}{81} \right\rfloor = \lfloor 0.61\dots \rfloor = 0 \).
• Since the term for \( k=4 \) is 0, all subsequent terms for \( k > 4 \) will also be 0 because the denominator \( 3^k \) will become larger than the numerator 50.

Final Computation & Result:

To find the total exponent \( n \), we sum the non-zero values calculated in the steps above.

\[ n = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor + \dots \]

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

This result means that \( 3^{22} \) is the highest power of 3 that is a factor of \( 50! \).

Thus, the largest \( n \in \mathbb{N} \) such that \( 3^n \) divides \( 50! \) is 22.