The largest n N such that 3^n divides 50! is: — Solution | JEE Mathematics

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

Answer

Correct answer:B

Standard Method

Given: We need the largest $n \in \mathbb{N}$ such that $3^n$ divides $50!$ .

Find: The highest power of $3$ in $50!$ .

Use Legendre's formula for the exponent of a prime $p$ in $n!$ :

\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor

For $n = 50$ and $p = 3$ ,

\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

Now calculate 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

So the total exponent is

16 + 5 + 1 = 22

Therefore, the largest $n$ such that $3^n$ divides $50!$ is $22$ . The correct option is B.

Legendre Formula Expansion

Given: We are asked to find the largest natural number $n$ such that $3^n$ divides $50!$ .

Find: The exponent of the prime $3$ in the prime factorization of $50!$ .

Concept Used: By Legendre's Formula, the exponent of a prime $p$ in $n!$ is

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

Here, $n = 50$ and $p = 3$ . So,

E_3(50!) = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{3^2} \right\rfloor + \left\lfloor \frac{50}{3^3} \right\rfloor + \left\lfloor \frac{50}{3^4} \right\rfloor + \dots

Evaluate term by term:

\left\lfloor \frac{50}{3} \right\rfloor = \lfloor 16.66\dots \rfloor = 16

This counts multiples of $3$ up to $50$ .

\left\lfloor \frac{50}{9} \right\rfloor = \lfloor 5.55\dots \rfloor = 5

This counts multiples of $9$ , contributing one extra factor of $3$ .

\left\lfloor \frac{50}{27} \right\rfloor = \lfloor 1.85\dots \rfloor = 1

This counts multiples of $27$ , contributing yet another factor of $3$ .

\left\lfloor \frac{50}{81} \right\rfloor = \lfloor 0.61\dots \rfloor = 0

Since the term for $3^4 = 81$ is $0$ , all later terms are also $0$ .

Hence,

E_3(50!) = 16 + 5 + 1 + 0 = 22

Thus, $3^{22}$ is the highest power of $3$ dividing $50!$ . Therefore, the largest $n$ is $22$ and the correct option is B.

A common mistake is counting only multiples of $3$ and stopping at $\left\lfloor \frac{50}{3} \right\rfloor = 16$ . This is wrong because multiples of $9, 27, \dots$ contribute extra factors of $3$ . Use Legendre's formula and keep adding higher-power terms until the quotient becomes $0$ .
Another mistake is using ordinary division instead of the floor function. Values like $\frac{50}{3} = 16.66\dots$ must be taken as $\left\lfloor \frac{50}{3} \right\rfloor = 16$ , not rounded to $17$ . Always apply the greatest integer function at every step.
Students may stop at $\left\lfloor \frac{50}{9} \right\rfloor$ and forget the $\left\lfloor \frac{50}{27} \right\rfloor$ term. This misses one additional factor of $3$ . Continue the series for all powers $3^k \le 50$ .

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