The largest n N, for which 7^n divides 101!, is: — Solution | JEE Mathematics

MCQEasyJEE 2026Applications of P&C

The largest $n \in \mathbb{N}$ , for which $7^n$ divides $101!$ , is :

Answer

Correct answer:C

Standard Method

Given: We need the largest $n \in \mathbb{N}$ such that $7^n$ divides $101!$ .

Find: The exponent of the prime $7$ in $101!$ .

Use Legendre's formula:

E_p(m!) = \left\lfloor \frac{m}{p} \right\rfloor + \left\lfloor \frac{m}{p^2} \right\rfloor + \dots

Here, $m = 101$ and $p = 7$ .

n = \left\lfloor \frac{101}{7} \right\rfloor + \left\lfloor \frac{101}{49} \right\rfloor + \left\lfloor \frac{101}{343} \right\rfloor + \dots

Now evaluate each term:

\left\lfloor \frac{101}{7} \right\rfloor = 14, \qquad \left\lfloor \frac{101}{49} \right\rfloor = 2, \qquad \left\lfloor \frac{101}{343} \right\rfloor = 0

So,

n = 14 + 2 = 16

Therefore, the largest $n$ is $16$ , so the correct option is C.

Using only $\left\lfloor \frac{101}{7} \right\rfloor$ and stopping there is incorrect because multiples of $49$ contribute an extra factor of $7$ . Always keep dividing by higher powers of the prime until the quotient becomes $0$ .
Applying the formula to a non-prime base would be wrong. Here the base is $7$ , which is prime, so Legendre's formula applies directly to the exponent of $7$ in $101!$ .
Treating $\left\lfloor \frac{101}{49} \right\rfloor$ as $1$ by rough estimation is incorrect. Compute the floor value carefully: $\frac{101}{49} > 2$ , so its floor is $2$ .

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