Algebra – Number Theory – JEE Main 21 Jan 2026 Shift 2

Solution:

To find the largest power of a prime $p$ that divides $n!$, we use Legendre’s Formula:
$$E_p(n!) = \left[ \frac{n}{p} \right] + \left[ \frac{n}{p^2} \right] + \left[ \frac{n}{p^3} \right] + \dots$$

Here $n = 101$ and $p = 7$.

$$E_7(101!) = \left[ \frac{101}{7} \right] + \left[ \frac{101}{7^2} \right] + \left[ \frac{101}{7^3} \right] + \dots$$

Calculate each term:
$$\left[ \frac{101}{7} \right] = [14.42…] = 14$$

$$\left[ \frac{101}{49} \right] = [2.06…] = 2$$

$$\left[ \frac{101}{343} \right] = 0$$

Sum of the exponents:
$$n = 14 + 2 + 0 = 16$$

So, the largest value of $n$ is 16.

Ans. (1)