Number Theory – Factorials – JEE Main 24 Jan 2026 Shift 2

Solution:

$$40^n = (2^3 \times 5)^n = 2^{3n} \times 5^n$$

Exponent of 2 in $60!$:
$$E_2(60!) = \left[\frac{60}{2}\right] + \left[\frac{60}{4}\right] + \left[\frac{60}{8}\right] + \left[\frac{60}{16}\right] + \left[\frac{60}{32}\right]$$
$$= 30 + 15 + 7 + 3 + 1 = 56$$

Exponent of 5 in $60!$:
$$E_5(60!) = \left[\frac{60}{5}\right] + \left[\frac{60}{25}\right] = 12 + 2 = 14$$

We need $3n \le 56 \Rightarrow n \le 18.66$ and $n \le 14$.

The limiting factor is the power of 5.

Max value of $n = 14$.

Ans. (4)