JEE Main 2025 — 2 April, Morning Shift — Question 34

The exponent of prime $p$ in $n!$ is given by Legendre's formula: $e_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots$. For $p = 3$ and $n = 50$: $\left\lfloor \frac{50}{3} \right\rfloor = 16$, $\left\lfloor \frac{50}{9} \right\rfloor = 5$, $\left\lfloor \frac{50}{27} \right\rfloor = 1$, $\left\lfloor \frac{50}{81} \right\rfloor = 0$. Summing: $16 + 5 + 1 = 22$. Thus the largest $n$ is $22$.