JEE Main 2026 — 24 January, Evening Shift — Question 10

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

$\mathrm{E}_{2}(60!)=\left[\frac{60}{2}\right]+\left[\frac{60}{2^{2}}\right]+\left[\frac{60}{2^{3}}\right]+\left[\frac{60}{2^{4}}\right]+\left[\frac{60}{2^{5}}\right]$

$=30+15+7+3+1=56$

$\mathrm{E}_{5}(60!)=\left[\frac{60}{5}\right]+\left[\frac{60}{5^{2}}\right]$

$=12+2=14$

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

$60!=2^{56} \times 5^{14} \ldots=2^{14} \cdot\left(2^{3} .5\right)^{14}$

∴ Maximum value of n is 14 .