Reply To: Question 42 — JEE Main 24 Jan 2026 S Ii
Solution
\[
40=8 \times 5=2^{3} \times 5
\]
Exponent of 2 in $60!=\left[\frac{60}{2}\right]+\left[\frac{60}{4}\right]+\left[\frac{60}{8}\right]+\left[\frac{60}{16}\right]+\left[\frac{60}{32}\right]+\left[\frac{60}{64}\right]+\cdots$
\[
=30+15+7+3+1+0+\cdots=56
\]
Exponent of 5 in $60!=\left[\frac{60}{5}\right]+\left[\frac{60}{25}\right]+\left[\frac{60}{125}\right]+\cdots=12+2+0=14$
\[
60!=2^{56} 5^{14}(\mathrm{~K})=2^{14}\left(2^{42}\right) 5^{14} \mathrm{~K}=2^{14}\left(2^{3} \times 5\right)^{14} \mathrm{~K}=2^{14}(40)^{14} \mathrm{~K}
\]
∴ Maximum value of n is 14
Answer (4)