The largest value of \(n\), for which \(40^n\) divides \(60!\), is [JEE Main 2026, 24 Jan (Shift 2)]

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

Now, highest power of 2 which divides \(60!\) is

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

\(=30+15+7+3+1=56\)

Now, highest power of 5 which divides \(60!\) is

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

\(=12+2=14\)

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

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

Maximum value of \(n\) is \(14\).