The largest value of n, for which $$40^{n}$$ divides 60! , is

The value $$40$$ upon factorisation gives, $$40\ =\ 2^3\times\ 5$$

So, we need to find the highest powers of $$2$$ and $$5$$ in $$60!$$ to find the highest power of $$40$$ in $$60!$$

The highest power of $$2$$ in $$60!$$ $$=\left[\dfrac{60}{2}\right]+\left[\dfrac{60}{2^2}\right]+\left[\dfrac{60}{2^3}\right]+...\ =\ 30+15+7+3+1=56$$

The highest power of $$5$$ in $$60!$$ $$=\left[\dfrac{60}{5}\right]+\left[\dfrac{60}{5^2}\right]+\left[\dfrac{60}{5^3}\right]+...\ =\ 12+2=14$$

We calculated the highest power of $$2$$ in $$60!$$ to be $$56$$, and the highest power of $$2^3$$ in $$60!$$ is $$\left[\dfrac{56}{3}\right]\ =\ 18$$

The highest power of $$40$$ is the minimum value among the highest power of $$5$$ and the highest power of $$2^3$$ in $$60!$$, which is $$\min(14,18)=14$$

So, the highest power $$40$$ in $$60!$$ is $$14$$.

Hence, the correct answer is option C.