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

$ 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 $

$ E_5(60!) = \left[ \frac{60}{5} \right] + \left[ \frac{60}{5^2} \right] $

$ = 12 + 2 = 14 $

$ \therefore n = \min \left( \frac{56}{3}, 14 \right) = 14 $