Question

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

$${E_{2}(60!) = \left\lbrack \frac{60}{2} \right\rbrack + \left\lbrack \frac{60}{2^{2}} \right\rbrack + \left\lbrack \frac{60}{2^{3}} \right\rbrack + \left\lbrack \frac{60}{2^{4}} \right\rbrack + \left\lbrack \frac{60}{2^{5}} \right\rbrack }{= 30 + 15 + 7 + 3 + 1 = 56 }{E_{5}(60!) = \left\lbrack \frac{60}{5} \right\rbrack + \left\lbrack \frac{60}{5^{2}} \right\rbrack }{= 12 + 2 = 14 }{40^{n} = \left( 2^{3} \right)^{n} \times 5^{n} = \left( 2^{3} \times 5 \right)^{n} }{60! = 2^{56} \times 5^{14}\ldots = 2^{14} \cdot \left( 2^{3} \cdot 5 \right)^{14} }$$∴ Maximum value of n is 14 .