The number of zero at the end of 126! is

Answers (1)

Solution:       E_2 = The exponent of 2 in (126!)

                       E_2=[frac1262]+[frac1262^2]+[frac1262^3]+[frac1262^4]+[frac1262^5]+[frac1262^6]

                     E_2=63+31+15+7+3+1=120.

                   E_5 =The exponent of 5 in (126!)

                  E_5=[frac1265]+[frac1265^2]+[frac1265^3]

                 E_5=25+5+1=31.

             	herefore       E_10 = The exponent of 10 in (126!)=Min120,31=31.

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