The number of zeros at the end of 70! is

70!= 2a×3b×5c×7d×... Now,     a= E2 (70!)  =  [102]+[7022]+[7024]+[7025]+[7026] ⇒   a=  35 +17+8+4+2+1 =67 And,   C = E5 (70!)= [705]+[7052]= 14+2=16 ∴   70! = 267 ×516×3b×7d×... ⇒    70! =  (2×5)16×251×3b×7d×... ⇒    70! = 1016×251 ×3b×74 ×... Thus, the number of zeros at the end of 70! is 16.