+ (49!×5×10)50!+(99!×100)100! = (12×10)5!+(9!×10)10! + (49!×5×10)50!+(99!×100)100! Now, (10)5! = (10)120 is a number that have 120 zeroes (10)10! is a number that have more than 120 zeroes at the end. (10)50! is a number that have more than 120 zeroes at the end. and (100)100! is a number that have more than 120 zeroes at the end. So, number of zeroes at the end of the sum = 120.