Let n = 60! + 55! + 50! the unit digit of n and a number of digits to the left of the units digits are consecutive zeros before we come to the first non-zero digit. how many such consecutive zeros are there until the first non-zero digit?

n=8320987125437793528348710310901843271933297732006335848276356027645952000000000000

So you can see there are 12 zeros before you hit the first non-zero, which is 2.

answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

hope this ^_^

it's d

step-by-step explanation: