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.

false

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so then what are you asking?

step-by-step explanation: