Question

please answer for the problem

Answer 1

The No. of consecutive zeros at the end is 3.
Since the prime factors of 10 are 5 and 2.
So above there it is
${5}^{3} \times {2}^{3}$
so It can be written as
${10}^{3}$
So the no. of zeros depend on this 10³.
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I Hope This Will Help You