Answer:

Here, Number 12

n

where nϵN

Now 12

n

=(2

2

×3)

n

Now, for 12

n

to end with 0, it should have 2 as well as 5 in its prime factors. Also to end with 5, of require at least a single multiple of 5 in its prime factors, so 12

n

cannot end with the digit 0 or 5.

Explanation:

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Answer:

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