show that 12^{n }cannot end with digit 0 or 5 for any natural number n

Here, Number=12^{n }where n stand for any natural number .

Now 12^{n}= (2^{2} x3)^{n}

Now , For 12^{n} to end with 0, it should have 2 as well as 5 in its Prime factors to end with 0, Also to end with 5 , it requires at least a single multiple of 5 in its Prime Factors, So 12^{n} cannot end with the digit 0 or 5.

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