Check whether 6" can end with the digit 0 for any natural number n.

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Let if possible ( 6^{n} ) ends with digit ( 0 . ) ( Rightarrow quad 6^{n}=10 times q Rightarrow 2^{n} times 3^{n}=2 times 5 times q ) ( Rightarrow 5 ) is a prime factor of ( 2^{n} times 3^{n} ) which is not possible because ( 2^{n} times 3^{n} ) can have only 2 and 3 prime factors. Hence, not possible.

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