Check whether\(6^{n}\)can end with the digit \(0\) for any natural number \(n\).

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Check whether\(6^{n}\)can end with the digit \(0\) for any natural number \(n\).

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If the number\(6^{n}\)ends with the digits zero. Then, it is divisible by \(5\). Therefore, the prime factorization of\(6^{n}\)contains the prime \(5\). This is not possible because the only primes in the factorization of\(6^{n}\)are \(2\) and \(3\) and the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other prime-in the factorization of\(6^{n}\).

So, there is no value of \(n\) in natural numbers for which\(6^{n}\)ends with the digit zero.

So, there is no value of \(n\) in natural numbers for which\(6^{n}\)ends with the digit zero.

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