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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 digit $0$, then it should be divisible by $2$ and $5$.
But prime factorisation of $6^n$ is $(2 \times 3)^n$.
$\therefore$ Prime factorisation of $6^n$ does not contain prime number $5$.
Hence, $6^n$ can't end with the digit $0$.
But prime factorisation of $6^n$ is $(2 \times 3)^n$.
$\therefore$ Prime factorisation of $6^n$ does not contain prime number $5$.
Hence, $6^n$ can't end with the digit $0$.