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

Answer

$6^n$$
\Rightarrow \quad(2 \times 3)^n
$
It can be observed that 5 is not in the prime factorisation of 6 . Hence for any value of $n, 6^n$ will not be divisible by 5 .
$\therefore 6^{ n }$ cannot end with 0 for any natural no. $n$.

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