Question

Show that ${12}^n$ cannot end with the digit $0$ or $5$ for any natural number n.

Answer 1

Here, Number ${12}^n$ where $nϵN$

${12}^n=(2^2\times 3{)}^n$

The factors of ${12}^n$ are $2$ and $3$.

Now, for ${12}^n$ to end with $0$, it should have $2$ as well as $5$ in its prime factors.

Also to end with $5$, it (i.e ${12}^n$) requires at least a single multiple of 5 in its prime factors, and there is not a single $5$ in the factors of ${12}^n$

So, ${12}^n$ cannot end with the digit $0$ or $5$.