Question

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

Answer 1

Fundamental Theorem of Arithmetic:

Any number ends with the digit $0$ which means it should be factors of both $2$ and $5$

Example:

$100=2\times 2\times 5\times 5$

$\begin{array}{rcl}& =& 2^2\times 5^2\\ & & \\ 50& =& 2\times 5\times 5\\ & =& 2^1\times 5^2\\ & & \\ 10& =& 2\times 5\\ & =& 2^1\times 5^1\end{array}$

Whereas,

$\begin{array}{rcl}{6}^{\mathrm{n}}& =& {\left(2\times 3\right)}^{\mathrm{n}}\\ & =& 2^{\mathrm{n}}\times 3^{\mathrm{n}}\end{array}$

$6^{\mathrm{n}}$ is a factor of $2$ but it's not a factor of $5$

Hence, $6^{\mathrm{n}}$ can't end with the digit $0$ for any natural number $\mathrm{n}$.