Can the number 8^n, n being a natural number, end with the digit 0? Give reasons.

\[8^{n}=(2 \times 2 \times 2)^{n} \text { or } 2^{3 n} \] A number ends with digit $\displaystyle 0$ if it has two prime factors $\displaystyle 2$ and $\displaystyle 5$ both. Since \(\displaystyle 8^{\mathrm{n}}\) does not have $\displaystyle 5$ as a prime factor, so it can't end with digit 0.