(A)OR(B)

Show that $\displaystyle 6^{\mathrm{n}}$ can not end with digit $\displaystyle 0$ for any natural number 'n'.

If \(\displaystyle 6^{\mathrm{n}}\) ends with digit $\displaystyle 0$, it would be divisible by 5. So, prime factorization of \(\displaystyle 6^{\mathrm{n}}\) would contain 5. But \(\displaystyle 6^{\mathrm{n}}=(2 \times 3)^{\mathrm{n}}\), the only prime factorization of \(\displaystyle 6^{\mathrm{n}}\) are $\displaystyle 2$ and $\displaystyle 3$ as per fundamental theorem of Arithmetic. There is no other prime in the factorization of \(\displaystyle 6^{\mathrm{n}}\). So, there is no natural number n for which \(\displaystyle 6^{\mathrm{n}}\) ends with digit zero.

\[\begin{aligned} & 72=2^{3} \times 3^{2} \\ & 120=2^{3} \times 3 \times 5 \\ & \mathrm{HCF}=24 \\ & \mathrm{LCM}=360 \end{aligned} \]