Question Number 170046 by Rasheed.Sindhi last updated on 14/May/22
$$\mathrm{PROVE}\:\mathrm{THAT} \\ $$$$\:\:\:\:\mathrm{6}\:\mid\:\left(\mathrm{8}^{{x}} −\mathrm{2}^{{x}} \right) \\ $$
Answered by JDamian last updated on 14/May/22
$$\left(\mathrm{8}^{{x}} −\mathrm{2}^{{x}} \right)\:\mathrm{mod}\:\mathrm{6}\:= \\ $$$$=\mathrm{8}^{{x}} \mathrm{mod}\:\mathrm{6}\:−\:\mathrm{2}^{{x}} \mathrm{mod}\:\mathrm{6}= \\ $$$$=\left(\mathrm{8}\:\mathrm{mod}\:\mathrm{6}\right)^{{x}} \:−\:\mathrm{2}^{{x}} \mathrm{mod}\:\mathrm{6}= \\ $$$$=\mathrm{2}^{{x}} \mathrm{mod}\:\mathrm{6}\:−\:\mathrm{2}^{{x}} \mathrm{mod}\:\mathrm{6}= \\ $$$$=\:\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 15/May/22
$$\mathcal{T}{han}\mathcal{X}\:{sir}! \\ $$
Answered by mr W last updated on 15/May/22
$$\mathrm{8}^{{n}} =\left(\mathrm{2}+\mathrm{6}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{6}^{{n}−{k}} \mathrm{2}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{6}^{{n}−{k}} \mathrm{2}^{{k}} +\mathrm{2}^{{n}} \\ $$$$\mathrm{8}^{{n}} −\mathrm{2}^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{6}^{{n}−{k}} \mathrm{2}^{{k}} \\ $$$$\Rightarrow\mathrm{6}\mid\mathrm{8}^{{n}} −\mathrm{2}^{{n}} \\ $$
Commented by Rasheed.Sindhi last updated on 15/May/22
$$\mathcal{T}{han}\mathcal{X}\:{sir}! \\ $$