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Question Number 170046 by Rasheed.Sindhi last updated on 14/May/22

PROVE THAT 6 ∣ (8^x −2^x )

$$\mathrm{PROVE}\:\mathrm{THAT} \\ $$$$\:\:\:\:\mathrm{6}\:\mid\:\left(\mathrm{8}^{{x}} −\mathrm{2}^{{x}} \right) \\ $$

Answered by JDamian last updated on 14/May/22

(8^x −2^x ) mod 6 = =8^x mod 6 − 2^x mod 6= =(8 mod 6)^x − 2^x mod 6= =2^x mod 6 − 2^x mod 6= = 0

$$\left(\mathrm{8}^{{x}} −\mathrm{2}^{{x}} \right)\:\mathrm{mod}\:\mathrm{6}\:= \\ $$$$=\mathrm{8}^{{x}} \mathrm{mod}\:\mathrm{6}\:−\:\mathrm{2}^{{x}} \mathrm{mod}\:\mathrm{6}= \\ $$$$=\left(\mathrm{8}\:\mathrm{mod}\:\mathrm{6}\right)^{{x}} \:−\:\mathrm{2}^{{x}} \mathrm{mod}\:\mathrm{6}= \\ $$$$=\mathrm{2}^{{x}} \mathrm{mod}\:\mathrm{6}\:−\:\mathrm{2}^{{x}} \mathrm{mod}\:\mathrm{6}= \\ $$$$=\:\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 15/May/22

ThanX sir!

$$\mathcal{T}{han}\mathcal{X}\:{sir}! \\ $$

Answered by mr W last updated on 15/May/22

8^n =(2+6)^n =Σ_(k=0) ^n C_k ^n 6^(n−k) 2^k =Σ_(k=0) ^(n−1) C_k ^n 6^(n−k) 2^k +2^n 8^n −2^n =Σ_(k=0) ^(n−1) C_k ^n 6^(n−k) 2^k ⇒6∣8^n −2^n

$$\mathrm{8}^{{n}} =\left(\mathrm{2}+\mathrm{6}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{6}^{{n}−{k}} \mathrm{2}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{6}^{{n}−{k}} \mathrm{2}^{{k}} +\mathrm{2}^{{n}} \\ $$$$\mathrm{8}^{{n}} −\mathrm{2}^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{6}^{{n}−{k}} \mathrm{2}^{{k}} \\ $$$$\Rightarrow\mathrm{6}\mid\mathrm{8}^{{n}} −\mathrm{2}^{{n}} \\ $$

Commented by Rasheed.Sindhi last updated on 15/May/22

ThanX sir!

$$\mathcal{T}{han}\mathcal{X}\:{sir}! \\ $$

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