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Question Number 148732 by ethiork last updated on 30/Jul/21

show that 8^n −3^n is divisible by 5 for all natural number.

$$\:\:\:\:\:\:{show}\:{that}\: \\ $$$$\mathrm{8}^{{n}} \:−\mathrm{3}^{{n}} \:{is}\:{divisible}\:{by}\:\mathrm{5}\:\:{for}\:{all}\:{natural} \\ $$$${number}. \\ $$

Answered by Rasheed.Sindhi last updated on 30/Jul/21

5 ∣ (8^n −3^n ) ⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣ ∵8≡3(mod 5) ∴8^n ≡3^n (mod 5)

$$\mathrm{5}\:\mid\:\left(\mathrm{8}^{{n}} −\mathrm{3}^{{n}} \right) \\ $$$$\smile\smile\smile\smile\smile\smile\smile\smile\smile\smile\smile\smile \\ $$$$\because\mathrm{8}\equiv\mathrm{3}\left({mod}\:\mathrm{5}\right) \\ $$$$\therefore\mathrm{8}^{{n}} \equiv\mathrm{3}^{{n}} \left({mod}\:\mathrm{5}\right) \\ $$

Answered by mr W last updated on 30/Jul/21

8^n =(5+3)^n =3^n +Σ_(k=1) ^n C_k ^n 3^(n−k) 5^k 8^n −3^n =Σ_(k=1) ^n C_k ^n 3^(n−k) 5^k ≡0 mod 5

$$\mathrm{8}^{{n}} =\left(\mathrm{5}+\mathrm{3}\right)^{{n}} =\mathrm{3}^{{n}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{3}^{{n}−{k}} \mathrm{5}^{{k}} \\ $$$$\mathrm{8}^{{n}} −\mathrm{3}^{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{3}^{{n}−{k}} \mathrm{5}^{{k}} \equiv\mathrm{0}\:{mod}\:\mathrm{5} \\ $$

Answered by physicstutes last updated on 31/Jul/21

let f(n) = 8^n −3^n f(1)= 8^1 −3^1 =5=5(1) ⇒ true for f(1) assume f(k) is true ⇒ 8^k −3^k =5n, n∈N now f(k+1)= 8^(k+1) −3^(k+1) =8(8^k )−3(3^k ) f(k+1)= 8(5n+3^k )−3(3^k ) = 5(8n)+5(3^k ) = 5(8n+3^k ) since k ∈N, 3^k ∈N and so( 8n + 3^k ) ∈ N ⇒ f(k+1)=5p p∈N ⇒ f(k+1) is true and so for all natural numbers the expression f(n) is divisible by 5.

$$\mathrm{let}\:{f}\left({n}\right)\:=\:\mathrm{8}^{{n}} −\mathrm{3}^{{n}} \\ $$$${f}\left(\mathrm{1}\right)=\:\mathrm{8}^{\mathrm{1}} −\mathrm{3}^{\mathrm{1}} =\mathrm{5}=\mathrm{5}\left(\mathrm{1}\right)\:\Rightarrow\:\mathrm{true}\:\mathrm{for}\:{f}\left(\mathrm{1}\right) \\ $$$$\mathrm{assume}\:{f}\left({k}\right)\:\mathrm{is}\:\mathrm{true}\:\Rightarrow\:\mathrm{8}^{{k}} −\mathrm{3}^{{k}} =\mathrm{5}{n},\:{n}\in\mathbb{N} \\ $$$$\mathrm{now}\:{f}\left({k}+\mathrm{1}\right)=\:\mathrm{8}^{{k}+\mathrm{1}} −\mathrm{3}^{{k}+\mathrm{1}} =\mathrm{8}\left(\mathrm{8}^{{k}} \right)−\mathrm{3}\left(\mathrm{3}^{{k}} \right) \\ $$$${f}\left({k}+\mathrm{1}\right)=\:\mathrm{8}\left(\mathrm{5}{n}+\mathrm{3}^{{k}} \right)−\mathrm{3}\left(\mathrm{3}^{{k}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{5}\left(\mathrm{8}{n}\right)+\mathrm{5}\left(\mathrm{3}^{{k}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{5}\left(\mathrm{8}{n}+\mathrm{3}^{{k}} \right)\:\mathrm{since}\:{k}\:\in\mathbb{N},\:\mathrm{3}^{{k}} \in\mathbb{N} \\ $$$$\mathrm{and}\:\mathrm{so}\left(\:\mathrm{8}{n}\:+\:\mathrm{3}^{{k}} \right)\:\in\:\mathbb{N}\:\Rightarrow\:{f}\left({k}+\mathrm{1}\right)=\mathrm{5}{p} \\ $$$${p}\in\mathbb{N}\:\Rightarrow\:{f}\left({k}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{true}\:\mathrm{and}\:\mathrm{so}\:\mathrm{for}\:\mathrm{all} \\ $$$$\mathrm{natural}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{expression}\:{f}\left({n}\right) \\ $$$$\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5}. \\ $$

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