Question Number 76228 by arkanmath7@gmail.com last updated on 25/Dec/19
$${prove}\:{that}\:\mathrm{6}^{{n}} \:\equiv\:\mathrm{6}\:\left({mod}\:\mathrm{10}\right),\:{for}\:{any}\:{n}\:\in\:{Z}^{\:+} \\ $$
Commented by turbo msup by abdo last updated on 25/Dec/19
![⇔6^n −6 ≡0[10] ⇔ 10 divide 6^n −6 let x_n =6^n −6 by recurrence n=1 x_1 =6−6=0 (true) let suppose x_n divided by 10 and x_(n+1) =6^(n+1) −6 =6^n .6−6 x_n =10k ⇒x_(n+1) =(10k+6)6−6 =6×10k+30 =10(6k+3)≡0[20]](https://www.tinkutara.com/question/Q76263.png)
$$\Leftrightarrow\mathrm{6}^{{n}} −\mathrm{6}\:\equiv\mathrm{0}\left[\mathrm{10}\right]\:\Leftrightarrow\:\mathrm{10}\:{divide}\:\mathrm{6}^{{n}} −\mathrm{6} \\ $$$${let}\:{x}_{{n}} =\mathrm{6}^{{n}} −\mathrm{6}\:\:{by}\:{recurrence} \\ $$$${n}=\mathrm{1}\:\:{x}_{\mathrm{1}} =\mathrm{6}−\mathrm{6}=\mathrm{0}\:\:\left({true}\right) \\ $$$${let}\:{suppose}\:\:{x}_{{n}} \:{divided}\:{by}\:\mathrm{10}\:{and} \\ $$$${x}_{{n}+\mathrm{1}} =\mathrm{6}^{{n}+\mathrm{1}} −\mathrm{6}\:=\mathrm{6}^{{n}} .\mathrm{6}−\mathrm{6} \\ $$$${x}_{{n}} =\mathrm{10}{k}\:\Rightarrow{x}_{{n}+\mathrm{1}} =\left(\mathrm{10}{k}+\mathrm{6}\right)\mathrm{6}−\mathrm{6} \\ $$$$=\mathrm{6}×\mathrm{10}{k}+\mathrm{30}\:=\mathrm{10}\left(\mathrm{6}{k}+\mathrm{3}\right)\equiv\mathrm{0}\left[\mathrm{20}\right] \\ $$
Commented by abdomathmax last updated on 25/Dec/19
![x_(n+1) ≡0[10]](https://www.tinkutara.com/question/Q76281.png)
$${x}_{{n}+\mathrm{1}} \equiv\mathrm{0}\left[\mathrm{10}\right] \\ $$
Answered by Rio Michael last updated on 25/Dec/19
$$\mathrm{L}{ets}\:{prove}\:{by}\:{induction}. \\ $$$${prove}\:{for}\:{n}=\mathrm{1},\:\mathrm{6}^{\mathrm{1}} \:\equiv\:\mathrm{6}\left({mod}\:\mathrm{10}\right) \\ $$$${as}\:\:\mathrm{6}−\mathrm{6}\:=\:\mathrm{0}\mid\mathrm{10} \\ $$$${Asume}\:{for}\:{n}\:=\:{k}\Rightarrow\: \\ $$$$\mathrm{6}^{{k}} \:\equiv\:\mathrm{6}\left({mod}\:\mathrm{10}\right)\:\Rightarrow\:\mathrm{10}\mid\left(\mathrm{6}^{{k}} −\mathrm{6}\right) \\ $$$$\mathrm{10}\mid\left(\mathrm{6}^{{k}} −\mathrm{6}\right)\:\Rightarrow\:\exists\:{m}\in\mathbb{Z}\::\:\mathrm{6}^{{k}} −\mathrm{6}\:=\:\mathrm{10}{m} \\ $$$${prove}\:{for}\:{n}\:=\:{k}+\mathrm{1} \\ $$$$\:\mathrm{6}^{{k}+\mathrm{1}} \equiv\:\mathrm{6}^{{k}} .\mathrm{6}^{\mathrm{1}} \equiv\:\mathrm{6}\left({mod}\:\mathrm{10}\right) \\ $$$$\mathrm{6}^{{k}} \:=\:\mathrm{10}{m}\:+\:\mathrm{6} \\ $$$$\mathrm{6}^{{k}+\mathrm{1}} \:\equiv\:\left(\mathrm{10}{m}\:+\:\mathrm{6}\right)\mathrm{6}\:\:{which}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{6} \\ $$$${hence}\:{it}\:{is}\:{true}\:{for}\:{n}\:=\:{k}+\mathrm{1} \\ $$$${thus}\:\forall{n}\in\mathbb{Z}^{+} \:\mathrm{6}^{{n}} \equiv\mathrm{6}\left({mod}\:\mathrm{10}\right) \\ $$$$ \\ $$
Commented by arkanmath7@gmail.com last updated on 25/Dec/19
$${thnx} \\ $$