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17-202-the-end-3-number-

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Question Number 126542 by MathSh last updated on 21/Dec/20
17^(202) the end 3 number?
$$\mathrm{17}^{\mathrm{202}} \\ $$$${the}\:{end}\:\:\mathrm{3}\:{number}? \\ $$
Commented by AlagaIbile last updated on 21/Dec/20
Answered by Olaf last updated on 21/Dec/20
17^0 = 1 17^1 = 17 → 7 17^2 : 7×7 = 49 → 9 17^3 : 7×9 = 63 → 3 17^4 : 7×3 = 21 → 1 ... The end digit is : 1 for 17^(4n) , n∈N 7 for 17^(4n+1) 9 for 17^(4n+2) 3 for 17^(4n+3) 202 = 4×50+2 ⇒ end digit is 9 Let u_n = ((17^(4n+2) −9)/(10)) u_(n+1) = ((17^(4(n+1)+2) −9)/(10)) u_(n+1) = ((17^(4.) 17^(4n+2) −9)/(10)) u_(n+1) = ((17^(4.) 10.((17^(4n+2) −9)/(10))+9.17^4 −9)/(10)) ((17^(4.) 10.u_n +9(17^4 −1))/(10)) End digit of 17^4 = 1 ⇒ end digit of u_(n+1) = end digit of u_n u_2 = 28 ⇒ end digit of u_(50) = 8 But u_(50) = ((17^(202) −9)/(10)) ⇒ second end digit of 17^(202) = 8 ...to be continued...
$$\mathrm{17}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$$$\mathrm{17}^{\mathrm{1}} \:=\:\mathrm{17}\:\rightarrow\:\mathrm{7} \\ $$$$\mathrm{17}^{\mathrm{2}} \::\:\mathrm{7}×\mathrm{7}\:=\:\mathrm{49}\:\rightarrow\:\mathrm{9} \\ $$$$\mathrm{17}^{\mathrm{3}} \::\:\mathrm{7}×\mathrm{9}\:=\:\mathrm{63}\:\rightarrow\:\mathrm{3} \\ $$$$\mathrm{17}^{\mathrm{4}} \::\:\mathrm{7}×\mathrm{3}\:=\:\mathrm{21}\:\rightarrow\:\mathrm{1} \\ $$$$… \\ $$$$\mathrm{The}\:\mathrm{end}\:\mathrm{digit}\:\mathrm{is}\:: \\ $$$$\mathrm{1}\:\mathrm{for}\:\mathrm{17}^{\mathrm{4}{n}} ,\:{n}\in\mathbb{N} \\ $$$$\mathrm{7}\:\mathrm{for}\:\mathrm{17}^{\mathrm{4}{n}+\mathrm{1}} \\ $$$$\mathrm{9}\:\mathrm{for}\:\mathrm{17}^{\mathrm{4}{n}+\mathrm{2}} \\ $$$$\mathrm{3}\:\mathrm{for}\:\mathrm{17}^{\mathrm{4}{n}+\mathrm{3}} \\ $$$$\mathrm{202}\:=\:\mathrm{4}×\mathrm{50}+\mathrm{2}\:\Rightarrow\:\mathrm{end}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{9} \\ $$$$\mathrm{Let}\:{u}_{{n}} \:=\:\frac{\mathrm{17}^{\mathrm{4}{n}+\mathrm{2}} −\mathrm{9}}{\mathrm{10}} \\ $$$${u}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{17}^{\mathrm{4}\left({n}+\mathrm{1}\right)+\mathrm{2}} −\mathrm{9}}{\mathrm{10}} \\ $$$${u}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{17}^{\mathrm{4}.} \mathrm{17}^{\mathrm{4}{n}+\mathrm{2}} −\mathrm{9}}{\mathrm{10}} \\ $$$${u}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{17}^{\mathrm{4}.} \mathrm{10}.\frac{\mathrm{17}^{\mathrm{4}{n}+\mathrm{2}} −\mathrm{9}}{\mathrm{10}}+\mathrm{9}.\mathrm{17}^{\mathrm{4}} −\mathrm{9}}{\mathrm{10}} \\ $$$$\frac{\mathrm{17}^{\mathrm{4}.} \mathrm{10}.{u}_{{n}} +\mathrm{9}\left(\mathrm{17}^{\mathrm{4}} −\mathrm{1}\right)}{\mathrm{10}} \\ $$$$\mathrm{End}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{17}^{\mathrm{4}} \:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{end}\:\mathrm{digit}\:\mathrm{of}\:{u}_{{n}+\mathrm{1}} \:=\:\mathrm{end}\:\mathrm{digit}\:\mathrm{of}\:{u}_{{n}} \\ $$$${u}_{\mathrm{2}} \:=\:\mathrm{28}\:\Rightarrow\:\mathrm{end}\:\mathrm{digit}\:\mathrm{of}\:{u}_{\mathrm{50}} \:=\:\mathrm{8} \\ $$$$\mathrm{But}\:{u}_{\mathrm{50}} \:=\:\frac{\mathrm{17}^{\mathrm{202}} −\mathrm{9}}{\mathrm{10}} \\ $$$$\Rightarrow\:\mathrm{second}\:\mathrm{end}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{17}^{\mathrm{202}} \:=\:\mathrm{8} \\ $$$$…\mathrm{to}\:\mathrm{be}\:\mathrm{continued}… \\ $$
Commented by MathSh last updated on 21/Dec/20
Thanks sir, answer sir
$${Thanks}\:{sir},\:{answer}\:{sir} \\ $$

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