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Question Number 195443 by MM42 last updated on 02/Aug/23
10^(10) +10^(10^2 ) +10^(10^3 ) +...+10^(10^(10) ) ≡^7 ?
$$\mathrm{10}^{\mathrm{10}} +\mathrm{10}^{\mathrm{10}^{\mathrm{2}} } +\mathrm{10}^{\mathrm{10}^{\mathrm{3}} } +…+\mathrm{10}^{\mathrm{10}^{\mathrm{10}} } \:\:\overset{\mathrm{7}} {\equiv}\:\:? \\ $$
Answered by BaliramKumar last updated on 02/Aug/23
5
$$\mathrm{5} \\ $$
Commented by MM42 last updated on 02/Aug/23
ok
$${ok} \\ $$
Answered by BaliramKumar last updated on 02/Aug/23
φ(7) = 6 ((10^N )/6) ≡ 4 10^(10) +10^(10^2 ) +10^(10^3 ) +...+10^(10^(10) ) ≡^7 ? 10^4 + 10^4 + ............... (10 time) ≡^7 ? 3^4 + 3^4 + ............... (10 time) ≡^7 ? 3^4 ×10 ≡^7 ? 3^4 ×3 ≡^7 ? 3^5 ≡^7 ? 9×9×3 ≡^7 ? 2×2×3 ≡^7 ? 12 ≡^7 ? 5
$$\phi\left(\mathrm{7}\right)\:=\:\mathrm{6} \\ $$$$\frac{\mathrm{10}^{\mathrm{N}} }{\mathrm{6}}\:\equiv\:\mathrm{4} \\ $$$$\mathrm{10}^{\mathrm{10}} +\mathrm{10}^{\mathrm{10}^{\mathrm{2}} } +\mathrm{10}^{\mathrm{10}^{\mathrm{3}} } +…+\mathrm{10}^{\mathrm{10}^{\mathrm{10}} } \:\:\overset{\mathrm{7}} {\equiv}? \\ $$$$\mathrm{10}^{\mathrm{4}} \:+\:\mathrm{10}^{\mathrm{4}} \:+\:……………\:\left(\mathrm{10}\:\mathrm{time}\right)\:\overset{\mathrm{7}} {\equiv}\:? \\ $$$$\mathrm{3}^{\mathrm{4}} \:+\:\mathrm{3}^{\mathrm{4}} \:+\:……………\:\left(\mathrm{10}\:\mathrm{time}\right)\:\overset{\mathrm{7}} {\equiv}\:? \\ $$$$\mathrm{3}^{\mathrm{4}} \:×\mathrm{10}\:\:\overset{\mathrm{7}} {\equiv}\:? \\ $$$$\mathrm{3}^{\mathrm{4}} \:×\mathrm{3}\:\:\overset{\mathrm{7}} {\equiv}\:? \\ $$$$\mathrm{3}^{\mathrm{5}} \:\overset{\mathrm{7}} {\equiv}\:? \\ $$$$\mathrm{9}×\mathrm{9}×\mathrm{3}\:\overset{\mathrm{7}} {\equiv}? \\ $$$$\mathrm{2}×\mathrm{2}×\mathrm{3}\:\overset{\mathrm{7}} {\equiv}? \\ $$$$\mathrm{12}\:\overset{\mathrm{7}} {\equiv}? \\ $$$$\mathrm{5} \\ $$
Commented by MM42 last updated on 03/Aug/23
your solution was beautiful another solution 10^n =3k+1 ; “k” is odd. 10≡_7 3 10^(10) +10^(10^2 ) +10^(10^3 ) +...+10^(10^(10) ) ≡_7 3^(10) +3^(10^2 ) +3^(10^3 ) +...+3^(10^(10) ) =3^(3k_1 +1) +3^(3k_2 +1) +3^(3k_3 +1) +...+3^(3k_(10) +1) ; “k_i ” is odd =3×(27)^k_1 +3×(27)^k_2 +3×(27)^k_3 +...+3×(27)^k_(10) ≡_7 −3−3−3−...−3=−30≡_7 5
$${your}\:{solution}\:{was}\:{beautiful} \\ $$$${another}\:{solution} \\ $$$$\mathrm{10}^{{n}} =\mathrm{3}{k}+\mathrm{1}\:\:;\:“{k}''\:{is}\:{odd}. \\ $$$$\mathrm{10}\underset{\mathrm{7}} {\equiv}\mathrm{3} \\ $$$$\mathrm{10}^{\mathrm{10}} +\mathrm{10}^{\mathrm{10}^{\mathrm{2}} } +\mathrm{10}^{\mathrm{10}^{\mathrm{3}} } +…+\mathrm{10}^{\mathrm{10}^{\mathrm{10}} } \\ $$$$\underset{\mathrm{7}} {\equiv}\mathrm{3}^{\mathrm{10}} +\mathrm{3}^{\mathrm{10}^{\mathrm{2}} } +\mathrm{3}^{\mathrm{10}^{\mathrm{3}} } +…+\mathrm{3}^{\mathrm{10}^{\mathrm{10}} } \\ $$$$=\mathrm{3}^{\mathrm{3}{k}_{\mathrm{1}} +\mathrm{1}} +\mathrm{3}^{\mathrm{3}{k}_{\mathrm{2}} +\mathrm{1}} +\mathrm{3}^{\mathrm{3}{k}_{\mathrm{3}} +\mathrm{1}} +…+\mathrm{3}^{\mathrm{3}{k}_{\mathrm{10}} +\mathrm{1}} \:\:\:\:;\:“{k}_{{i}} ''\:\:{is}\:\:{odd} \\ $$$$=\mathrm{3}×\left(\mathrm{27}\right)^{{k}_{\mathrm{1}} } +\mathrm{3}×\left(\mathrm{27}\right)^{{k}_{\mathrm{2}} } +\mathrm{3}×\left(\mathrm{27}\right)^{{k}_{\mathrm{3}} } +…+\mathrm{3}×\left(\mathrm{27}\right)^{{k}_{\mathrm{10}} } \\ $$$$\underset{\mathrm{7}} {\equiv}−\mathrm{3}−\mathrm{3}−\mathrm{3}−…−\mathrm{3}=−\mathrm{30}\underset{\mathrm{7}} {\equiv}\:\mathrm{5} \\ $$$$\: \\ $$

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