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Question Number 182718 by mnjuly1970 last updated on 13/Dec/22

If , 7^( n) ≡^(10) 7^( 19) then find the 1st digit of the numer , 8^( n+4) .

$$ \\ $$$$\:\:\:\:\:\:\mathrm{If}\:,\:\:\:\:\mathrm{7}^{\:{n}} \:\overset{\mathrm{10}} {\equiv}\:\mathrm{7}^{\:\mathrm{19}} \\ $$$$\:\:\:\:\:\:\:{then}\:\:{find}\:{the}\:\:\mathrm{1}{st}\:{digit} \\ $$$$\:\:\:\:{of}\:\:{the}\:{numer}\:\:,\:\:\:\mathrm{8}^{\:{n}+\mathrm{4}} \:.\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Commented by JDamian last updated on 13/Dec/22

"1st digit" is the right-most one?

Commented by mnjuly1970 last updated on 13/Dec/22

yes sir example... 1234≡^(10) 4

$$\:\:{yes}\:\:{sir}\:\: \\ $$$$\:\:\:{example}...\:\:\:\:\mathrm{1234}\overset{\mathrm{10}} {\equiv}\mathrm{4} \\ $$$$\:\:\:\: \\ $$

Commented by Rasheed.Sindhi last updated on 13/Dec/22

What′s relation between m & n ? Is m=n ?

$${What}'{s}\:{relation}\:{between}\:{m}\:\&\:{n}\:? \\ $$$${Is}\:{m}={n}\:? \\ $$

Commented by mnjuly1970 last updated on 13/Dec/22

yes

$$\:{yes} \\ $$

Answered by TheSupreme last updated on 13/Dec/22

n=19 a_n ={a:8^n ≡_(10) a} 8^1 =_(10) 8 8^2 =_(10) 4 8^3 =_(10) 2 8^4 =_(10) 6 8^5 =_(10) 8 8^(5k+i) ={8,4,2,6,8} 8^(19+4) =_(10) 2

$${n}=\mathrm{19}\: \\ $$$${a}_{{n}} =\left\{{a}:\mathrm{8}^{{n}} \equiv_{\mathrm{10}} {a}\right\} \\ $$$$\mathrm{8}^{\mathrm{1}} =_{\mathrm{10}} \mathrm{8} \\ $$$$\mathrm{8}^{\mathrm{2}} =_{\mathrm{10}} \mathrm{4} \\ $$$$\mathrm{8}^{\mathrm{3}} =_{\mathrm{10}} \mathrm{2} \\ $$$$\mathrm{8}^{\mathrm{4}} =_{\mathrm{10}} \mathrm{6} \\ $$$$\mathrm{8}^{\mathrm{5}} =_{\mathrm{10}} \mathrm{8} \\ $$$$\mathrm{8}^{\mathrm{5}{k}+{i}} =\left\{\mathrm{8},\mathrm{4},\mathrm{2},\mathrm{6},\mathrm{8}\right\} \\ $$$$\mathrm{8}^{\mathrm{19}+\mathrm{4}} =_{\mathrm{10}} \mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 14/Dec/22

MOD 10 _(−) 7^2 =49⇒7^2 ≡9 7^2 .7≡9.7=63⇒ determinant (((7^3 ≡3)))...i 7^3 .7≡3.7=21⇒7^4 ≡1 (7^4 )^k ≡1^k ⇒ determinant (((7^(4k) ≡1)))....ii i×ii: determinant (((7^(4k+3) ≡3))) 7^n ≡7^(19) =7^(4(4)+3) ≡3≡7^(4k+3) n=4k+3 8^(n+4) =8^(4k+3+4) =8^(4k+7) =?_(−) 8^2 ≡4 8^3 ≡2....iii 8^4 ≡6....iv 8^(4k) ≡6^k ≡6^★ ....v iii×iv: 8^7 ≡2...vi v×vi: 8^(4k+7) ≡2 determinant ((( 8^(n+4) ≡2))) determinant (((^★ 6^k ≡6 (mod 10))))

$$\:\:\:\:\:\:\:\:\underset{−} {\:\:\:\boldsymbol{\mathrm{MOD}}\:\mathrm{10}\:\:} \\ $$$$\mathrm{7}^{\mathrm{2}} =\mathrm{49}\Rightarrow\mathrm{7}^{\mathrm{2}} \equiv\mathrm{9} \\ $$$$\mathrm{7}^{\mathrm{2}} .\mathrm{7}\equiv\mathrm{9}.\mathrm{7}=\mathrm{63}\Rightarrow\begin{array}{|c|}{\mathrm{7}^{\mathrm{3}} \equiv\mathrm{3}}\\\hline\end{array}...{i} \\ $$$$\mathrm{7}^{\mathrm{3}} .\mathrm{7}\equiv\mathrm{3}.\mathrm{7}=\mathrm{21}\Rightarrow\mathrm{7}^{\mathrm{4}} \equiv\mathrm{1} \\ $$$$\left(\mathrm{7}^{\mathrm{4}} \right)^{\boldsymbol{\mathrm{k}}} \equiv\mathrm{1}^{\boldsymbol{\mathrm{k}}} \Rightarrow\begin{array}{|c|}{\mathrm{7}^{\mathrm{4}\boldsymbol{\mathrm{k}}} \equiv\mathrm{1}}\\\hline\end{array}....{ii} \\ $$$${i}×{ii}:\:\:\:\:\begin{array}{|c|}{\mathrm{7}^{\mathrm{4}\boldsymbol{\mathrm{k}}+\mathrm{3}} \equiv\mathrm{3}}\\\hline\end{array} \\ $$$$\mathrm{7}^{{n}} \equiv\mathrm{7}^{\mathrm{19}} =\mathrm{7}^{\mathrm{4}\left(\mathrm{4}\right)+\mathrm{3}} \equiv\mathrm{3}\equiv\mathrm{7}^{\mathrm{4}\boldsymbol{\mathrm{k}}+\mathrm{3}} \\ $$$${n}=\mathrm{4k}+\mathrm{3} \\ $$$$\underset{−} {\mathrm{8}^{\mathrm{n}+\mathrm{4}} =\mathrm{8}^{\mathrm{4k}+\mathrm{3}+\mathrm{4}} =\mathrm{8}^{\mathrm{4k}+\mathrm{7}} =?} \\ $$$$\mathrm{8}^{\mathrm{2}} \equiv\mathrm{4} \\ $$$$\mathrm{8}^{\mathrm{3}} \equiv\mathrm{2}....{iii} \\ $$$$\mathrm{8}^{\mathrm{4}} \equiv\mathrm{6}....{iv} \\ $$$$\mathrm{8}^{\mathrm{4}\boldsymbol{\mathrm{k}}} \equiv\mathrm{6}^{\boldsymbol{\mathrm{k}}} \equiv\mathrm{6}^{\bigstar} ....{v} \\ $$$${iii}×{iv}:\:\mathrm{8}^{\mathrm{7}} \equiv\mathrm{2}...{vi} \\ $$$${v}×{vi}:\:\mathrm{8}^{\mathrm{4}\boldsymbol{\mathrm{k}}+\mathrm{7}} \equiv\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{\:\mathrm{8}^{\boldsymbol{\mathrm{n}}+\mathrm{4}} \equiv\mathrm{2}}\\\hline\end{array}\: \\ $$$$ \\ $$$$\begin{array}{|c|}{\:^{\bigstar} \mathrm{6}^{\boldsymbol{\mathrm{k}}} \equiv\mathrm{6}\:\left({mod}\:\mathrm{10}\right)}\\\hline\end{array} \\ $$

Commented by mnjuly1970 last updated on 14/Dec/22

thx alot sir Rasheed

$${thx}\:{alot}\:{sir}\:{Rasheed} \\ $$

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