Question Number 182718 by mnjuly1970 last updated on 13/Dec/22
$$ \\ $$$$\:\:\:\:\:\:\mathrm{If}\:,\:\:\:\:\mathrm{7}^{\:{n}} \:\overset{\mathrm{10}} {\equiv}\:\mathrm{7}^{\:\mathrm{19}} \\ $$$$\:\:\:\:\:\:\:{then}\:\:{find}\:{the}\:\:\mathrm{1}{st}\:{digit} \\ $$$$\:\:\:\:{of}\:\:{the}\:{numer}\:\:,\:\:\:\mathrm{8}^{\:{n}+\mathrm{4}} \:.\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$
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
$$\:\:{yes}\:\:{sir}\:\: \\ $$$$\:\:\:{example}…\:\:\:\:\mathrm{1234}\overset{\mathrm{10}} {\equiv}\mathrm{4} \\ $$$$\:\:\:\: \\ $$
Commented by Rasheed.Sindhi last updated on 13/Dec/22
$${What}'{s}\:{relation}\:{between}\:{m}\:\&\:{n}\:? \\ $$$${Is}\:{m}={n}\:? \\ $$
Commented by mnjuly1970 last updated on 13/Dec/22
$$\:{yes} \\ $$
Answered by TheSupreme last updated on 13/Dec/22
$${n}=\mathrm{19}\: \\ $$$${a}_{{n}} =\left\{{a}:\mathrm{8}^{{n}} \equiv_{\mathrm{10}} {a}\right\} \\ $$$$\mathrm{8}^{\mathrm{1}} =_{\mathrm{10}} \mathrm{8} \\ $$$$\mathrm{8}^{\mathrm{2}} =_{\mathrm{10}} \mathrm{4} \\ $$$$\mathrm{8}^{\mathrm{3}} =_{\mathrm{10}} \mathrm{2} \\ $$$$\mathrm{8}^{\mathrm{4}} =_{\mathrm{10}} \mathrm{6} \\ $$$$\mathrm{8}^{\mathrm{5}} =_{\mathrm{10}} \mathrm{8} \\ $$$$\mathrm{8}^{\mathrm{5}{k}+{i}} =\left\{\mathrm{8},\mathrm{4},\mathrm{2},\mathrm{6},\mathrm{8}\right\} \\ $$$$\mathrm{8}^{\mathrm{19}+\mathrm{4}} =_{\mathrm{10}} \mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Dec/22
$$\:\:\:\:\:\:\:\:\underset{−} {\:\:\:\boldsymbol{\mathrm{MOD}}\:\mathrm{10}\:\:} \\ $$$$\mathrm{7}^{\mathrm{2}} =\mathrm{49}\Rightarrow\mathrm{7}^{\mathrm{2}} \equiv\mathrm{9} \\ $$$$\mathrm{7}^{\mathrm{2}} .\mathrm{7}\equiv\mathrm{9}.\mathrm{7}=\mathrm{63}\Rightarrow\begin{array}{|c|}{\mathrm{7}^{\mathrm{3}} \equiv\mathrm{3}}\\\hline\end{array}…{i} \\ $$$$\mathrm{7}^{\mathrm{3}} .\mathrm{7}\equiv\mathrm{3}.\mathrm{7}=\mathrm{21}\Rightarrow\mathrm{7}^{\mathrm{4}} \equiv\mathrm{1} \\ $$$$\left(\mathrm{7}^{\mathrm{4}} \right)^{\boldsymbol{\mathrm{k}}} \equiv\mathrm{1}^{\boldsymbol{\mathrm{k}}} \Rightarrow\begin{array}{|c|}{\mathrm{7}^{\mathrm{4}\boldsymbol{\mathrm{k}}} \equiv\mathrm{1}}\\\hline\end{array}….{ii} \\ $$$${i}×{ii}:\:\:\:\:\begin{array}{|c|}{\mathrm{7}^{\mathrm{4}\boldsymbol{\mathrm{k}}+\mathrm{3}} \equiv\mathrm{3}}\\\hline\end{array} \\ $$$$\mathrm{7}^{{n}} \equiv\mathrm{7}^{\mathrm{19}} =\mathrm{7}^{\mathrm{4}\left(\mathrm{4}\right)+\mathrm{3}} \equiv\mathrm{3}\equiv\mathrm{7}^{\mathrm{4}\boldsymbol{\mathrm{k}}+\mathrm{3}} \\ $$$${n}=\mathrm{4k}+\mathrm{3} \\ $$$$\underset{−} {\mathrm{8}^{\mathrm{n}+\mathrm{4}} =\mathrm{8}^{\mathrm{4k}+\mathrm{3}+\mathrm{4}} =\mathrm{8}^{\mathrm{4k}+\mathrm{7}} =?} \\ $$$$\mathrm{8}^{\mathrm{2}} \equiv\mathrm{4} \\ $$$$\mathrm{8}^{\mathrm{3}} \equiv\mathrm{2}….{iii} \\ $$$$\mathrm{8}^{\mathrm{4}} \equiv\mathrm{6}….{iv} \\ $$$$\mathrm{8}^{\mathrm{4}\boldsymbol{\mathrm{k}}} \equiv\mathrm{6}^{\boldsymbol{\mathrm{k}}} \equiv\mathrm{6}^{\bigstar} ….{v} \\ $$$${iii}×{iv}:\:\mathrm{8}^{\mathrm{7}} \equiv\mathrm{2}…{vi} \\ $$$${v}×{vi}:\:\mathrm{8}^{\mathrm{4}\boldsymbol{\mathrm{k}}+\mathrm{7}} \equiv\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{\:\mathrm{8}^{\boldsymbol{\mathrm{n}}+\mathrm{4}} \equiv\mathrm{2}}\\\hline\end{array}\: \\ $$$$ \\ $$$$\begin{array}{|c|}{\:^{\bigstar} \mathrm{6}^{\boldsymbol{\mathrm{k}}} \equiv\mathrm{6}\:\left({mod}\:\mathrm{10}\right)}\\\hline\end{array} \\ $$
Commented by mnjuly1970 last updated on 14/Dec/22
$${thx}\:{alot}\:{sir}\:{Rasheed} \\ $$