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Question Number 172798 by mnjuly1970 last updated on 01/Jul/22

a^( 2) −1 ≡^(10) 14a +6 a^( 2) +a ≡^(10) ? choices: 2 3 7 8

$$ \\ $$$$\:\:\:\:\:\mathrm{a}^{\:\mathrm{2}} −\mathrm{1}\:\overset{\mathrm{10}} {\equiv}\:\mathrm{14a}\:+\mathrm{6} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}^{\:\mathrm{2}} \:+\mathrm{a}\:\overset{\mathrm{10}} {\equiv}\:?\:\:\:\: \\ $$$$\:\:\mathrm{choices}:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\mathrm{8} \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 01/Jul/22

a^( 2) −1 ≡^(10) 14a +6 a^2 −14a−7≡0(mod 10) a^2 −14a−7+10×2≡0(mod 10) a^2 −14a+13≡0(mod 10) (a−1)(a−13)≡0(mod 10) { ((a−1≡0(mod 10))),((a−13≡0(mod 10))) :} { ((a≡1(mod 10))),((a≡13(mod 10))) :} { ((a^2 +a≡1^2 +1≡2(mod 10))),((a^2 +a≡13^2 +13=182≡2(mod 10))) :}

$$\:\:\:\:\:\mathrm{a}^{\:\mathrm{2}} −\mathrm{1}\:\overset{\mathrm{10}} {\equiv}\:\mathrm{14a}\:+\mathrm{6} \\ $$$$\:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{14a}−\mathrm{7}\equiv\mathrm{0}\left({mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{14a}−\mathrm{7}+\mathrm{10}×\mathrm{2}\equiv\mathrm{0}\left({mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{14a}+\mathrm{13}\equiv\mathrm{0}\left({mod}\:\mathrm{10}\right) \\ $$$$\:\:\:\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{a}−\mathrm{13}\right)\equiv\mathrm{0}\left({mod}\:\mathrm{10}\right) \\ $$$$\begin{cases}{\mathrm{a}−\mathrm{1}\equiv\mathrm{0}\left({mod}\:\mathrm{10}\right)}\\{\mathrm{a}−\mathrm{13}\equiv\mathrm{0}\left({mod}\:\mathrm{10}\right)}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{a}\equiv\mathrm{1}\left({mod}\:\mathrm{10}\right)}\\{\mathrm{a}\equiv\mathrm{13}\left({mod}\:\mathrm{10}\right)}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{a}^{\mathrm{2}} +\mathrm{a}\equiv\mathrm{1}^{\mathrm{2}} +\mathrm{1}\equiv\mathrm{2}\left({mod}\:\mathrm{10}\right)}\\{\mathrm{a}^{\mathrm{2}} +\mathrm{a}\equiv\mathrm{13}^{\mathrm{2}} +\mathrm{13}=\mathrm{182}\equiv\mathrm{2}\left({mod}\:\mathrm{10}\right)}\end{cases} \\ $$

Commented by mnjuly1970 last updated on 01/Jul/22

thanks alot sir

$$\mathrm{thanks}\:\mathrm{alot}\:\mathrm{sir} \\ $$

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