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


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

$a^{2} - 1 \overset{10}{\equiv} 14 a + 6$ $a^{2} + a \overset{10}{\equiv} ?$ $choices : 2 3 7 8$
	Answered by Rasheed.Sindhi last updated on 01/Jul/22
	$a^( 2) −1 ≡^(10) 14a +6 a^2 −14a−7≡0(mod 10) a^2 −14a−7+10×2≡0(mod 10) a^2 −14a+13≡0(mod 10) (a−1)(a−13)≡0(mod 10) { ((a−1≡0(mod 10))),((a−13≡0(mod 10))) :} { ((a≡1(mod 10))),((a≡13(mod 10))) :} { ((a^2 +a≡1^2 +1≡2(mod 10))),((a^2 +a≡13^2 +13=182≡2(mod 10))) :}$
	$a^{2} - 1 \overset{10}{\equiv} 14 a + 6$ $a^{2} - 14 a - 7 \equiv 0 (m o d 10)$ $a^{2} - 14 a - 7 + 10 \times 2 \equiv 0 (m o d 10)$ $a^{2} - 14 a + 13 \equiv 0 (m o d 10)$ $(a - 1) (a - 13) \equiv 0 (m o d 10)$ ${\begin{cases} a - 1 \equiv 0 (m o d 10) \\ a - 13 \equiv 0 (m o d 10) \end{cases}$ ${\begin{cases} a \equiv 1 (m o d 10) \\ a \equiv 13 (m o d 10) \end{cases}$ ${\begin{cases} a^{2} + a \equiv 1^{2} + 1 \equiv 2 (m o d 10) \\ a^{2} + a \equiv 13^{2} + 13 = 182 \equiv 2 (m o d 10) \end{cases}$
	Commented by mnjuly1970 last updated on 01/Jul/22

	$thanks alot sir$
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