Find remainder when 27^(40) is divided by 12 ?


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Question Number 48675 by rahul 19 last updated on 27/Nov/18

$F i n d r e m a i n d e r w h e n 27^{40} i s d i v i d e d$ $b y 12 ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18
	$(24+3)^(40) =f(24)+3^(40) 3^(40) =3^(39) ×3 (27)^(13) ×3 (24+3)^(13) ×3 ={g(24)+3^(13) }×3 =3×g(24)+3^(14) (3^3 )^3 ×3^5 (24+3)^3 ×3^5 {h(24)+3^3 }×3^5 (3^3 )^2 ×3^2 (24+3)^2 ×3^2 {24^2 +2×24×3+3^2 }×3^2 3^4 =(24+3)×3=72+9 so remainder is 9 pls check...$
	${(24 + 3)}^{40} = f (24) + 3^{40}$ $3^{40} = 3^{39} \times 3$ ${(27)}^{13} \times 3$ ${(24 + 3)}^{13} \times 3$ $= {g (24) + 3^{13}} \times 3$ $= 3 \times g (24) + 3^{14}$ ${(3^{3})}^{3} \times 3^{5}$ ${(24 + 3)}^{3} \times 3^{5}$ ${h (24) + 3^{3}} \times 3^{5}$ ${(3^{3})}^{2} \times 3^{2}$ ${(24 + 3)}^{2} \times 3^{2}$ ${24^{2} + 2 \times 24 \times 3 + 3^{2}} \times 3^{2}$ $3^{4} = (24 + 3) \times 3 = 72 + 9$ $s o r e m a i n d e r i s 9 p l s c h e c k . . .$
	Commented by rahul 19 last updated on 27/Nov/18
	wow �� thanks sir ��
	Answered by MJS last updated on 27/Nov/18

	$n \in N^{★} : 3^{2 n - 1} \equiv 3 \mod 12; 3^{2 n} \equiv 9 \mod 12$ $27^{2 n - 1} = 3^{3 (2 n - 1)} \Rightarrow 27^{2 n - 1} \equiv 3 \mod 12$ $27^{2 n} = 3^{3 (2 n)} \Rightarrow 27^{2 n} \equiv 9 \mod 12$ $27^{40} = 27^{2 \times 20} \Rightarrow 27^{40} \equiv 9 \mod 12$
	Commented by rahul 19 last updated on 27/Nov/18

	$w h a t i s \equiv m o d . . . ?$ ${c a n}^{'} t u n d e r s t a n d a n y t h i n g . . .$
	Commented by MJS last updated on 27/Nov/18

	$a \equiv b \mod c means a has the same remainder$ $as b when divided by c .$
	Commented by rahul 19 last updated on 27/Nov/18
	thanks sir ��
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