2009^3^(2016n+2013) +2010^2^(2016n+2013) ≡x mod(11) where n is any integer ≥0


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Question Number 192275 by York12 last updated on 13/May/23

$2009^{3^{2016 n + 2013}} + 2010^{2^{2016 n + 2013}} \equiv x m o d (11) w h e r e n i s a n y i n t e g e r \geq 0$
	Answered by BaliramKumar last updated on 14/May/23

	$0$
	Answered by BaliramKumar last updated on 14/May/23

	$ϕ (11) = 10 \Rightarrow ϕ (10) = 4$ $\frac{2009}{11} = 7 rem . \Rightarrow \frac{3}{10} = 3 rem . \Rightarrow \frac{2016 n + 2013}{4} = 1 rem .$ $\frac{2010}{11} = 8 rem . \Rightarrow \frac{2}{10} = 2 rem . \Rightarrow \frac{2016 n + 2013}{4} = 1 rem .$ $\frac{7^{3^{1}}}{11} + \frac{8^{2^{1}}}{11} = \frac{343}{11} + \frac{64}{11} = \frac{2}{11} + \frac{- 2}{11} = \frac{0}{11} = 0 Remainder$
	Commented by York12 last updated on 14/May/23
	$2009 ≡ 7 mod (11) → 2009^(10) ≡ 7^(10) mod (11) ∵ 11 is a prime number → we can apply fermat′s little theorm ∴ 7^(10) ≡ 1 mod (11) → 7^(10k) ≡ 1 mod (11) where k is a an integer ≥ 0 2009^3^(2016n+2013) ≡ 7^3^(2016n+2013) mod (11) 3^(2016n+2013) can be written as 3^(4(504n+503)+1) → The Unit digit would be 3 ∴ 3^(2016n+2013) ≡ 3 mod (10) → 3^(2016n+2013) can be written as 10k +3 7^(10k) ≡ 1 mod (11) , 7^3 ≡ 2 mod (11) ∴ 7^(10k+3) ≡ 2 mod (11) ∴ 2009^3^(2016n+2013) ≡ 2 mod (11) → {I} 2010 ≡ 8 mod (11) → 2010^(10) ≡ 8^(10) mod (11) By using fermat′s little theorm: we can state the following 8^(10 ) ≡ 1 mod (11) → 8^(10m) ≡ 1 mod (11) 2010^2^(2016n+2013) ≡ 8^2^(2016n+2013) mod(11) 2^(2016n+2013) can be written as 2^(4(504n+503)+1) ∴The Unit digit of 2^(2016n+2013) is 2 ∴2^(2016n+2013) ≡ 2 mod (10) → can bd written as 10m+2 8^(10m) ≡ 1 mod (11) , 8^2 ≡ 9 mod (11) ∴ 8^(10m+2) ≡ 9 mod (11) ∴ 2010^2^(2016n+2013) ≡ 9 mod (11) → {II} from I and II we conclude that: 2009^3^(2016n+2013) + 2010^2^(2016n+2013) ≡ 0 mod (11) ∴ x = 0 (That′s it) [BY YORK]$
	$2009 \equiv 7 m o d (11) \to 2009^{10} \equiv 7^{10} m o d (11)$ $∵ 11 i s a p r i m e n u m b e r \to w e c a n a p p l y {f e r m a t}^{'} s l i t t l e t h e o r m$ $∴ 7^{10} \equiv 1 m o d (11) \to 7^{10 k} \equiv 1 m o d (11)$ $w h e r e k i s a a n i n t e g e r \geq 0$ $2009^{3^{2016 n + 2013}} \equiv 7^{3^{2016 n + 2013}} m o d (11)$ $3^{2016 n + 2013} c a n b e w r i t t e n a s 3^{4 (504 n + 503) + 1} \to T h e U n i t d i g i t w o u l d b e 3$ $∴ 3^{2016 n + 2013} \equiv 3 m o d (10) \to 3^{2016 n + 2013} c a n b e w r i t t e n a s 10 k + 3$ $7^{10 k} \equiv 1 m o d (11), 7^{3} \equiv 2 m o d (11)$ $∴ 7^{10 k + 3} \equiv 2 m o d (11)$ $∴ 2009^{3^{2016 n + 2013}} \equiv 2 m o d (11) \to {I}$ $2010 \equiv 8 m o d (11) \to 2010^{10} \equiv 8^{10} m o d (11)$ $B y u s i n g {f e r m a t}^{'} s l i t t l e t h e o r m : w e c a n s t a t e t h e f o l l o w i n g$ $8^{10} \equiv 1 m o d (11) \to 8^{10 m} \equiv 1 m o d (11)$ $2010^{2^{2016 n + 2013}} \equiv 8^{2^{2016 n + 2013}} m o d (11)$ $2^{2016 n + 2013} c a n b e w r i t t e n a s 2^{4 (504 n + 503) + 1}$ $∴ T h e U n i t d i g i t o f 2^{2016 n + 2013} i s 2$ $∴ 2^{2016 n + 2013} \equiv 2 m o d (10) \to c a n b d w r i t t e n a s 10 m + 2$ $8^{10 m} \equiv 1 m o d (11), 8^{2} \equiv 9 m o d (11)$ $∴ 8^{10 m + 2} \equiv 9 m o d (11)$ $∴ 2010^{2^{2016 n + 2013}} \equiv 9 m o d (11) \to {II}$ $f r o m I a n d II w e c o n c l u d e t h a t :$ $2009^{3^{2016 n + 2013}} + 2010^{2^{2016 n + 2013}} \equiv 0 m o d (11)$ $∴ x = 0 (T {h a t}^{'} s i t)$ $[BY Y O R K]$
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