2009-3-2016n-2013-2010-2-2016n-2013-x-mod-11-where-n-is-any-integer-0-

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 \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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