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Algebra

Cool goIITian

Joined:	22 May 2009
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20 Aug 2009 16:48:03 IST

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547

binomial

None

what is the remainder when 2^2009 is divided by 17??

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kabi

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Joined: 11 Jan 2009

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20 Aug 2009 21:07:13 IST

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factor of 2009 = 7² .41

2 ²⁰⁰⁹=2 ^7.7.41 = (128)²⁸⁷=(126+2)²⁸⁷ = ²⁸⁷C₀(126) +²⁸⁷C₁(126)¹2+²⁸⁷C₂*4* (126)²...............................+2

126=2*9*7

so we can take 7 common from first 287 numbers .

it can be wriiten as 7 m +2

where m is integer so reminder is 2 .

If u know number theory u can solve it more easily .

Sagar Saxena

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Joined: 8 Oct 2008

Posts: 7221

20 Aug 2009 21:28:04 IST

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hello dear

2²⁰⁰⁹ = 2 . 2²⁰⁰⁸

= 2. [ 1 6 ] ⁵⁰²

= 2 . [17 -1]⁵⁰²

expand the above expression binomially.hence divide it by 17 hence,remainder is 2

®µD®A

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21 Aug 2009 00:03:58 IST

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$2^{16}\equiv1\mod17\Rightarrow 2^{2000}\equiv1\mod17\Rightarrow 2^{2009}\equiv512\equiv2\mod17$

Deepak Aggarwal

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21 Aug 2009 14:21:46 IST

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2²⁰⁰⁹ = 2 . 2²⁰⁰⁸ = 2 (16)⁵⁰² = 2 ( 17 - 1)⁵⁰² = 2 ( 1 - 17 )⁵⁰² = 2 ( C₀ - C₁(17) + C₂ (17)² - .......... )

= 2 (C₀) + 17 (k) where k = 2 ( - C₁ + C₂ (17) - .......... ) = 2 + 17k

So, when it is divided by 17, the remainder is 2.

jagdish singh

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Joined: 19 Jan 2008

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25 Aug 2009 11:47:23 IST

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ans: (2)⁴ =(-1)mod(17).

so (16)⁵⁰² =(-1)502mod(17)..........(1) here (2)⁴=16

(2)¹=(1)mod(17)....................(2)

multiply (1) and (2)

(2)²⁰⁰⁹ =(2)mod(17)

means remainder =2

using the formula of a = (b)mod(c) means when we divide a by c we get a remainder =b.

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