Remainder

AIEEE 2015 Answers › Algebra questions › Remainder

(1) Find The Remainder when (32) ^{(32)^(}³²⁾
is divided by 7 (using no. Theory).

#Mathematics #Algebra

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6 Answers

62

Lokesh Verma ·2009-10-10 08:55:31

Hint:

2³â‰¡1 (mod 7)

So 2^3kâ‰¡1 (mod 7)
So 2^3k+1â‰¡2 (mod 7)
So 2^3k+2â‰¡4 (mod 7)

So you have to find the remainder of 32^32 on divison by 3

32â‰¡(4) mod 7

so 32³²â‰¡(2)¹⁶⁰â‰¡4⁸⁰â‰¡1 mod 3

The the remainder will be teh same as

32^3k+1 mod 7

Hence the remainder is 4

Sorry for the earlier mistake :(

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19

Debotosh.. ·2009-10-10 08:58:09

given no. =( 2¹⁶⁰)³²
we know that ( 32)³² = 1+3M
thus, (( 32)³²)³² = 32^1+3M = 32 .32 ^3M =2⁵ .2^15M = (2³)^5M . 32 =32(1+7)^5M =32 +32.7M =32+ 7Î»

THUS, 32+ 7Î» = 28+ 4 + 7Î» 7(4) +7Î» +4 = 7Î»' +4
...THUS,REMAINDER =4

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1

$ourav @@@ -- WILL Never give ·2009-10-10 19:50:38

sir u wrote,32â‰¡(-1) mod 7...it will be (-3) mod 7

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62

Lokesh Verma ·2009-10-10 22:55:54

yes.. my wrong..

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341

Hari Shankar ·2009-10-10 23:02:14

i guess u intended to write32â‰¡-1 mod 3

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62

Lokesh Verma ·2009-10-10 23:31:59

yup that is what i did ... but then i as always .. :( :D

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