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What is the remainder of 2 ^51 divided by 7

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What is the remainder of 2 ^51 divided by 7 [#permalink] New post 01 Jul 2006, 09:26
What is the remainder of 2 ^51 divided by 7 ?
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 [#permalink] New post 01 Jul 2006, 10:47
1 it is...


2^1's remainder = 2
2^2 --> 4
2^3 --> 1
2^4 --> 2
2^5 --> 4
2^6 --> 1

2,4,1 / 2,4,1 / ......
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Re: Remainder Theorm [#permalink] New post 01 Jul 2006, 19:34
briozeal wrote:
What is the remainder of 2^51 divided by 7 ?

agree with 1.

2^1 / 7 has reminder 2
2^2 / 7 has reminder 4
2^3 / 7 has reminder 1

2^4 / 7 has reminder 2
2^5 / 7 has reminder 4
2^6 / 7 has reminder 1

2^7 / 7 has reminder 2
2^8 / 7 has reminder 4
2^9 / 7 has reminder 1

this pattren seems trepeting after every 3rd term. so 2^51 has 1 reminder when it is divided by 7.
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 [#permalink] New post 02 Jul 2006, 15:07
OA is 1.
Thanks for answer guys.
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 [#permalink] New post 06 Jul 2006, 02:55
2^3 has 8 in the unit digit.
Hence 2^51 (51 is divisble by 3) as unit digit 8

Divided by 7 will give remainder 1
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 [#permalink] New post 06 Jul 2006, 03:08
2^3 = 7(1)+1

so (2^3)^17= (7(1)+1)^17= (many terms that are multiples of 7) +1

so 2^51 is a multiple of 7 +1 === remainder when 2^51 is divided by 7 is 1
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 [#permalink] New post 06 Jul 2006, 06:37
kevincan wrote:
2^3 = 7(1)+1

so (2^3)^17= (7(1)+1)^17= (many terms that are multiples of 7) +1

so 2^51 is a multiple of 7 +1 === remainder when 2^51 is divided by 7 is 1


Nice method kevin. :good
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 [#permalink] New post 06 Jul 2006, 19:02
Good alternate explanation.
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  [#permalink] 06 Jul 2006, 19:02
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