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Find the remainder of 2^256 divided by 17 a)1 b)7 c)11 d)13

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Find the remainder of 2^256 divided by 17 a)1 b)7 c)11 d)13 [#permalink] New post 23 Jan 2004, 07:54
Find the remainder of 2^256 divided by 17

a)1
b)7
c)11
d)13
e)15
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 [#permalink] New post 23 Jan 2004, 08:20
D) 13
units digit is 4 and only possible remainder is 13
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 [#permalink] New post 23 Jan 2004, 10:39
Just a word of advice: know how to do these kind of remainder problems as they will certainly show up in the actual test.
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Why is the unit digit 4? [#permalink] New post 23 Jan 2004, 12:44
2, 4, 8, 16, 32, 64..
------------

Freq 4, so divide 256 by 4..
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 [#permalink] New post 23 Jan 2004, 12:58
And the answer is ... :musband A) You are correct Giddi77!
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 [#permalink] New post 23 Jan 2004, 13:14
Hey, Paul--

Why are you peeing in my pool?
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 [#permalink] New post 23 Jan 2004, 13:22
Hmm, this feels like a double edged sword question. I am the ghost of Paul coming back to haunt you. Now tell us the answer please!
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 [#permalink] New post 23 Jan 2004, 13:35
The answer is indeed A.
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 [#permalink] New post 23 Jan 2004, 13:38
Sorry I stole your thunder Stoolfi. But it's really fun some time. You should try it too :wink:
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 [#permalink] New post 23 Jan 2004, 13:42
i didnt understand..how answer is A.. i am getting 13 :cry:
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 [#permalink] New post 23 Jan 2004, 14:02
For any "a",

(a^n-1) is divisible by (a+1), provided "n" is even

For eg (a^2-1) = (a+1) (a-1) and hence divisible by (a+1)

Similarly you can see (a^4-1), (a^6-1) ....... are all divisible by (a+1)

In this problem a = 16 and n= 64.

So 16^64-1 is divisible by (16+1 = 17). Hence You will have 1 as the remainder when you divide 16^64 by 17.
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 [#permalink] New post 23 Jan 2004, 14:03
shubhangi--

You ought to ask Paul, since he's a smarty-pants who seems to know everything.
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 [#permalink] New post 23 Jan 2004, 14:10
Shoot! Giddi stole my thunder.
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 [#permalink] New post 23 Jan 2004, 14:16
Paul wrote:
Shoot! Giddi stole my thunder.


Sorry Paul. But you were a litte late :lol:
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 [#permalink] New post 23 Jan 2004, 19:40
A? I did this way, the remainder when

2^1/17=2, 2^2/17=4, 2^3/17=8, 2^4/17=16, 2^5/17=15, 2^6/17=13, 2^7/17=9, 2^8/17=1, 2^9/17=2 so the remainder repeats after 2^8 so 256 is exactly divisible by 8 so remainder is 1. How is that
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 [#permalink] New post 24 Jan 2004, 07:23
yeah! Thats gut :P rakesh.
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 [#permalink] New post 24 Jan 2004, 15:37
:lol: :lol:
  [#permalink] 24 Jan 2004, 15:37
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