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# What is the remainder of 2 to the power of k after divided

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Director
Joined: 20 Apr 2005
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What is the remainder of 2 to the power of k after divided [#permalink]

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28 May 2005, 13:11
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148. What is the remainder of 2 to the power of k after divided by 10?

(1) k is divisible by 10

(2) k is divisible by 4

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Director
Joined: 18 Apr 2005
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28 May 2005, 13:42
E

1) insufficient k=0 r=9, k=10 r =4, k=20 r=6
2)insufficient k=0 r=9, and then r=6 for all others k's divisible by 4

1)+2) k =0, 20, 40, 60, ... so reminder is either 9 or 6 => insufficient

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Current Student
Joined: 28 Dec 2004
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28 May 2005, 16:21

my thinking is

I) is sufficient if we only allow positive numbers...greater than 0! but its insuff if zero is allowed...
II) should be sufficinet?

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Director
Joined: 18 Apr 2005
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28 May 2005, 17:13
>I) is sufficient if we only allow positive numbers...greater than 0! but its insuff if zero is allowed...

and zero is allowed => insufficient

>II) should be sufficinet?

nope, reminders follow a repetetive pattern of 4 starting with k=4 and on, not a pattern of 10

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Manager
Joined: 28 Aug 2004
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28 May 2005, 21:18
I think it's D.

case of zero is out since it is not divisible by 10 (or any other number) - though theoritically, it could be.

unless of course there is a specific example from OG which says otherwise.

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Intern
Joined: 30 Jan 2005
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28 May 2005, 21:23
(E)

The power of 2 will be only a multiple of 2 and they do not follow any regular pattern.
say
1, 2, 4 ,8, 16, 32, 64, 126, 256, 512, 1024

Now you can see by divind by 10 you will have diffenet figures all the time in remainder so none is sufficent.
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Senior Manager
Joined: 21 Mar 2004
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29 May 2005, 08:54
Dan wrote:
I think it's D.

case of zero is out since it is not divisible by 10 (or any other number) - though theoritically, it could be.

unless of course there is a specific example from OG which says otherwise.

E it is.

0 is divisible any all real numbers.

0/x = 0
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ash
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I'm crossing the bridge.........

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Senior Manager
Joined: 21 Mar 2004
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29 May 2005, 08:59
To explain my answers , i forgot.

First remember that 2^4 always end in 6.

A. k =0,10,20,30

remainders = 0,4,6,4

B. k=0,4,8,12

remainders = 0,6,6,6,6,6

C. k=0,20,40,60

remainder = 0,6,6

Ans is E
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ash
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I'm crossing the bridge.........

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29 May 2005, 08:59
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