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

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Director
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What is the remainder of 2 to the power of k after divided [#permalink] New post 28 May 2005, 13:11
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A
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D
E

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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
Director
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 [#permalink] New post 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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 [#permalink] New post 28 May 2005, 16:21
I am confused about this...

please show workings....

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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 [#permalink] New post 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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 [#permalink] New post 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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Answer E [#permalink] New post 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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 [#permalink] New post 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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 [#permalink] New post 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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  [#permalink] 29 May 2005, 08:59
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