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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
Posts: 586
Followers: 2

Kudos [?]: 61 [0], given: 0

What is the remainder of 2 to the power of k after divided [#permalink]  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
Director
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [0], given: 0

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
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

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?
Director
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [0], given: 0

>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
Manager
Joined: 28 Aug 2004
Posts: 205
Followers: 1

Kudos [?]: 1 [0], given: 0

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.
Intern
Joined: 30 Jan 2005
Posts: 37
Location: INDIA
Followers: 0

Kudos [?]: 5 [0], given: 0

(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
Posts: 444
Location: Cary,NC
Followers: 2

Kudos [?]: 20 [0], given: 0

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
_________________

ash
________________________
I'm crossing the bridge.........

Senior Manager
Joined: 21 Mar 2004
Posts: 444
Location: Cary,NC
Followers: 2

Kudos [?]: 20 [0], given: 0

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
_________________

ash
________________________
I'm crossing the bridge.........

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