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Director
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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, 14:11
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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: 545
Location: Canuckland

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
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Location: New York City
Schools: Wharton'11 HBS'12

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?



Director
Joined: 18 Apr 2005
Posts: 545
Location: Canuckland

>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

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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Joined: 30 Jan 2005
Posts: 37
Location: INDIA

Answer E [#permalink]
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28 May 2005, 22:23
(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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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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Senior Manager
Joined: 21 Mar 2004
Posts: 445
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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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