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Senior Manager
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Exponent [#permalink] New post 13 Sep 2006, 12:02
If 2^20-n is dividable by 3, which of the following could be value of n?
I. 0
II. 1
III. 4
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 [#permalink] New post 13 Sep 2006, 13:52
Agree with 2 &3.


I. 2^20 - n = will always yield an even integer not divisible by 3. (2,4,8,16,32,64,128, etc)

II. 2^4 - 1 = 15. Divisible by 3.

III. 2^6 - 4 = 60. Divisible by 3.
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 [#permalink] New post 13 Sep 2006, 15:02
positive soul wrote:
Agree with 2 &3.


I. 2^20 - n = will always yield an even integer not divisible by 3. (2,4,8,16,32,64,128, etc)

II. 2^4 - 1 = 15. Divisible by 3.

III. 2^6 - 4 = 60. Divisible by 3.


Can you please explain the reasoning behind 2^4-1=15, and 2^6-4=60, I don't follow.

thanks :(
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 [#permalink] New post 13 Sep 2006, 15:03
positive soul wrote:
Agree with 2 &3.


I. 2^20 - n = will always yield an even integer not divisible by 3. (2,4,8,16,32,64,128, etc)

II. 2^4 - 1 = 15. Divisible by 3.

III. 2^6 - 4 = 60. Divisible by 3.


Can you please explain the reasoning behind 2^4-1=15, and 2^6-4=60, I don't follow.

thanks :(
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 [#permalink] New post 13 Sep 2006, 15:59
I think the basis of positive soul's reasoning is that 2^n follows a recurring set of 4; 2,4,8,16, 2^20 cycles through this 5 times.

2^4 = 16 16 - 1 = 15 which is divisible by 3

16 - 4 = 12 which is divisible by 3.

Another cycle of 4 ---> 2^8 = 256 - 1 = 255 which is divisble by 3

256 - 4 = 252 which is divisible by 3.

Another way to look at this is that 2^20 = 16^5
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 [#permalink] New post 13 Sep 2006, 16:24
Given
2^20 -n = 3k where k is some integer.

or 2^20 = 3k + n

Question is basically asking what is the remainder when 2^20 divided by 3.

For any power of 2, when divided by 3, the remainder is either 1 or 2.

Therefore ,
I. n=0 is out.
II. n =1 is correct.
III. n = 4
If 2^20 = 3k + 1, then 2^20 = 3(k-1) + 4

If 2^20 = 3k + 2, then 2^20 = 3(k-1) +5

n = 4 is possible.

Answer: II & III.










I
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 [#permalink] New post 13 Sep 2006, 16:27
errr, that's what I meant. I think!?!

haas_mba07 you've got skills!

In fact - care to lend them to this problem?

http://www.gmatclub.com/phpbb/viewtopic.php?t=35048
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 [#permalink] New post 14 Sep 2006, 00:22
Haas , ya method say it all.... brilliant :lol:
  [#permalink] 14 Sep 2006, 00:22
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