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2^20 - n is divisible by 3, what can n be? A. 0 B. 1 C. 2 D.

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2^20 - n is divisible by 3, what can n be? A. 0 B. 1 C. 2 D. [#permalink]

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13 Apr 2006, 10:02
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2^20 - n is divisible by 3, what can n be?

A. 0
B. 1
C. 2
D. 3
E. None of these
SVP
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13 Apr 2006, 10:14
It's B)

(2^n)/3 => remainder 1 for n - even & remainder 2 for n - odd
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13 Apr 2006, 10:14
2^20 - n is divisible by 3, what can n be?

Let's look at the pattern.

2
4 - 1 is divisible by 3
8 - 2 is divisible by 3
6 - 1 is divisible by 3
2 - 2 is divisible by 3
4 - 1 is divisible by 3

... and so on.

Since this pattern continues, we can extrapolate this to find what needs to be subtracted from 2^20 to have it divisible by 3. 2^20 ends in a 6. Therefore, we must subtract 1 to have it divisible by 3.

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13 Apr 2006, 14:07
agree with B. The same logic as goodchild.
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13 Apr 2006, 16:22
Why is n=1?

I understand the part where 2^20's last digit is 6. Why can't n=0?
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13 Apr 2006, 18:13
Its 1..

Cannot be zero.. It means that 2^20 is divisible by 3... Which is never possible cos, no power of 2 is divisible by 3. Similarly it cannot be 3 either! and neither can it be E!
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13 Apr 2006, 20:23
vivek123 wrote:
It's B)

(2^n)/3 => remainder 1 for n - even & remainder 2 for n - odd

I got B though vivek's way of solving is superior...
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14 Apr 2006, 06:18
vivek123 wrote:
It's B)

(2^n)/3 => remainder 1 for n - even & remainder 2 for n - odd

beautiful job, buddy!!
14 Apr 2006, 06:18
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