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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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SVP
Joined: 24 Sep 2005
Posts: 1893
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Kudos [?]: 139 [0], given: 0

2^20 - n is divisible by 3, what can n be? A. 0 B. 1 C. 2 D. [#permalink]  13 Apr 2006, 09:02
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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
Joined: 14 Dec 2004
Posts: 1707
Followers: 1

Kudos [?]: 50 [0], given: 0

It's B)

(2^n)/3 => remainder 1 for n - even & remainder 2 for n - odd
Manager
Joined: 09 Feb 2006
Posts: 129
Location: New York, NY
Followers: 1

Kudos [?]: 3 [0], given: 0

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.

Manager
Joined: 13 Aug 2005
Posts: 136
Followers: 1

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

agree with B. The same logic as goodchild.
Manager
Joined: 08 Feb 2006
Posts: 128
Followers: 1

Kudos [?]: 6 [0], given: 0

Why is n=1?

I understand the part where 2^20's last digit is 6. Why can't n=0?
VP
Joined: 29 Apr 2003
Posts: 1405
Followers: 2

Kudos [?]: 18 [0], given: 0

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!
Senior Manager
Joined: 05 Jan 2006
Posts: 383
Followers: 1

Kudos [?]: 30 [0], given: 0

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...
SVP
Joined: 24 Sep 2005
Posts: 1893
Followers: 11

Kudos [?]: 139 [0], given: 0

vivek123 wrote:
It's B)

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

beautiful job, buddy!!
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