It is currently 26 Sep 2017, 07:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
SVP
Joined: 24 Sep 2005
Posts: 1884

Kudos [?]: 361 [0], given: 0

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

### Show Tags

13 Apr 2006, 10:02
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

2^20 - n is divisible by 3, what can n be?

A. 0
B. 1
C. 2
D. 3
E. None of these

Kudos [?]: 361 [0], given: 0

SVP
Joined: 14 Dec 2004
Posts: 1685

Kudos [?]: 167 [0], given: 0

### Show Tags

13 Apr 2006, 10:14
It's B)

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

Kudos [?]: 167 [0], given: 0

Manager
Joined: 09 Feb 2006
Posts: 129

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

Location: New York, NY

### Show Tags

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.

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

Manager
Joined: 13 Aug 2005
Posts: 133

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

### Show Tags

13 Apr 2006, 14:07
agree with B. The same logic as goodchild.

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

Manager
Joined: 08 Feb 2006
Posts: 124

Kudos [?]: 12 [0], given: 0

### Show Tags

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?

Kudos [?]: 12 [0], given: 0

VP
Joined: 29 Apr 2003
Posts: 1403

Kudos [?]: 29 [0], given: 0

### Show Tags

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!

Kudos [?]: 29 [0], given: 0

Senior Manager
Joined: 05 Jan 2006
Posts: 381

Kudos [?]: 92 [0], given: 0

### Show Tags

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...

Kudos [?]: 92 [0], given: 0

SVP
Joined: 24 Sep 2005
Posts: 1884

Kudos [?]: 361 [0], given: 0

### Show Tags

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!!

Kudos [?]: 361 [0], given: 0

14 Apr 2006, 06:18
Display posts from previous: Sort by