It is currently 18 Oct 2017, 08:07

# Live Now:

### 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

# Find the remainder of the division 2^87/9 a. 8 b. 1 c. -1 d.

Author Message
Senior Manager
Joined: 30 Aug 2003
Posts: 329

Kudos [?]: 25 [0], given: 0

Location: BACARDIVILLE
Find the remainder of the division 2^87/9 a. 8 b. 1 c. -1 d. [#permalink]

### Show Tags

07 Jan 2004, 14:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Find the remainder of the division 2^87/9

a. 8
b. 1
c. -1
d. None of these.

***I got answer choice (c) -1.
The official answer was (a). Whose correct? any idea
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Kudos [?]: 25 [0], given: 0

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4285

Kudos [?]: 527 [0], given: 0

### Show Tags

07 Jan 2004, 14:33
A is the answer. Look at the pattern:
2^1 = 2 2^5 = 32
2^2 = 4 2^6 = 64
2^3 = 8 2^7 = 128
2^4 = 16 2^8 = 256

Look at the pattern of the units digits. It repeats after every 4. Therefore, the exponent, 87/4 gives 21 remainder 3. A remainder of 3 means that the unit digit will be an 8 at the end. When you divide that unit's digit by 9, you will get a remainder of 8. Hope it helps
_________________

Best Regards,

Paul

Last edited by Paul on 07 Jan 2004, 14:45, edited 1 time in total.

Kudos [?]: 527 [0], given: 0

Senior Manager
Joined: 23 Sep 2003
Posts: 292

Kudos [?]: 4 [0], given: 0

Location: US

### Show Tags

07 Jan 2004, 14:38
Yep, 8 it is.

2^1 / 9 is 0 remainder 2
2^2 / 9 is 0 remainder 4
2^3 / 9 is 0 remainder 8 .....

for 2^1 to 2^6, the respective remainders are 2,4,8,7,5,1
after the sixth exponent, you find that you have yourself a cycle.

So, 87/6 is 14 remainder 3. Excluding 14 complete cycles, the third position is 8 which is the remainder.

Don't know if I've made sense but...

Kudos [?]: 4 [0], given: 0

Director
Joined: 13 Nov 2003
Posts: 958

Kudos [?]: 172 [0], given: 0

Location: Florida

### Show Tags

07 Jan 2004, 14:43
ndidi204 wrote:
Yep, 8 it is.

2^1 / 9 is 0 remainder 2
2^2 / 9 is 0 remainder 4
2^3 / 9 is 0 remainder 8 .....

for 2^1 to 2^6, the respective remainders are 2,4,8,7,5,1
after the sixth exponent, you find that you have yourself a cycle.

So, 87/6 is 14 remainder 3. Excluding 14 complete cycles, the third position is 8 which is the remainder.

Don't know if I've made sense but...

perfect...I did it up till getting the 2 to 1 cycle, but couldn't figured out what to do next.. yep 8 it is.
good work..!

hey, sunniboy, how did you get -1 as a remainder..

Kudos [?]: 172 [0], given: 0

Senior Manager
Joined: 30 Aug 2003
Posts: 329

Kudos [?]: 25 [0], given: 0

Location: BACARDIVILLE

### Show Tags

07 Jan 2004, 14:47
k, this was my reasoning:

Using the formula for remainder theorem: I concluded that

9 = 2^3 + 1

2^87 = (2^3)^29

Therefore, let 2^3 = x

So, 9 = x +1; f(-1)

(-1)^29 = -1
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Kudos [?]: 25 [0], given: 0

Director
Joined: 13 Nov 2003
Posts: 958

Kudos [?]: 172 [0], given: 0

Location: Florida

### Show Tags

07 Jan 2004, 14:51
sunniboy007 wrote:
k, this was my reasoning:

Using the formula for remainder theorem: I concluded that

9 = 2^3 + 1

2^87 = (2^3)^29

Therefore, let 2^3 = x

So, 9 = x +1; f(-1)

(-1)^29 = -1

I don't see any point for considering it as a negative function. Can very well be +ve.

Kudos [?]: 172 [0], given: 0

07 Jan 2004, 14:51
Display posts from previous: Sort by