What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 05:18

# STARTING SOON:

Live Chat with Admission Manager and Current Student of NUS SIngapore - Join Chat Room to Participate.

### 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

# What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is

Author Message
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 67

Kudos [?]: 862 [0], given: 781

What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is [#permalink]

### Show Tags

07 Oct 2003, 16:03
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.

What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is divided by 6?

1) 3
2) 2
3) 0
4) 5

Last edited by Praetorian on 07 Oct 2003, 18:40, edited 1 time in total.
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
Followers: 3

Kudos [?]: 73 [0], given: 0

### Show Tags

07 Oct 2003, 18:07
From my lengthy calculation, I get 4 as the remainder. But it's not in any of the answers
Intern
Joined: 12 Sep 2003
Posts: 31
Location: Peru
Followers: 0

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

Re: PS : Remainder # 2 [#permalink]

### Show Tags

07 Oct 2003, 21:39
Even multiples of 9 divided by 6 leave 0 as their remainder
Odd multiples of 9 divided by 6 leave 3 as their remainder

There are eight terms in the addition, each one odd, but as the number of terms is even, the result is even, so 0 should be the remainder.
SVP
Joined: 03 Feb 2003
Posts: 1603
Followers: 8

Kudos [?]: 245 [0], given: 0

### Show Tags

07 Oct 2003, 22:23
9^N=6z+R (0<=R<=5) both parts should be divisible by 3, so R=3

9^1 mod 6 leaves 3
9^2 mod 6 leaves 3
9^3 mod 6 leaves 3
...... and so on

the sum of remainders=8*3=24, divisible by 6 evenly, i.e. the final remainder is 0.
07 Oct 2003, 22:23
Display posts from previous: Sort by