Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 18:15

### 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 reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is

Author Message
VP
Joined: 18 Nov 2004
Posts: 1436
Followers: 2

Kudos [?]: 39 [0], given: 0

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

### Show Tags

18 Mar 2005, 13:57
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 reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6

Last edited by banerjeea_98 on 18 Mar 2005, 14:36, edited 1 time in total.
Manager
Joined: 11 Jan 2005
Posts: 57
Location: Mexico City
Followers: 1

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

### Show Tags

18 Mar 2005, 14:29
i get 2

It took me a minute to get to 950 as the sum but after that it is easy.

Add the digits of 950 so 9+5+0=14 divide by = 2 with remainder 2.

Remainders when the sum of the digits is divided by 3 and 6 are the same as the actual number. so 452: 4+5+2= 11 divide 11 by 3 and get 3 with 2 remainder.

divide 452 get 150 with 2 remainder
Intern
Joined: 23 Jan 2005
Posts: 31
Location: Ft. Worth, TX
Followers: 0

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

### Show Tags

18 Mar 2005, 14:36
You don't need to add the numbers individually.

if you list them out individually:

91 92 93 94 95 96 97 98 99

you can see the average of these numbers is 95. since there are 10 numbers total, 95*10 = 950.

950 / 6 = 158, remainder 2.
_________________

thanks,
hornsfan2005

VP
Joined: 18 Nov 2004
Posts: 1436
Followers: 2

Kudos [?]: 39 [0], given: 0

### Show Tags

18 Mar 2005, 14:37
guys try it again, I have updated the ques, there was an error before.
Senior Manager
Joined: 15 Mar 2005
Posts: 418
Location: Phoenix
Followers: 2

Kudos [?]: 27 [0], given: 0

### Show Tags

18 Mar 2005, 15:25
banerjeea_98 wrote:
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6

Let A = 9^9 + 9^8 + ... + 9^1

9A = 9^10 + (9^9 + 9^8 + ... + 9^2 + 9^1) - 9^1
9A = 9^10 + A - 9
8A = 9^10 - 9
A = (9^10 - 9)/8
= 9( 9^9 -1)/8

Since 9^9 = (9^3)^3

A = 9 ((9^3)^3 - 1)/8
= 9 (729^3 -1)/8

Since a^3 - b^3 = (a-b)^3 + 3ab(a-b)

A = 9/8[(729-1)^3 + 3*729(729-1)]
= 9/8[728^3 + 3*729*728]
= 9[91*728^2 + 3*729*91]

Dividing this by 6, we have this remainder
9[ 1*2*2 + 3*3*1 ] = 9*13 = 117 which leaves 3 as the remainder.

Thus I'd go with 3 as the answer.

Hope that helps.
_________________

Who says elephants can't dance?

VP
Joined: 25 Nov 2004
Posts: 1486
Followers: 7

Kudos [?]: 104 [0], given: 0

### Show Tags

18 Mar 2005, 22:41
banerjeea_98 wrote:
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6

=9^1 + 9^2 + 9^3 + ...... + 9^9
sum of 1st term and 2nd term is even and when divided by 6 leaves 0 as reminder. similarly, sum of 3rd and 4th also result in even number which is divided by 6 and leaves 0 as reminders. so on for 5th and 6th, and 7th and 8th term. the only remaining term is 9th. the reminder is 3 no mater the value of 9^n.
Re: PS   [#permalink] 18 Mar 2005, 22:41
Display posts from previous: Sort by