Powers : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 09:31

# STARTING SOON:

Open Admission Chat with MBA Experts of Personal MBA Coach - 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

# Powers

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 536
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 74

Kudos [?]: 2956 [0], given: 217

### Show Tags

26 Dec 2011, 10:19
00:00

Difficulty:

(N/A)

Question Stats:

100% (03:25) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

1^1+2^2+3^3+...+10^10 is divided by 5. What is the remainder?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Any idea how to solve this and the answer please?
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Intern
Joined: 29 Jun 2011
Posts: 5
Followers: 0

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

### Show Tags

26 Dec 2011, 20:52
To find the remiander we need to know the number at the units digit of the series
The units digit for individual squares the upper series would be

1+4+9+6+5+6+9+4+1+0
now if we divide the summation of these digits by 5 we will get the remiander

Remainder = (1+4+9+6+5+6+9+4+1+0)/5=0

Manager
Joined: 26 Apr 2011
Posts: 228
Followers: 2

Kudos [?]: 2 [0], given: 14

### Show Tags

26 Dec 2011, 21:53
The sum of the squares of the first n natural numbers is:
S = {n(n+1)(2n+1)}/6
=(10*11*21)/6 = 2310/6=385

S/5=385/5=77
remainder is 0
Intern
Joined: 29 Jun 2011
Posts: 5
Followers: 0

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

### Show Tags

27 Dec 2011, 01:26

my bad!! i read the question wrong.

The question is:
1^1+2^2+3^3+...+10^10

We can write the integers having power greater than 4 in the form of
5 ^(4n+1) +6^(4n+2)+7^(4n+3)+….+10(4n+2)

Any number whose power will be greater than 4 the units digit will start repeating after 4
E.g.
2^1=2, 2^2=4, 2^3=8 , 2^4=16 (just the units digit)
2^5=32 2^6=64 ,2^7=128
Similarly, we find that units digit for
1=>1
2=>4
3=>7
4=>6
5=>5^(4n+1)=>5^1 (units digit)=>5
6=>6^(4n+1)=>6^2 (units digit)=>6
7=>3
8=>6
9=>9
10=>0
Therefore,
Remainder=(1+4+7+6+5+6+3+6+9+0)/5=2

Manager
Joined: 26 Apr 2011
Posts: 228
Followers: 2

Kudos [?]: 2 [0], given: 14

### Show Tags

27 Dec 2011, 02:17
I think that we can solve this question very easily by using formula to sum squares on n natural no's
http://www.trans4mind.com/personal_deve ... quares.htm
Intern
Joined: 29 Jun 2011
Posts: 5
Followers: 0

Kudos [?]: 4 [1] , given: 0

### Show Tags

27 Dec 2011, 02:21
1
KUDOS
Sandeep-Even i ahd assumed that the question is about sum of squares of natural numbers but if we look at it carefully the question is about summation of n^n form not summation n^2.
Hope this helps.
Manager
Joined: 26 Apr 2011
Posts: 228
Followers: 2

Kudos [?]: 2 [0], given: 14

### Show Tags

27 Dec 2011, 02:33
OMG!!!!!
You are right hyena1986......my apology i also read the question wrong
Re: Powers   [#permalink] 27 Dec 2011, 02:33
Similar topics Replies Last post
Similar
Topics:
2 Factorial Powers 5 26 Dec 2011, 10:29
3 Powers and Roots 11 17 Dec 2011, 11:21
1 Power Trouble 7 18 Sep 2011, 05:13
7 Find the power of 80 in 40!??? 12 18 Oct 2010, 17:17
Powers problem 1 14 Jul 2010, 12:24
Display posts from previous: Sort by

# Powers

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.