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# Powers

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Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 539
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)

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26 Dec 2011, 11:19
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Question Stats:

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

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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

_________________

Best Regards,
E.

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

Intern
Joined: 29 Jun 2011
Posts: 5

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26 Dec 2011, 21:52
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

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26 Dec 2011, 22:53
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

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27 Dec 2011, 02:26

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

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27 Dec 2011, 03:17
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

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27 Dec 2011, 03:21
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

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27 Dec 2011, 03:33
OMG!!!!!
You are right hyena1986......my apology i also read the question wrong
Re: Powers   [#permalink] 27 Dec 2011, 03:33
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