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VP
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Did we discuss this one? 1^1+2^2+3^3+...+10^10 is divided [#permalink]
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11 May 2006, 21:27
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Did we discuss this one?
1^1+2^2+3^3+...+10^10 is divided by 5. What is the remainder?
A) 1
B) 2
C) 3
D) 4
E) 0



VP
Joined: 21 Sep 2003
Posts: 1057
Location: USA

Prof,
Is it E? Just used brute force way.. Don't know any short cut
1. Find all the last digits of 1^1, 2^2, 3^3... 10^10
2. Find remainder of sum of all the last digits when divided by 5.
1^1 > 1
2^2 > 4
3^3 > 7
4^4 > 6
5^5 > 5
6^6 > 6
7^7 > 3
8^8 > 6
9^9 > 9
10^10 > 0
Remainder of sum of last digits when div by 5 = 0
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Intern
Joined: 04 May 2006
Posts: 49

I think its 2. But wanna know the best approach.
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Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

The best approach is when you add the numbers 1,2,3....10.Their sum is (11*10)/2=55 which is divisible by 5 so the reminder is 0 or E)



Intern
Joined: 04 May 2006
Posts: 49

why should I add 0,1,2...10. We're supposed to add the unit's digit of 1^1,2^2,3^3...10^10.
That comes out to be 1+4+7+6+5+6+3+6+9+0=47.
so 47/5 gives remainder as 2.
Please correct if I'm wrong.
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If A equals success, then the formula is: A = X + Y + Z, X is work. Y is play. Z is keep your mouth shut.
Albert Einstein



Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

You add the DIGITS not the NUMBERS. The numbers that are bases of the powers.



Director
Joined: 24 Oct 2005
Posts: 659
Location: London

The units digit for each are
1^1 = 1
2^2 = 4
3^3 = 7
4^4 = 6
5^5 = 5
6^6 = 6
7^7 = 3
8^8 = 6
9^9 = 9
10^10 = 0
Add all these up and we get units digit = 7. 7/5 gives remainder = 2
Ans= 2



VP
Joined: 29 Dec 2005
Posts: 1341

giddi77 wrote: Is it E? Just used brute force way.. Don't know any short cut 1. Find all the last digits of 1^1, 2^2, 3^3... 10^10 2. Find remainder of sum of all the last digits when divided by 5. 1^1 > 1 2^2 > 4 3^3 > 7 4^4 > 6 5^5 > 5 6^6 > 6 7^7 > 3 8^8 > 6 9^9 > 9 10^10 > 0 Remainder of sum of last digits when div by 5 = 0
this is best approach but the reminder is 2 not 0.



Director
Joined: 16 Aug 2005
Posts: 938
Location: France

Answer is E (remainder is 0).
To get to it, you add all the units' digits of all the squares.
1.....1
2.....4
3.....9
4.....6
5.....5
6.....6
7.....9
8.....4
9.....1
10...0
So the total ends in 5, remainder 0



VP
Joined: 29 Dec 2005
Posts: 1341

gmatmba wrote: Answer is E (remainder is 0).
To get to it, you add all the units' digits of all the squares.
1.....1 2.....4 3.....9 4.....6 5.....5 6.....6 7.....9 8.....4 9.....1 10...0 So the total ends in 5, remainder 0
you are doing only squares. but the question is not only about sqares. its 1^1+2^2+.........10^10



Director
Joined: 16 Aug 2005
Posts: 938
Location: France

hmm...my bad. silly mistake
I assumed squares. Thanks



VP
Joined: 06 Jun 2004
Posts: 1053
Location: CA

I like Giddi's approach but got 2 as the remainder. Please clarify that adding the last digits are enough?
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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

1^1 = 1
2^2 = 4
3^3 = 27
4^4 = 256
5^1 = 5, 5^2 = 25, 5^3 = 125... > units digit always 5
6^1 = 6, 6^2 = 36, 6^3 = 216... > units digit always 6
7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5 = 16807 > 7^7 should have 3 as units digit
8^1 = 8, 8^2 = 64, 8^3 = 512, 8^4 = 4096, 8^5 = 32768 > 8^8 should have 6 as units digit
9^1 = 9, 9^2 = 81, 9^3 = 729 ... > 9^9 should have 9 as units digit
10^10 > 0 as units digit
sum of units digit = 47 > keep 7, carry 4 over.
The remainder when divided by 5 should be 2










