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# Did we discuss this one? 1^1+2^2+3^3+...+10^10 is divided

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VP
Joined: 29 Dec 2005
Posts: 1341
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

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11 May 2006, 22:42
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

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11 May 2006, 22:44
I think its 2. But wanna know the best approach.
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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.
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Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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12 May 2006, 02:31
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

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12 May 2006, 02:44
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.

_________________

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

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12 May 2006, 02:47
You add the DIGITS not the NUMBERS. The numbers that are bases of the powers.
Director
Joined: 24 Oct 2005
Posts: 659
Location: London

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12 May 2006, 03:04
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

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14 May 2006, 08:46
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

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14 May 2006, 13:43
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

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14 May 2006, 16:00
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

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14 May 2006, 16:10
I assumed squares. Thanks
VP
Joined: 06 Jun 2004
Posts: 1053
Location: CA

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14 May 2006, 21:01
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

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14 May 2006, 22:24
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
14 May 2006, 22:24
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