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

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Did we discuss this one? 1^1+2^2+3^3+...+10^10 is divided [#permalink] New post 11 May 2006, 20:27
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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
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 [#permalink] New post 11 May 2006, 21:42
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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 [#permalink] New post 11 May 2006, 21:44
I think its 2. But wanna know the best approach.
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 [#permalink] New post 12 May 2006, 01:31
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)
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 [#permalink] New post 12 May 2006, 01:44
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.

Please correct if I'm wrong.
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 [#permalink] New post 12 May 2006, 01:47
You add the DIGITS not the NUMBERS. The numbers that are bases of the powers.
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 [#permalink] New post 12 May 2006, 02:04
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
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 [#permalink] New post 14 May 2006, 07:46
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.
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 [#permalink] New post 14 May 2006, 12:43
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
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 [#permalink] New post 14 May 2006, 15:00
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
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 [#permalink] New post 14 May 2006, 15:10
hmm...my bad. silly mistake 8-)
I assumed squares. Thanks
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 [#permalink] New post 14 May 2006, 20:01
I like Giddi's approach but got 2 as the remainder. Please clarify that adding the last digits are enough?
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 [#permalink] New post 14 May 2006, 21:24
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
  [#permalink] 14 May 2006, 21:24
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