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

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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, 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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VP
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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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Intern
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11 May 2006, 21:44
I think its 2. But wanna know the best approach.
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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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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.

_________________

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
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12 May 2006, 01:47
You add the DIGITS not the NUMBERS. The numbers that are bases of the powers.

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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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VP
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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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Director
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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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VP
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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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Director
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14 May 2006, 15:10
I assumed squares. Thanks

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VP
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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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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

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14 May 2006, 21:24
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