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Did we discuss this one? 1^1+2^2+3^3+...+10^10 is divided [#permalink]
 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
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VP
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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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I think its 2. But wanna know the best approach.
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Director
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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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Intern
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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
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Director
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You add the DIGITS not the NUMBERS. The numbers that are bases of the powers.
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Director
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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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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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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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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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hmm...my bad. silly mistake 
I assumed squares. Thanks
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VP
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I like Giddi's approach but got 2 as the remainder. Please clarify that adding the last digits are enough?
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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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