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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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23 Oct 2014, 00:43
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Tough and Tricky questions: Remainders. 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4
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Re: 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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23 Oct 2014, 01:57
Bunuel wrote: Tough and Tricky questions: Remainders. 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 a number is divisible by 5, if its last digit is divisible by 5 let's look into the sum of last digits of each term of the given expression 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 adding all these numbers we get 47 which gives a remainder of 2 when divided by 5. so answer must be 2. bunuel, can you please confirm the answer of this question.



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Re: 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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23 Oct 2014, 02:06
manpreetsingh86 wrote: Bunuel wrote: Tough and Tricky questions: Remainders. 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 a number is divisible by 5, if its last digit is divisible by 5 let's look into the sum of last digits of each term of the given expression 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 adding all these numbers we get 47 which gives a remainder of 2 when divided by 5. so answer must be 2. bunuel, can you please confirm the answer of this question. Yes, the OA is C. Clicked the wrong button when posting. Edited. Thank you.
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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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16 Dec 2015, 06:28
It's not necessary to count 10^10, 5^5, and 1^1 2^2 (1 + 4 is divisible by 5). Realize this thing can save time ! 3^3, 4^4, 6^6, 7^7, 8^8, 9^9 plus the last units of them together and get the answer.



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Re: 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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05 May 2016, 21:24
Bunuel wrote: manpreetsingh86 wrote: Bunuel wrote: Tough and Tricky questions: Remainders. 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 a number is divisible by 5, if its last digit is divisible by 5 let's look into the sum of last digits of each term of the given expression 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 adding all these numbers we get 47 which gives a remainder of 2 when divided by 5. so answer must be 2. bunuel, can you please confirm the answer of this question. Yes, the OA is C. Clicked the wrong button when posting. Edited. Thank you. I have also done the same way. Although i also tried out by calculating individual remainders i.e 1=1 4=4 27=2 and so on. And then I am adding these remainders and then dividing by 5. I am getting the correct ans. But is this approach correct?



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Re: 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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05 May 2016, 21:35
tallyho_88 wrote: I have also done the same way. Although i also tried out by calculating individual remainders i.e 1=1 4=4 27=2 and so on.
And then I am adding these remainders and then dividing by 5. I am getting the correct ans. But is this approach correct?
Hi, if the terms are adding as in this Q, you can add up all individual remainders as correctly done by you.. But may be calculating units digit saves time.. say 7^7.. i don't have to get in finding what is 7^7.. I know 7 gives a pattern 7,9,3,1.. so 7th will give same as 4+3rd.. and 3rd is 3 so remainder = 3..
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Re: 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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05 May 2016, 21:54
tallyho_88 wrote: I have also done the same way. Although i also tried out by calculating individual remainders i.e 1=1 4=4 27=2 and so on.
And then I am adding these remainders and then dividing by 5. I am getting the correct ans. But is this approach correct? Yes, you can always do that but division by 2, 5 or 10 is special. All you need is the last digit in these cases. Here is why it is so: http://www.veritasprep.com/blog/2015/12 ... questions/http://www.veritasprep.com/blog/2015/12 ... nspart2/Using your method, you would have spend a considerable amount of time. Using cylicity would be faster.
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Re: 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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01 Dec 2016, 05:24
Bunuel wrote: Tough and Tricky questions: Remainders. 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Different approach with application of modular arithmetic. \(2^2 = 4 = 1 (mod_5)\) \(3^3 = 2 (mod_5)\) \(6 = 1 (mod_5)\) \(7 = 2 (mod_5)\) \(8 = 3 (mod_5)\) \(9 = 1 (mod_5)\) \(5, 10 = 0 (mod_5)\) \(\frac{1  1 + 2 + (1)^4 + 0 + 1^6 + 2^7 + 3^8 + (1)^9 + 0}{5}\) \(3^2 = (1) (mod_5)\) \(\frac{1  1 + 2 + 1 + 1 + 1 + (2^2)^3*2 + (3^2)^4 1}{5}\) \(= \frac{4 – 2 + 1 – 1}{5} = \frac{2}{5}\) Remainder 2



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Re: 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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01 Dec 2016, 09:26
vitaliyGMAT wrote: Different approach with application of modular arithmetic.
\(2^2 = 4 = 1 (mod_5)\) \(3^3 = 2 (mod_5)\) \(6 = 1 (mod_5)\) \(7 = 2 (mod_5)\) \(8 = 3 (mod_5)\) \(9 = 1 (mod_5)\) \(5, 10 = 0 (mod_5)\)
\(\frac{1  1 + 2 + (1)^4 + 0 + 1^6 + 2^7 + 3^8 + (1)^9 + 0}{5}\)
\(3^2 = (1) (mod_5)\)
\(\frac{1  1 + 2 + 1 + 1 + 1 + (2^2)^3*2 + (3^2)^4 1}{5}\)
\(= \frac{4 – 2 + 1 – 1}{5} = \frac{2}{5}\)
Remainder 2 Modular arithmetic is definitely not within the scope of GMAT QA , however knowledge of MODULO ARITHMATIC really helps !! Kudos for introducing the discussion...
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Re: 1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder?
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01 Dec 2016, 10:06
Abhishek009 wrote: vitaliyGMAT wrote: Different approach with application of modular arithmetic.
\(2^2 = 4 = 1 (mod_5)\) \(3^3 = 2 (mod_5)\) \(6 = 1 (mod_5)\) \(7 = 2 (mod_5)\) \(8 = 3 (mod_5)\) \(9 = 1 (mod_5)\) \(5, 10 = 0 (mod_5)\)
\(\frac{1  1 + 2 + (1)^4 + 0 + 1^6 + 2^7 + 3^8 + (1)^9 + 0}{5}\)
\(3^2 = (1) (mod_5)\)
\(\frac{1  1 + 2 + 1 + 1 + 1 + (2^2)^3*2 + (3^2)^4 1}{5}\)
\(= \frac{4 – 2 + 1 – 1}{5} = \frac{2}{5}\)
Remainder 2 Modular arithmetic is definitely not within the scope of GMAT QA , however knowledge of MODULO ARITHMATIC really helps !! Kudos for introducing the discussion...I agree with you. For many questions, the concept of cyclicity is working pretty good and it’s more than enough. However, GMAT has some questions, which require finding remainders for numbers with composite powers. If we apply cyclicity in this case we’ll need to identify cycles for base as well as for composite power. This is very time consuming and arithmetic error prone approach, taking into consideration time pressure. In this case, modulo arithmetic and theorems of number theory can significantly simplify our lives. Thanks for kudos buddy



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Re: 1^1+2^2+3^3+...+10^10 is divided by 5. What is the
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