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805+ (Hard)|   Exponents|   Remainders|      
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Bunuel
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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 23 Oct 2014, 00:43
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

43% (02:28) correct 57% (02:02) wrong based on 659 sessions

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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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C
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manpreetsingh86
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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 23 Oct 2014, 01:57
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Bunuel

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
Show more

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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Bunuel
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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 23 Oct 2014, 02:06
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manpreetsingh86
Bunuel

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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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? 16 Dec 2015, 06:28
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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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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 05 May 2016, 21:24
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Bunuel
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Bunuel

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.
Show more

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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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 05 May 2016, 21:35
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tallyho_88


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?
Show more

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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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? Updated on: 30 Nov 2023, 09:18
Originally posted by KarishmaB on 05 May 2016, 21:54.
Last edited by KarishmaB on 30 Nov 2023, 09:18, edited 1 time in total.
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tallyho_88

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?
Show more

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.

Using your method, you would have spend a considerable amount of time. Using cylicity would be faster.
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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 01 Dec 2016, 05:24
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Bunuel

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
Show more

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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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 01 Dec 2016, 09:26
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vitaliyGMAT
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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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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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 01 Dec 2016, 10:06
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vitaliyGMAT
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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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 :-D
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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 05 Sep 2023, 00:27
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Bunuel
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
Show more

These are my favorite kind of problems.
All you need to know is cyclicity of each number ie find the last digit.
You need to add the unit digit of last number and check for remainder when divided with 5

1 = 1
2 = 2,4,8,6 Therefore 2^2 = 4
3 = 3,9,7,1 Therefore 3^3 = 7
4 = 4,6,4,6 Therefore 4^4 = 6
5 = 5 Therefore 5^5 = 5
6 = 6 Therefore 6^6 = 6
7 = 7,9,3,1 Therefore 7^7 = 3
8 = 8,4,2,6 Therefore 8^8 = 6
9 = 9,1,9,1 Therefore 9^9 = 9
0 = 0 Therefore 10^10 = 10
Adding all digits we get 7

7/5 = Remainder 2
Hence C
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1^1 + 2^2 + 3^3 + ... + 10^10 is divided by 5. What is the remainder? 14 Aug 2025, 02:15
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Just find units digit, add, the last digit is 7.. when divided by 5 will always throw 2 as remainder.. hence C
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