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What is the remainder when 3^0 + 3^1 + 3^2 + ... + 3^2009 is divided 8

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What is the remainder when 3^0 + 3^1 + 3^2 + ... + 3^2009 is divided 8  [#permalink]

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New post 29 Mar 2019, 01:10
2
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

34% (02:28) correct 66% (02:14) wrong based on 32 sessions

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What is the remainder when \(3^0 + 3^1 + 3^2 + ... + 3^{2009}\) is divided by 8?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6
Spoiler: OA

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Re: What is the remainder when 3^0 + 3^1 + 3^2 + ... + 3^2009 is divided 8  [#permalink]

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New post 29 Mar 2019, 01:35
1

Solution


To find:
    • The remainder, when 3^0 + 3^1 + 3^2 + ... +3^2009 is divided by 8

Approach and Working:
    • Rem (3^0/8) = 1
    • Rem (3^1/8) = 3
    • Rem (3^2/8) = 1
    • Rem (3^3/8) = 3
    • Rem (3^4/8) = 1
    • Rem (3^5/8) = 3 and so on…

We can observe that,
    • For even powers of 3, the remainder is 1
    • And for odd powers of 3, the remainder is 3.

Therefore, we can rewrite the given series, in terms of its remainders, as
    = 1 + 3 + 1 + 3 + 1 + 3 + … + 3 = 1005 pairs of (1 + 3) = 1005 x 4 = 4020
    • Rem (4020/8) = 4

Hence, the correct answer is option D.

Answer: D

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Re: What is the remainder when 3^0 + 3^1 + 3^2 + ... + 3^2009 is divided 8  [#permalink]

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New post 29 Mar 2019, 01:49
call A = 3^0 + 3^1 + ... + 3^2009
3A = 3^1 + 3^2 + ... + 3^2010
3A - A = 3^2010 - 3^0
2A = 3^2010 - 1
A = (3^2010 - 1) / 2
A = [ (3^2)^1005 -1 ] / 2
A = [(8 + 1)^1005 - 1] / 2
A = [8^1004 + 8^1003 + ... + 8^1 + 1 -1] / 2
A = [8^1003 * 4 + 8^1002 * 4 + ... + 8*4 + 4]
A divides 8 and have remainder of 4. D is the answer
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Re: What is the remainder when 3^0 + 3^1 + 3^2 + ... + 3^2009 is divided 8  [#permalink]

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New post 29 Mar 2019, 07:56
Bunuel wrote:
What is the remainder when \(3^0 + 3^1 + 3^2 + ... + 3^{2009}\) is divided by 8?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6


slightly different approach and method i used
cyclicity of 3 ;
3^0=1 , 3^1 = 3, 3^2 = 9; 3^3= 7 ; 3^4 = 1
so for set of every 4 no we get a repeat of unit digits
1+3+9+7 ; 20 now 2009 times would be 20*100= 2000+ 9 units more i.e two more cycles of 3 ; 20+20 and 1 of 3^1= 43
our last digits would be ~ 2000+43; 2043
when 2043 divided by 8 ; 255*8 ; 2040 ; ~ 3 remainder ; best answer is 4
IMO D
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Re: What is the remainder when 3^0 + 3^1 + 3^2 + ... + 3^2009 is divided 8  [#permalink]

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New post 30 Mar 2019, 03:25
NeoNguyen1989 wrote:
call A = 3^0 + 3^1 + ... + 3^2009
3A = 3^1 + 3^2 + ... + 3^2010
3A - A = 3^2010 - 3^0
2A = 3^2010 - 1
A = (3^2010 - 1) / 2
A = [ (3^2)^1005 -1 ] / 2
A = [(8 + 1)^1005 - 1] / 2
A = [8^1004 + 8^1003 + ... + 8^1 + 1 -1] / 2
A = [8^1003 * 4 + 8^1002 * 4 + ... + 8*4 + 4]
A divides 8 and have remainder of 4. D is the answer


Are you sure the Newton's Binomial formula is correct here, and there shouldn't be a 8^1005 in it as well? (8+1)^1005 = (8+1)*(8+1)^1004 = 8*8^1004+...=8^1005.
Though it doesn't alter the logic of the solution to the original question.
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Re: What is the remainder when 3^0 + 3^1 + 3^2 + ... + 3^2009 is divided 8   [#permalink] 30 Mar 2019, 03:25
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