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# If n is an integer such that n > 9, which of the following could be

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Intern
Joined: 18 Jun 2017
Posts: 27
Location: Mexico
Concentration: Technology, Strategy
If n is an integer such that n > 9, which of the following could be  [#permalink]

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27 May 2020, 16:49
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65% (hard)

Question Stats:

52% (01:43) correct 48% (02:06) wrong based on 77 sessions

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If n is an integer such that n > 9, which of the following could be the remainder when $$(2 + 2^2 + 2^3 +2^4 + .... +2^n)$$ is divided by 3 ?
I. 0
II. 1
III. 2

(A) Only I
(B) Only II
(C) Only III
(D) Only I and III
(E) I, II and III
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Posts: 268
Concentration: General Management, Technology
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Re: If n is an integer such that n > 9, which of the following could be  [#permalink]

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27 May 2020, 17:39
I think Ans is D .

If n is an integer such that n > 9, which of the following could be the remainder when (2+2^2+2^3+2^4+....+2^n) is divided
by 3 ?

it is an GP. with common ratio 2.
so Sum = a (r^n-1) / (r-1) = 2 (2^n - 1) / (2-1) = 2 ^ (n+1) - 2 .

from here on we can test value of n = 10 , 11 , 12 , 13 .14

for n = 10 .
2^11 - 2 / 3 = 4 (9-1)^3 - 2 / 3 = final term contributing to reminder from binomial theorem 4(......+(-1)^3) -2 = -4 -2 = -6 which is divisible by 3 . So reminder is zero when n=10.

for n = 11 .
2 ^ 12 - 2 / 3 = (9-1)^4 -2 /3 = final term contributing to reminder from binomial theorem (.... + (-1)^4 -2) = 1-2 = -1 , which will produce reminder of -1 +3 = 2 .

similarly n=12 . reminder 0
n = 13.. rem 2
n = 14 .. rem 0 ..

so only possible reminders are 0 and 2 depending on the value of n .
So ans is D.
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Ashish A Das.

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Re: If n is an integer such that n > 9, which of the following could be  [#permalink]

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27 May 2020, 19:56
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Bunuel, need an explanation for this answer. Kindly help
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If n is an integer such that n > 9, which of the following could be  [#permalink]

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30 May 2020, 21:34
3
Looking at the series I figured that:
2 divided by 3 leaves a remainder 2
2+2^2 divided by 3 leaves a remainder 0
2+2^2+2^3 divided by 3 leaves a remainder 2
2+2^2+2^3+2^4 divided by 3 leaves a remainder 0
.
.
and so on...

Based on this I marked D as an answer.

Is it safe to make such estimations, that this pattern will follow for all values of n?
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Joined: 18 Jun 2017
Posts: 27
Location: Mexico
Concentration: Technology, Strategy
Re: If n is an integer such that n > 9, which of the following could be  [#permalink]

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30 May 2020, 21:37
1
ShreyKapil08 yes, the estimation & the solution is right.
Math Expert
Joined: 02 Aug 2009
Posts: 8753
Re: If n is an integer such that n > 9, which of the following could be  [#permalink]

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30 May 2020, 21:53
1
ShreyKapil08 wrote:
Looking at the series I figured that:
2 divided by 3 leaves a remainder 2
2+2^2 divided by 3 leaves a remainder 0
2+2^2+2^3 divided by 3 leaves a remainder 2
2+2^2+2^3+2^4 divided by 3 leaves a remainder 0
.
.
and so on...

Based on this I marked D as an answer.

Is it safe to make such estimations, that this pattern will follow for all values of n?

Yes, the method is correct and one of the right ways to do it.
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Joined: 02 Aug 2009
Posts: 8753
Re: If n is an integer such that n > 9, which of the following could be  [#permalink]

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30 May 2020, 22:03
1
1
ogmega wrote:
If n is an integer such that n > 9, which of the following could be the remainder when $$(2 + 2^2 + 2^3 +2^4 + .... +2^n)$$ is divided by 3 ?
I. 0
II. 1
III. 2

(A) Only I
(B) Only II
(C) Only III
(D) Only I and III
(E) I, II and III

Now 3 can leave a remainder as 0, 1 or 2.
But we have to concentrate on power of 2s.
$$2^{Odd}.....2^1=2=0+2.......2^3=8=6+2.....2^5=32=30+2$$
So, in every case, the remainder is 2
$$2^{Even}.....2^2=4=3+1.......2^4=16=15+1.....2^6=64=63+1$$
So, in every case, the remainder is 1

Thus, the remainders move in pattern for $$2^1, 2^2, 2^3...$$ as $$2,1,2,1,2.....$$

$$(2 + 2^2 + 2^3 +2^4 + .... +2^n)$$
So if n is odd
say n=1, then remainder = 2
if n=3, then remainder=2+1+2=5 = 3+2, so 2 again
But if n is even
say n=2, then remainder = 2+1=3, so 0
if n=4, then remainder=2+1+2+1=6, so 0 again

D
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Re: If n is an integer such that n > 9, which of the following could be   [#permalink] 30 May 2020, 22:03