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Math Expert V
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Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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Question Stats: 74% (01:25) correct 26% (01:36) wrong based on 217 sessions

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Which of the following must be subtracted from 2^526 so that the resulting integer will be a multiple of 3?

A. 1
B. 2
C. 3
D. 5
E. 6

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Math Expert V
Joined: 02 Aug 2009
Posts: 7957
Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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1
stonecold wrote:
Here 2^(any power ≥ 11) is a factor of three.
Hence using the rule => Multiple - multiple is always a multiple
We must subtract a multiple of 3
I.e we must subtract 3
Smash that C for me.

I hope i am not missing anything Hi,
this is not correct..
$$2^{11} or 2^{254312}$$will have ONLY 2 as prime factor

A simple way would be $$2^1 =2$$ , which requires 1 to be added for it to be div by 3..
$$2^2 = 4$$ requires 1 to be subtracted ..
and so on..
basically an ODD power of 2 requires 1 to be added to the number to be div by 3..
and any EVEN power would require 1 to be subtracted..

here 256 is EVEN, so it requires 1 to be subtracted
so $$2^{256} - 1$$ is div by 3
ans A
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GRE 1: Q169 V154 Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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Here 2^(any power ≥ 11) is a factor of three.
Hence using the rule => Multiple - multiple is always a multiple
We must subtract a multiple of 3
I.e we must subtract 3
Smash that C for me.

I hope i am not missing anything _________________
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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1
stonecold wrote:
Here 2^(any power ≥ 11) is a factor of three.
Hence using the rule => Multiple - multiple is always a multiple
We must subtract a multiple of 3
I.e we must subtract 3
Smash that C for me.

I hope i am not missing anything I may be missing something here, but are you certain that your power of 2 rule checks out? I tried 2^15 and others. Unless I'm missing something, I don't think this would give the correct answer.
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Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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2
stonecold wrote:
Here 2^(any power ≥ 11) is a factor of three.
Hence using the rule => Multiple - multiple is always a multiple
We must subtract a multiple of 3
I.e we must subtract 3
Smash that C for me.

I hope i am not missing anything Your statement above (in red) is NOT correct.

Important point, 2^number where number >11 can not be a FACTOR of 3 but will be a MULTIPLE of 3.

2^12 is definitely NOT a multiple of 3 as 2^12 will only have 2s in it.

Coming back to the question,

2^1 leaves a remainder of 2 when divided by 3
2^2 leaves a remainder of 1 when divided by 3
2^3 leaves a remainder of 2 when divided by 3
2^4 leaves a remainder of 1 when divided by 3... etc. and the cyclicity continues.

Thus, $$2^{526}$$ will leave a remainder of 1 when divided by 3. Thus you must subtract 1 from $$2^{526}$$ to make it divisible by 3.

A is thus the correct answer.

Hope this helps.
Current Student D
Joined: 12 Aug 2015
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GRE 1: Q169 V154 Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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I made a HUGE error.!!!!!!!!
What was i thinking...!!!!!!!! offcourse 2^anything can never be divisible by 3..!!!
I think i need a break Thanks chetan2 and Engr2012

Regards
StoneCold
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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Anyway binomial expansion would help solve this? Basically upon opening the binomial expansion, everything but the last digit would be divisible by 3. (3-1)^n

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Current Student D
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GRE 1: Q169 V154 Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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1
1
rahulkashyap wrote:
Anyway binomial expansion would help solve this? Basically upon opening the binomial expansion, everything but the last digit would be divisible by 3. (3-1)^n

Posted from my mobile device

Hey rahulkashyap
Yes I think we can use that here
2^526 = (3-1)^526 => expanding using (a-b)^n => all terms will have 3 but the last
Expanding the same => 3^526 * .........(-1)^526 => 3p+1 for some integer p
hence the remainder will be 1

P.S => Thank you for reminding me that Binomial can be a Saviour on the GMAT find the remainder Question..!!
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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1
stonecold wrote:
rahulkashyap wrote:
Anyway binomial expansion would help solve this? Basically upon opening the binomial expansion, everything but the last digit would be divisible by 3. (3-1)^n

Posted from my mobile device

Hey rahulkashyap
Yes I think we can use that here
2^526 = (3-1)^526 => expanding using (a-b)^n => all terms will have 3 but the last
Expanding the same => 3^526 * .........(-1)^526 => 3p+1 for some integer p
hence the remainder will be 1

P.S => Thank you for reminding me that Binomial can be a Saviour on the GMAT find the remainder Question..!!

Most, if not all, remainder questions in GMAT can be solved by cyclicity. So, yes, use Binomial theorem if you are comfortable with it but do not forget about cyclicity (I did not use Binomial Theorem but used cyclicity instead).
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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Bunuel wrote:
Which of the following must be subtracted from 2^526 so that the resulting integer will be a multiple of 3?

A. 1
B. 2
C. 3
D. 5
E. 6

No need to go till the fourth power of 2 ; second power of 2 is sufficient

$$2^{526}$$ = $${2}^{2*128}$$

$${2}^{2*128}$$ = $$4^{128}$$

4/3 will produce 1 as remainder

1^128 = 1

Hence remainder will be 1

PS : Binomial Theorem is not in GMAT Syllabus , however if one is through with it , its his/her choice - Objective is to get the answer correct in whatever method you are confident with in least amount of time...

Abhishek
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Current Student D
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GRE 1: Q169 V154 Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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Abhishek009 wrote:
Bunuel wrote:
Which of the following must be subtracted from 2^526 so that the resulting integer will be a multiple of 3?

A. 1
B. 2
C. 3
D. 5
E. 6

No need to go till the fourth power of 2 ; second power of 2 is sufficient

$$2^{526}$$ = $${2}^{2*128}$$

$${2}^{2*128}$$ = $$4^{128}$$

4/3 will produce 1 as remainder

1^128 = 1

Hence remainder will be 1

PS : Binomial Theorem is not in GMAT Syllabus , however if one is through with it , its his/her choice - Objective is to get the answer correct in whatever method you are confident with in least amount of time...

Abhishek

Hey Just one doubt i have => when you say 4/3 gives one as the remainder ; you are essentially writing 4 as 3P+1 so (3p+1)^128
Isn't that binomial ??
Clearly that isnt cyclicity

Regards
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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2
stonecold wrote:

Hey Just one doubt i have => when you say 4/3 gives one as the remainder ; you are essentially writing 4 as 3P+1 so (3p+1)^128
Isn't that binomial ??
Clearly that isnt cyclicity

Regards
Stonecold[/quote]

{4^1} / 3 =4/3 remainder 1
{4^2} / 3 = 16/3 remainder 1
{4^3} / 3 = 64/3 remainder 1
{4^4} / 3 = 256/3 remainder 1

Actually the same remainder keeps repeating .....

Try with a diff no, say 2

{2^1}/3 = remainder 2
{2^2}/3 = remainder 1
{2^3}/3 = remainder 2
{2^4}/3 = remainder 1

Se there is a cyclic pattern....

Hope this helps , we are using this logic to solve the problem and/ or any other problem involving remainder.
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Intern  Joined: 24 Mar 2016
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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Bunuel wrote:
Which of the following must be subtracted from 2^526 so that the resulting integer will be a multiple of 3?

A. 1
B. 2
C. 3
D. 5
E. 6

The Answer is A) as a lot of people have written before me. But what I want to add about this question, is that if you don't have an approach on how to solve it, it is very easy to see that only option A) and B) could be true. If it would be C, well then the resulting integer would already be a multiple of 3. If it would be D), then it would also be true for B) since 5-2=3. The same counts for E) 6-3=3 which is C).
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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Just follow the pattern:

2^2 = 4 (-1 to be multiple of 3)
2^3 = 8 (-2 to be multiple of 3)
2^4 = 16 (-1 to be multiple of 3)
2^5 = 32 (-2 to be multiple of 3)

The pattern is subtracting 1 or 2. 2^256 is even, thus you must subtract 1 for a multiple of 3.
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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stonecold, chetan2u, Abhishek009

Thank you for the binomial method, indeed very helpful. I was just wondering what happens if the last term in the binomial is not even, e.g. if it would be xp-1, would the remainder then be x-1? Or can we only use the binomial method if the last term is positive (hence with even exponents)?
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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ghnlrug wrote:
stonecold, chetan2u, Abhishek009

Thank you for the binomial method, indeed very helpful. I was just wondering what happens if the last term in the binomial is not even, e.g. if it would be xp-1, would the remainder then be x-1? Or can we only use the binomial method if the last term is positive (hence with even exponents)?

The xp-1 not understood. Please give some values ..
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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chetan2u wrote:
ghnlrug wrote:
stonecold, chetan2u, Abhishek009

Thank you for the binomial method, indeed very helpful. I was just wondering what happens if the last term in the binomial is not even, e.g. if it would be xp-1, would the remainder then be x-1? Or can we only use the binomial method if the last term is positive (hence with even exponents)?

The xp-1 not understood. Please give some values ..

Thanks for the quick reply, so assuming the value would have been $$4^{525}$$ and we want to know the remainder when divided by 3, we could then say $$(4-1)^{525}$$, and the remainder would be (3-1)=2, is that correct? And can we do that in any such case?
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Re: Which of the following must be subtracted from 2^526 so that the resul  [#permalink]

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ghnlrug wrote:
chetan2u wrote:
ghnlrug wrote:
stonecold, chetan2u, Abhishek009

Thank you for the binomial method, indeed very helpful. I was just wondering what happens if the last term in the binomial is not even, e.g. if it would be xp-1, would the remainder then be x-1? Or can we only use the binomial method if the last term is positive (hence with even exponents)?

The xp-1 not understood. Please give some values ..

Thanks for the quick reply, so assuming the value would have been $$4^{525}$$ and we want to know the remainder when divided by 3, we could then say $$(4-1)^{525}$$, and the remainder would be (3-1)=2, is that correct? And can we do that in any such case?

If it is $$4^{525}$$, we can write it as $$(3+1)^{525}$$... so the remainder will be $$1^{525}$$, which is 1..
But If it is $$2^{525}$$, we can write it as $$(3-1)^{525}$$... so the remainder will be $$(-1)^{525}$$, which is -1. But the remainder cannot be negative, so remainder will be 3-1 or 2.

This would be the case even when $$4^{525}$$ is divided by 5, we can write it as $$(5-1)^{525}$$... so the remainder will be $$(-1)^{525}$$, which is -1. Thus the remainder here will be 5-1 or 4.
_________________ Re: Which of the following must be subtracted from 2^526 so that the resul   [#permalink] 28 Feb 2019, 09:28
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