For a nonnegative integer n, if the remainder is 1 when 2^n : GMAT Problem Solving (PS)
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# For a nonnegative integer n, if the remainder is 1 when 2^n

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For a nonnegative integer n, if the remainder is 1 when 2^n [#permalink]

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05 Nov 2010, 19:38
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59% (02:26) correct 41% (01:39) wrong based on 483 sessions

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For a nonnegative integer n, if the remainder is 1 when 2^n is divided by 3, then which of the following must be true?

I. n is greater than zero.
II. 3^n = (-3)^n
III. √2^n is an integer.

A. I only
B. II only
C. I and II
D. I and III
E. II and III
[Reveal] Spoiler: OA

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Last edited by Bunuel on 18 Jun 2013, 13:07, edited 1 time in total.
Edited the question.
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Re: For a nonnegative integer n [#permalink]

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05 Nov 2010, 20:32
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monirjewel wrote:
For a nonnegative integer n, if the remainder is 1 when 2^n is divided by 3, then which of the following must be true?
I. n is greater than zero.
II. 3^n = (-3)^n
III. √2^n is an integer.
A. I only
B. II only
C. I and II
D. I and III
E. II and III

Please check the questions when posting.

Given: $$n=integer\geq{0}$$ and $$2^n=3q+1$$, for some non-negative integer $$q$$:

If $$n=0=even$$ --> $$2^0=1$$ --> remainder upon division 1 by 3 is 1 - OK;
If $$n=1=odd$$ --> $$2^1=2$$ --> remainder upon division 2 by 3 is 2 - not OK;
If $$n=2=even$$ --> $$2^2=4$$ --> remainder upon division 4 by 3 is 1 - OK;
If $$n=3=odd$$ --> $$2^3=8$$ --> remainder upon division 8 by 3 is 2 - not OK;
...

So we can see the pattern of reminders 1-2-1-2-.... --> given condition that the remainder is 1 when $$2^n$$ is divided by 3 holds true when $$n=even$$. So $$n$$ must be non-negative even number: 0, 2, 4, ...

I. $$n$$ is greater than zero --> not necessarily true, as $$n$$ can be zero;
II. $$3^n = (-3)^n$$ --> as $$n$$ is even then this statement is always true;
III. $$\sqrt{2}^n=integer$$ --> as $$n$$ is non-negative even number then this statement is always true.

Answer: E (II and III only).
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Re: For a nonnegative integer n [#permalink]

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16 Feb 2011, 15:58
Thank you, Bunuel! I checked a lot of sources, but your explanation is the best, as usual!=)
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Re: For a nonnegative integer n [#permalink]

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16 Mar 2012, 03:17
Bunuel, Your explanation is always the BEST! THANK U.......
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Re: For a nonnegative integer For a nonnegative integer n, if [#permalink]

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21 Mar 2012, 20:09
just a question on this is it (√2)^n or (√2^n).

Just have some confusion on this.
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Re: For a nonnegative integer For a nonnegative integer n, if [#permalink]

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22 Mar 2012, 00:08
rggoel9 wrote:
just a question on this is it (√2)^n or (√2^n).

Just have some confusion on this.

They are the same: $$(\sqrt{2})^n=\sqrt{2^n}$$.
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Re: For a nonnegative integer n, if the remainder is 1 when 2n [#permalink]

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26 Mar 2012, 01:31
Tried solving using the 1st post and got mightily confused....Thx Bunuel!!
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Re: For a nonnegative integer n, if the remainder is 1 when 2n [#permalink]

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06 Jun 2013, 05:32
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: For a nonnegative integer n [#permalink]

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06 Jun 2013, 09:19
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Bunuel wrote:
monirjewel wrote:
For a nonnegative integer n, if the remainder is 1 when 2^n is divided by 3, then which of the following must be true?
I. n is greater than zero.
II. 3^n = (-3)^n
III. √2^n is an integer.
A. I only
B. II only
C. I and II
D. I and III
E. II and III

We know that $$x^n-a^n$$ is divisible by (x-a) AND (x+a) ONLY when n is an even integer.

Now , from the given problem, $$2^n$$ = 3q+1--> $$\frac{2^n-1^n}{3}$$ = q(An integer)--> $$\frac{2^n-1^n}{(2+1)}$$ = q.
Thus, n HAS to be even. Now, only options II and III stand for n= even.
E.
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Re: For a nonnegative integer n, if the remainder is 1 when 2n [#permalink]

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18 Jun 2013, 13:03
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Question is not typed clearly...solved it as 2n and not 2^n
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Re: For a nonnegative integer n, if the remainder is 1 when 2n [#permalink]

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18 Jun 2013, 13:08
putronix wrote:
Question is not typed clearly...solved it as 2n and not 2^n

Edited the original post.
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Re: For a nonnegative integer n [#permalink]

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28 Jun 2013, 08:34
Bunuel wrote:
monirjewel wrote:
For a nonnegative integer n, if the remainder is 1 when 2^n is divided by 3, then which of the following must be true?
I. n is greater than zero.
II. 3^n = (-3)^n
III. √2^n is an integer.
A. I only
B. II only
C. I and II
D. I and III
E. II and III

Please check the questions when posting.

Given: $$n=integer\geq{0}$$ and $$2^n=3q+1$$, for some non-negative integer $$q$$:

If $$n=0=even$$ --> $$2^0=1$$ --> remainder upon division 1 by 3 is 1 - OK;
If $$n=1=odd$$ --> $$[color=#ff0000][b]2^2=2$$ [/color][/b]--> remainder upon division 2 by 3 is 2 - not OK;
If $$n=2=even$$ --> $$2^2=4$$ --> remainder upon division 4 by 3 is 1 - OK;
If $$n=3=odd$$ --> $$2^3=8$$ --> remainder upon division 8 by 3 is 2 - not OK;
...

So we can see the pattern of reminders 1-2-1-2-.... --> given condition that the remainder is 1 when $$2^n$$ is divided by 3 holds true when $$n=even$$. So $$n$$ must be non-negative even number: 0, 2, 4, ...

I. $$n$$ is greater than zero --> not necessarily true, as $$n$$ can be zero;
II. $$3^n = (-3)^n$$ --> as $$n$$ is even then this statement is always true;
III. $$\sqrt{2}^n=integer$$ --> as $$n$$ is non-negative even number then this statement is always true.

Answer: E (II and III only).

Dont you think this should be 2^1= 2 and not 2^2=2???
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Re: For a nonnegative integer n [#permalink]

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28 Jun 2013, 08:39
prateekbhatt wrote:
Bunuel wrote:
monirjewel wrote:
For a nonnegative integer n, if the remainder is 1 when 2^n is divided by 3, then which of the following must be true?
I. n is greater than zero.
II. 3^n = (-3)^n
III. √2^n is an integer.
A. I only
B. II only
C. I and II
D. I and III
E. II and III

Please check the questions when posting.

Given: $$n=integer\geq{0}$$ and $$2^n=3q+1$$, for some non-negative integer $$q$$:

If $$n=0=even$$ --> $$2^0=1$$ --> remainder upon division 1 by 3 is 1 - OK;
If $$n=1=odd$$ --> $$[color=#ff0000][b]2^2=2$$ [/color][/b]--> remainder upon division 2 by 3 is 2 - not OK;
If $$n=2=even$$ --> $$2^2=4$$ --> remainder upon division 4 by 3 is 1 - OK;
If $$n=3=odd$$ --> $$2^3=8$$ --> remainder upon division 8 by 3 is 2 - not OK;
...

So we can see the pattern of reminders 1-2-1-2-.... --> given condition that the remainder is 1 when $$2^n$$ is divided by 3 holds true when $$n=even$$. So $$n$$ must be non-negative even number: 0, 2, 4, ...

I. $$n$$ is greater than zero --> not necessarily true, as $$n$$ can be zero;
II. $$3^n = (-3)^n$$ --> as $$n$$ is even then this statement is always true;
III. $$\sqrt{2}^n=integer$$ --> as $$n$$ is non-negative even number then this statement is always true.

Answer: E (II and III only).

Dont you think this should be 2^1= 2 and not 2^2=2???

Yes. Typo edited. Thank you.
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Re: For a nonnegative integer n [#permalink]

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28 Jun 2013, 09:12
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Re: For a nonnegative integer n [#permalink]

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28 Jun 2013, 09:14
prateekbhatt wrote:

$$\sqrt{2}^0$$ equals 1 or $$\sqrt{2}$$ its self

$$\sqrt{2}^0=1$$ .
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Re: For a nonnegative integer n, if the remainder is 1 when 2^n [#permalink]

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27 Oct 2014, 22:20
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Re: For a nonnegative integer n, if the remainder is 1 when 2^n [#permalink]

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31 Aug 2016, 06:36
Hello from the GMAT Club BumpBot!

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Re: For a nonnegative integer n, if the remainder is 1 when 2^n [#permalink]

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30 Nov 2016, 06:41
monirjewel wrote:
For a nonnegative integer n, if the remainder is 1 when 2^n is divided by 3, then which of the following must be true?

I. n is greater than zero.
II. 3^n = (-3)^n
III. √2^n is an integer.

A. I only
B. II only
C. I and II
D. I and III
E. II and III

We have following question:

$$\frac{2^n}{3} = 1 (mod 3)$$

$$2 = -1 (mod 3)$$ and our question boils down to the following:

$$\frac{(-1)^n}{3} = \frac{1}{3}$$

This is only possible when n is even. N=0, 2, 4 …..

I. n is greater than zero. No. n can be equal to 0 because 0 is even
II. 3^n = (-3)^n True when n is even
III. √2^n is an integer. Also true when n is even. $$\sqrt{2^0} = \sqrt{1}= 1$$

Options II and III only.

Re: For a nonnegative integer n, if the remainder is 1 when 2^n   [#permalink] 30 Nov 2016, 06:41
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