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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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Updated on: 18 Jun 2013, 14:07
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64% (02:05) correct 36% (02:19) wrong based on 618 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

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Originally posted by monirjewel on 05 Nov 2010, 20:38.
Last edited by Bunuel on 18 Jun 2013, 14:07, 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, 21:32
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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, 16:58
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, 04:17
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, 21:09
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, 01:08
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, 02:31
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, 06:32
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, 10:19
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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  [#permalink]

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28 Jun 2013, 10:12
Bunuel wrote:

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

Yes. Typo edited. Thank you.[/quote]

$$\sqrt{2}^0$$ equals 1 or $$\sqrt{2}$$ its self
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Re: For a nonnegative integer n  [#permalink]

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28 Jun 2013, 10:14
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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30 Nov 2016, 07:41
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.

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

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24 Sep 2018, 01:46
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

n is nonnegative integer - Immediately think of n = 0
When 2^0 = 1 is divided by 3, remainder is 1.
When n = 1, 2^1 divided by 3 gives remainder 2.
When n = 2, 2^2 = 4 divided by 3 gives remainder 1.

When a number is divided by 3, the remainder must be 0/1/2.
So here is the pattern: 2^n divided by 3 will not give 0 as remainder since 2^n has no 3s in it.
2^n will give remainder 1 when n is even.
2^n will give remainder 2 when n is odd.

Since we want the remainder to be 1, n MUST be even.

I. n is greater than zero.
Not necessary since n can be 0.

II. 3^n = (-3)^n
Since n is even, this must be true.

III. √2^n is an integer.
n is 0/2/4/... - In each case, this will be an integer.

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Re: For a nonnegative integer n, if the remainder is 1 when 2^n   [#permalink] 24 Sep 2018, 01:46
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