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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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New post Updated on: 18 Jun 2013, 14:07
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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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New post 05 Nov 2010, 21:32
10
3
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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New post 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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New post 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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New post 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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New post 22 Mar 2012, 01:08
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Re: For a nonnegative integer n, if the remainder is 1 when 2n  [#permalink]

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New post 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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New post 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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New post 06 Jun 2013, 10:19
3
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.
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Re: For a nonnegative integer n  [#permalink]

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New post 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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New post 28 Jun 2013, 10:14
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Re: For a nonnegative integer n, if the remainder is 1 when 2^n  [#permalink]

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New post 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.

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

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New post 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.

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