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Last edited by Bunuel on 18 Jun 2013, 13:07, edited 1 time in total.

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.

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.
_________________

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???
_________________

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???

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!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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.

Answer E.

gmatclubot

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

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