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Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

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

Re: For a nonnegative integer n [#permalink]
05 Nov 2010, 20:32

7

This post received KUDOS

Expert's post

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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.

Re: For a nonnegative integer n [#permalink]
06 Jun 2013, 09:19

2

This post received KUDOS

Expert's post

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. _________________

Re: For a nonnegative integer n [#permalink]
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??? _________________

Re: For a nonnegative integer n [#permalink]
28 Jun 2013, 08:39

Expert's post

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???

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

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