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M28-26

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M28-26  [#permalink]

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New post 16 Sep 2014, 01:29
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If \(n\) is a non-negative integer and the remainder when \(3^n\) is divided by 4 is a multiple of 3, then which of the following must be true?
I. \(n^2\) divided by 4 yields the reminder of 1

II. \((-2)^n\) is less than 0

III. \(n\) is a prime number


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(II\) only
E. \(II\) and \(III\) only

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Re M28-26  [#permalink]

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New post 16 Sep 2014, 01:29
Official Solution:

If \(n\) is a non-negative integer and the remainder when \(3^n\) is divided by 4 is a multiple of 3, then which of the following must be true?
I. \(n^2\) divided by 4 yields the reminder of 1

II. \((-2)^n\) is less than 0

III. \(n\) is a prime number


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(II\) only
E. \(II\) and \(III\) only


\(3^0=1\).The remainder when 1 is divided by 4 is 1;

\(3^1=3\). The remainder when 3 is divided by 4 is 3;

\(3^2=9\). The remainder when 9 is divided by 4 is 1;

\(3^3=27\). The remainder when 27 is divided by 4 is 3;

...

We can see that in order the condition to hold true \(n\) must be odd.

I. \(n^2\) divided by 4 yields the reminder of 1. Since \(odd^2\) divided by 4 always yields the reminder of 1, then this statement must be true.

II. \((-2)^n\) is less than 0. Since \((-2)^{odd} \lt 0\), then this statement must be true.

III. \(n\) is a prime number. Not necessarily true.


Answer: D
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Re: M28-26  [#permalink]

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New post 20 Sep 2014, 00:45
My apologies for a simple question, but how is it possible to prove that odd^2 divided by 4 always yields the reminder of 1?

Is it correct based on (o/2)*(o/2)=(q*2+1)(q*2+1), so after multiplication we can conclude that R=1

Or based on (odd^2-1+1)/4, where odd^2-1 is always divisible by 4

Or is it a rule that i missed?

My concern aroused cause originally i came to the conclusion base on several numbers substitution.
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Re: M28-26  [#permalink]

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New post 20 Sep 2014, 13:27
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Boycot wrote:
My apologies for a simple question, but how is it possible to prove that odd^2 divided by 4 always yields the reminder of 1?

Is it correct based on (o/2)*(o/2)=(q*2+1)(q*2+1), so after multiplication we can conclude that R=1

Or based on (odd^2-1+1)/4, where odd^2-1 is always divisible by 4

Or is it a rule that i missed?

My concern aroused cause originally i came to the conclusion base on several numbers substitution.


An odd number can be represented as 2k + 1 --> (2k + 1)^2 = 4k^2 + 4k + 1 --> first two terms are divisible by 4, so the remainder would come only from the third term --> 1 divided by 4 yields the remainder of 1.

Check it with numbers:
1^2 = 1 yields the remainder of 1 when divided by 4;
3^2 = 9 yields the remainder of 1 when divided by 4;
5^2 = 25 yields the remainder of 1 when divided by 4;
...

Hope it's clear.
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Re: M28-26  [#permalink]

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New post 02 Jul 2015, 05:48
Bunuel wrote:
Official Solution:

If \(n\) is a non-negative integer and the remainder when \(3^n\) is divided by 4 is a multiple of 3, then which of the following must be true?
I. \(n^2\) divided by 4 yields the reminder of 1

II. \((-2)^n\) is less than 0

III. \(n\) is a prime number


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(II\) only
E. \(II\) and \(III\) only


\(3^0=1\).The remainder when 1 is divided by 4 is 1;

\(3^1=3\). The remainder when 3 is divided by 4 is 3;

\(3^2=9\). The remainder when 9 is divided by 4 is 1;

\(3^3=27\). The remainder when 27 is divided by 4 is 3;

...

We can see that in order the condition to hold true \(n\) must be odd.

I. \(n^2\) divided by 4 yields the reminder of 1. Since \(odd^2\) divided by 4 always yields the reminder of 1, then this statement must be true.

II. \((-2)^n\) is less than 0. Since \((-2)^{odd} \lt 0\), then this statement must be true.

III. \(n\) is a prime number. Not necessarily true.


Answer: D


Hi Bunuel,

The first conclusion made was that N must be odd because of the pattern of 3^n, but is there another easy way to find this conclusion? My doubt is it is neccessary to do the numbers in order to see a pattern...

Thanks a lot.

Regards.

Luis Navarro
Looking for 700
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M28-26  [#permalink]

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New post 02 Jul 2015, 05:52
1
luisnavarro wrote:
Bunuel wrote:
Official Solution:

If \(n\) is a non-negative integer and the remainder when \(3^n\) is divided by 4 is a multiple of 3, then which of the following must be true?
I. \(n^2\) divided by 4 yields the reminder of 1

II. \((-2)^n\) is less than 0

III. \(n\) is a prime number


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(II\) only
E. \(II\) and \(III\) only


\(3^0=1\).The remainder when 1 is divided by 4 is 1;

\(3^1=3\). The remainder when 3 is divided by 4 is 3;

\(3^2=9\). The remainder when 9 is divided by 4 is 1;

\(3^3=27\). The remainder when 27 is divided by 4 is 3;

...

We can see that in order the condition to hold true \(n\) must be odd.

I. \(n^2\) divided by 4 yields the reminder of 1. Since \(odd^2\) divided by 4 always yields the reminder of 1, then this statement must be true.

II. \((-2)^n\) is less than 0. Since \((-2)^{odd} \lt 0\), then this statement must be true.

III. \(n\) is a prime number. Not necessarily true.


Answer: D


Hi Bunuel,

The first conclusion made was that N must be odd because of the pattern of 3^n, but is there another easy way to find this conclusion? My doubt is it is neccessary to do the numbers in order to see a pattern...

Thanks a lot.

Regards.

Luis Navarro
Looking for 700


Yes, it's the easiest ad not hard at all way.

Check Units digits, exponents, remainders problems directory in our Special Questions Directory.
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Re: M28-26  [#permalink]

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New post 02 Jul 2015, 06:23
Thanks¡¡
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Re M28-26  [#permalink]

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New post 26 Nov 2016, 15:30
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. II. (−2)n=1 when n=0?
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New post 27 Nov 2016, 01:44
Jesus Amado wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. II. (−2)n=1 when n=0?


(-2)^0 = 1 because (non-zero number)^0 = 1.


Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html

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Re: M28-26  [#permalink]

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New post 25 Dec 2016, 12:08
If we take ,for example, n=6: We get 3^6= 4*180 + 9 ( the remainder 9 is a multiple of 3 ).
And yet for n=6, none of the three conditions has to be true:
- 6²= 36. and dividing 36 by 4 doesn't yield a remainder of 1.
- (-2)^6 is positive.
- 6 is not a prime number.
Can someone please tell me where I'm mistaken ? because for me, none of the conditions has to be true, thus none of the answers is correct.
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New post 26 Dec 2016, 10:47
LamyaaBezoui wrote:
If we take ,for example, n=6: We get 3^6= 4*180 + 9 ( the remainder 9 is a multiple of 3 ).
And yet for n=6, none of the three conditions has to be true:
- 6²= 36. and dividing 36 by 4 doesn't yield a remainder of 1.
- (-2)^6 is positive.
- 6 is not a prime number.
Can someone please tell me where I'm mistaken ? because for me, none of the conditions has to be true, thus none of the answers is correct.


The remainder of 3^6/4 is 1 not 9. The reminder upon division by 4 cannot possible be 9 it can be 0, 1, 2, or 3. The remainder is always less than the divisor.
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Re: M28-26  [#permalink]

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New post 01 Jan 2017, 03:45
Hu Buenel,

II. (−2)^n is less than 0.

What about when n = 0?

This doesnt satisfy.

Please help
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New post 01 Jan 2017, 08:16
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New post 25 Oct 2017, 20:13
But for the case, when n =0, which hold true for original condition, then choice I does not work. 0 when
divided by 4 leaves 4 as remainder. Am I missing sth ?
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Re: M28-26  [#permalink]

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New post 25 Oct 2017, 22:33
shohm wrote:
But for the case, when n =0, which hold true for original condition, then choice I does not work. 0 when
divided by 4 leaves 4 as remainder. Am I missing sth ?


0 is divisible by all integers (except 0 itself). So, 0/4 leaves the remainder of 0.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.


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Re: M28-26 &nbs [#permalink] 25 Oct 2017, 22:33
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