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# M28-26

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Math Expert
Joined: 02 Sep 2009
Posts: 49968

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16 Sep 2014, 01:29
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Difficulty:

65% (hard)

Question Stats:

69% (01:39) correct 31% (01:43) wrong based on 101 sessions

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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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Joined: 02 Sep 2009
Posts: 49968

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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.

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Joined: 14 Jan 2012
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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.
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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20 Sep 2014, 13:27
2
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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Joined: 24 Jun 2015
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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.

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
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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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.

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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Joined: 24 Jun 2015
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02 Jul 2015, 06:23
Thanks¡¡
Intern
Joined: 17 Nov 2016
Posts: 1

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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?
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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27 Nov 2016, 01:44
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

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

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Joined: 10 Dec 2016
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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.
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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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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01 Jan 2017, 03:45
Hu Buenel,

II. (−2)^n is less than 0.

What about when n = 0?

This doesnt satisfy.

Math Expert
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01 Jan 2017, 08:16
dsheth7 wrote:
Hu Buenel,

II. (−2)^n is less than 0.

What about when n = 0?

This doesnt satisfy.

Have you read the actual solution above? ...in order the condition to hold true n must be odd. 0 is even not odd.
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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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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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.

Check below for more:
ALL YOU NEED FOR QUANT ! ! !

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

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