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# Remainder and Divisible

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Remainder and Divisible [#permalink]  29 Sep 2009, 02:44
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33% (02:23) correct 66% (06:13) wrong based on 0 sessions
If {n} denote the remainder when 3n is divided by 2 then which of the following is equal to 1 for all positive integers n?

I/ {2n+1}
II/ {2n}+1
III/ 2{n+1}

A/ I only
B/ II only
C/ I and II
D/ III only
E/ II and III

I don't really understand what it asking for in the question. Could somebody help?
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Re: Remainder and Divisible [#permalink]  29 Sep 2009, 03:42
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{n}=3n mod 2.

options
1) {2n+1)=6n+3 mod 2 that is always equal 1.
2) {2n}+1= 6n mod 2 + 1 that is always equal 1.
3) 2{n+1}=2.(3n+3 mod 2) it can be either 0 or 2.

C)
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Re: Remainder and Divisible [#permalink]  29 Sep 2009, 04:41
Solution is in better shape then question... i could not get the question. and solution can we a bit easier way.
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Re: Remainder and Divisible [#permalink]  02 Oct 2009, 05:47
maliyeci wrote:
{n}=3n mod 2.

options
1) {2n+1)=6n+3 mod 2 that is always equal 1.
2) {2n}+1= 6n mod 2 + 1 that is always equal 1.
3) 2{n+1}=2.(3n+3 mod 2) it can be either 0 or 2.

C)

I can only understand the question after reading this solution

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Re: Remainder and Divisible [#permalink]  27 Oct 2009, 17:45
maliyeci wrote:
{n}=3n mod 2.

options
1) {2n+1)=6n+3 mod 2 that is always equal 1.
2) {2n}+1= 6n mod 2 + 1 that is always equal 1.
3) 2{n+1}=2.(3n+3 mod 2) it can be either 0 or 2.

C)

I couldn't get a single point of it. I think this is something modular arithmetic, which I have no idea. Is there any other way of solving the question, but the problem is that I didn't pick the question either.
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Re: Remainder and Divisible [#permalink]  28 Oct 2009, 09:24
andrewng wrote:
If {n} denote the remainder when 3n is divided by 2 then which of the following is equal to 1 for all positive integers n?

I/ {2n+1}
II/ {2n}+1
III/ 2{n+1}

A/ I only
B/ II only
C/ I and II
D/ III only
E/ II and III

I don't really understand what it asking for in the question. Could somebody help?

Substitute 2 for each answer and solve..as a check do the same with value of 1
{2} remainder = 0
{1} remainder = 1

I. {2(2) + 1} = {5} plug in 5 into 3n/2 and remainder = 1
II. {2(2} + 1 = {4} remainder = 0 and add 1 so answer = 1
III. 2{2+1} = {3} remainder = 1 but then multiply that by 2 and answer is 2

I. {2(1) +1} = {3} remainder = 1
II. {2(1} + 1 = {2} remainder = 0 and add 1 so answer is 1
III. 2{1+1} = {2} remainder = 0 then multiply by two answer is 0

someone double-check my logic...not 100% sure if correct
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Re: Remainder and Divisible [#permalink]  29 Oct 2009, 12:33
Hussain15 wrote:
maliyeci wrote:
{n}=3n mod 2.

options
1) {2n+1)=6n+3 mod 2 that is always equal 1.
2) {2n}+1= 6n mod 2 + 1 that is always equal 1.
3) 2{n+1}=2.(3n+3 mod 2) it can be either 0 or 2.

C)

I couldn't get a single point of it. I think this is something modular arithmetic, which I have no idea. Is there any other way of solving the question, but the problem is that I didn't pick the question either.

Yes. It is about modular arithmetic. I think one must learn this before taking gmat.
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Re: Remainder and Divisible [#permalink]  29 Oct 2009, 17:20
1
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This is another way of solving the problem.

It is given that when 3n is divided by 2, the remainder is denoted by {n}. for all positive integers of n. This mean, {n} can be either 0 or 1 depending on whether n is even or odd. Now we need to find which of the following remainders result in 1.

This is like saying if f(n) is the remainder when 3n/2, then what values of the function results in 1.

1. {2n+1} is the remainder obtained when 3(2n+1) is divided by 2. ==> Obviously this results in 1.
2. {2n}+1 can be treated as the remainder obtained when 3(2n) is divided by 2 + adding 1 ==> Obviously this results in 1.
3. 2{n+1} can be obtained when 2 * 3(n+1) is divided by 2. This results in 0.

Hence my take will be C.
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Re: Remainder and Divisible [#permalink]  31 Jan 2010, 06:05
maliyeci wrote:
Yes. It is about modular arithmetic. I think one must learn this before taking gmat.

Do you recommend any books which could teach this concept..... I am still not able to understand ur solution
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Re: Remainder and Divisible [#permalink]  31 Jan 2010, 06:23
maliyeci wrote:
{n}=3n mod 2.

options
1) {2n+1)=6n+3 mod 2 that is always equal 1.
2) {2n}+1= 6n mod 2 + 1 that is always equal 1.
3) 2{n+1}=2.(3n+3 mod 2) it can be either 0 or 2.

C)

How do you get {2n+1} = 6n+3 mod 2 ...... Am confused!
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Re: Remainder and Divisible [#permalink]  31 Jan 2010, 07:48
jeeteshsingh wrote:
maliyeci wrote:
{n}=3n mod 2.

options
1) {2n+1)=6n+3 mod 2 that is always equal 1.
2) {2n}+1= 6n mod 2 + 1 that is always equal 1.
3) 2{n+1}=2.(3n+3 mod 2) it can be either 0 or 2.

C)

How do you get {2n+1} = 6n+3 mod 2 ...... Am confused!

Given {n} = remainder, when 3n is divide by 2.

Which of the following equals to 1:

I. {2n+1} = remainder, when 3(2n+1)=6n+3 is divide by 2. 6n+3 divided by 2 will always give remainder of 1;
II. {2n}+1 = remainder, when 3(2n)=6n is divide by 2 plus 1. Remainder when 6n divided by 2 is 0 plus 1=1;
III. 2{n+1} = 2*(remainder, when 3(n+1)=3n+3 is divide by 2). 3n+3 divided by 2 will give different remainders.

So only I and II equals to 1.
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Re: Remainder and Divisible [#permalink]  12 Feb 2010, 18:49
C - I and II

3n/2 = 0 or 1
I just substituted 1 in the equations and solved.
Re: Remainder and Divisible   [#permalink] 12 Feb 2010, 18:49
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