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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.
We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.
Re: What is the largest integer n such that 1/2^n > 0 ?
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11 Feb 2014, 13:24
1
Took me a while but i think it should be 1/2^n > 1/100 so we need to compare really the maximum value of 2^n such that it is less than 100, because if it is more than 100 then 1/2^n becomes less than 1/100
2^5 = 32 2^6 = 64 2^7 = 128
So n cannot be 7 and the largest value it can have while still keeping 1/2^n > 1/100 will be 6 B
Re: What is the largest integer n such that 1/2^n > 0 ?
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10 Jul 2014, 10:40
Bunuel wrote:
SOLUTION
What is the largest integer n such that \(\frac{1}{2^n}> 0.01\) ?
(A) 5 (B) 6 (C) 7 (D) 10 (E) 51
[m]\frac{1}{2^n}> 0.01[/m];
\(\frac{1}{2^n}> \frac{1}{100}\);
\(2^n<100\);
\(2^6=64<100\) and \(2^7=128>100\), therefore, 6 is the largest integer such that \(\frac{1}{2^n}> 0.01\).
Answer: B.
Hi
I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know i may be missing some basic concept here..Please advice
Re: What is the largest integer n such that 1/2^n > 0 ?
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10 Jul 2014, 10:40
Bunuel wrote:
SOLUTION
What is the largest integer n such that \(\frac{1}{2^n}> 0.01\) ?
(A) 5 (B) 6 (C) 7 (D) 10 (E) 51
[m]\frac{1}{2^n}> 0.01[/m];
[color=#ff0000]\(\frac{1}{2^n}> \frac{1}{100}\);
\(2^n<100\); [/color] \(2^6=64<100\) and \(2^7=128>100\), therefore, 6 is the largest integer such that \(\frac{1}{2^n}> 0.01\).
Answer: B.
Hi
I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know i may be missing some basic concept here..Please advice
Re: What is the largest integer n such that 1/2^n > 0 ?
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10 Jul 2014, 11:01
GGMAT760 wrote:
Bunuel wrote:
SOLUTION
What is the largest integer n such that \(\frac{1}{2^n}> 0.01\) ?
(A) 5 (B) 6 (C) 7 (D) 10 (E) 51
[m]\frac{1}{2^n}> 0.01[/m];
\(\frac{1}{2^n}> \frac{1}{100}\);
\(2^n<100\);
\(2^6=64<100\) and \(2^7=128>100\), therefore, 6 is the largest integer such that \(\frac{1}{2^n}> 0.01\).
Answer: B.
Hi
I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know i may be missing some basic concept here..Please advice
Thanks Komal
Cross-multiply \(\frac{1}{2^n}> \frac{1}{100}\) to get \(2^n<100\).
_________________
Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.
We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.
Re: What is the largest integer n such that 1/2^n > 0 ?
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23 Jun 2017, 13:08
\(\frac{1}{2^n}> 0.01\)
\(\frac{1}{2^n}> 1/100\)
\(2 ^n < 100\)
We need to find largest value of n which satisfies the above value.
\(2^6 = 64\) is the maximum value of n that satisfies above equation.
Hence, Answer is B = 6 _________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."
Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475
Re: What is the largest integer n such that 1/2^n > 0 ?
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28 Feb 2018, 21:49
Hi All,
Since the answers to this question are numbers, we can use them to our advantage and TEST THE ANSWERS.
We're told that N has to be an INTEGER and we want to make N as LARGE as possible so that 1/(2^N) > .01
Since this inequality uses a fraction on one side and a decimal on the other, I'm going to convert the decimal to a fraction. This gives us....
1/(2^N) > 1/100
We want to make N as LARGE as possible while still maintaining the inequality. This means that we have to make 2^N as BIG as possible BUT it still has to be less than 100.
One of the 5 answer choices MUST be correct, so let's TEST THE ANSWERS....
If N = 5, 2^5 = 32 1/32 > 1/100 If N = 6, 2^6 = 64 1/64 > 1/100 If N = 7, 2^7 = 128 1/128 is NOT > 1/100
Re: What is the largest integer n such that 1/2^n > 0 ?
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23 Mar 2018, 04:29
Bunuel wrote:
GGMAT760 wrote:
Bunuel wrote:
SOLUTION
What is the largest integer n such that \(\frac{1}{2^n}> 0.01\) ?
(A) 5 (B) 6 (C) 7 (D) 10 (E) 51
[m]\frac{1}{2^n}> 0.01[/m];
\(\frac{1}{2^n}> \frac{1}{100}\);
\(2^n<100\);
\(2^6=64<100\) and \(2^7=128>100\), therefore, 6 is the largest integer such that \(\frac{1}{2^n}> 0.01\).
Answer: B.
Hi
I could not understand highlighted step.. How 2^1/n >1/100 .....turned in to 2^n<100???? I know i may be missing some basic concept here..Please advice
Thanks Komal
Cross-multiply \(\frac{1}{2^n}> \frac{1}{100}\) to get \(2^n<100\).
Why are we changing inequility sign direction after we cross multiply ? there are no negatives sign to cause the inequlity sign change we only change sign when we divide or multiply by negative value, right ?
What is the largest integer n such that 1/2^n > 0 ?
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23 Mar 2018, 04:39
1
dave13 wrote:
Why are we changing inequility sign direction after we cross multiply ? there are no negatives sign to cause the inequlity sign change we only change sign when we divide or multiply by negative value, right ?
To recall above steps, i.e. flip inequality while reversing the fraction, do I need to take simple examples as 2 < 3 hence 1/2 > 1/3 or there is a better approach?
Why does sign of 'n' does not matter?
Bunuel can you please add relevant discussion in this post?
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To recall above steps, i.e. flip inequality while reversing the fraction, do I need to take simple examples as 2 < 3 hence 1/2 > 1/3 or there is a better approach?
Why does sign of 'n' does not matter?
Bunuel can you please add relevant discussion in this post?
You an simply cross-multiply \(\frac{1}{2^n}> \frac{1}{100}\) to get \(100<2^n\). We can safely do that because cross-multiplication means multiplying by 100 and by 2^n, both of which are positive (2^n > 0 for any value of n).
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71% (00:55) correct 
we only change sign when we divide or multiply by negative value, right ? 
