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What is the largest integer n such that 1/2^n > 0.001?
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Updated on: 25 Jun 2014, 09:13
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What is the largest integer n such that \(\frac{1}{2^n}> 0.01\) ? (A) 5 (B) 6 (C) 7 (D) 10 (E) 51 Here is what I did:
[1/(2^n)] > 1/100 Hence, (2^n) > 100 but the book says it'll be (2^n) < 100
Can someone tell me why did the sign change?
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Originally posted by bfman on 04 Jul 2010, 18:52.
Last edited by Bunuel on 25 Jun 2014, 09:13, edited 1 time in total.
Renamed the topic and edited the question.



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Re: What is the largest integer n such that 1/2^n > 0.001?
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04 Jul 2010, 19:19
bfman wrote: Page 73. Q: 86
What is the largest integer n such that [1/(2^n)] > 0.01 ?
A. 5 B. 6 C. 7 D. 10 E. 51
Here is what I did:
[1/(2^n)] > 1/100 Hence, (2^n) > 100 but the book says it'll be (2^n) < 100
Can someone tell me why did the sign change?
Thanks! 1/2 > 1/3 => 3>2, but according to you, 2>3 1/x > 1/y if both x and y are +ve cross multiple which gives y>x if there is ve sign then also inequality sign is reversed. PS: If you have problem with quants, do buy number prop and inequality books by Mgmat
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Re: What is the largest integer n such that 1/2^n > 0.001?
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04 Jul 2010, 19:56
ohhh cross multipyyyyy.... I was taking the denominators as is... I thought it was ok to do that.



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Re: What is the largest integer n such that 1/2^n > 0.001?
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10 Aug 2010, 18:36
In inequalities,
if both x and y are positive and \((1/x) > (1/y)\) then, \(x < y\)
In the question given
\((1/2^n) > 0.01\)
\((1/2^n) > 1/100\)
\(2^n < 100\)
\(2^6 = 64\)
\(2^7 = 128\) which is greater than 100.
\(Thus, 2^6 < 100\)
Hence \(n=6\)



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Re: What is the largest integer n such that 1/2^n > 0.001?
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10 Aug 2010, 18:57
Quote: Page 73. Q: 86
What is the largest integer n such that [1/(2^n)] > 0.01 ?
A. 5 B. 6 C. 7 D. 10 E. 51
Guess it is easy for people who deal with bits and bytes to crack this within 30 secs... \(1/(2^n) > 1/100 ==> 1/(2^6) > 1/100\) and \(1/(2^7) < 1/100\). Hence largest integer n would be 6.
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Re: What is the largest integer n such that 1/2^n > 0.001?
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10 Aug 2010, 19:22
1/(2^n) > 1/100 => 100 > 2^n
the only max n possible is 6 from the choices



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Re: What is the largest integer n such that 1/2^n > 0.001?
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30 Sep 2011, 14:05
86) What is the largest integer N such that 1/2^n greater than 0.01?
a. 5 b.6 c. 7 d. 10 e. 51
I understand a little bit of the explanation from the guide book, but why/ how did they remove the fraction? and why is 6 the answer, why couldn't it be 2 to the 7th , 10th or 51st. all of those answers should be greater than 100.



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Re: What is the largest integer n such that 1/2^n > 0.001?
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30 Sep 2011, 14:27
Slash2009 wrote: 86) What is the largest integer N such that 1/2^n greater than 0.01?
a. 5 b.6 c. 7 d. 10 e. 51
I understand a little bit of the explanation from the guide book, but why/ how did they remove the fraction? and why is 6 the answer, why couldn't it be 2 to the 7th , 10th or 51st. all of those answers should be greater than 100. Bc we know that \(2^n\) and 100 are +ve, we can cross multiply. and we will receive  \(2^n<100\) we know that \(2^6 = 64\), if we multiply 64 in 2 again  we will have a number greater than 100. so it must be that n=6 (max) You can find the a full existing discussion here: http://gmatclub.com/forum/fromquantogpowersquestion96753.html
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Re: What is the largest integer n such that 1/2^n > 0.001?
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30 Sep 2011, 14:49
Thanks Yo, I need to pay attention to the greater than , less than signs



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Re: What is the largest integer n such that 1/2^n > 0.001?
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30 Sep 2011, 17:57
1/(2^n) > 1/(10^2)
multiplying by 2^n on both sides we have (as 2^n is always positive ,we don't have change sign)
1 > 2^n / 10^2
multiplying by 10^2 on both sides we have
100 > 2^n
In other words 2^n < 100
plugging in values we can see that max value of n that satisfy the above condition is 6.
Answer is B.



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Re: What is the largest integer n such that 1/2^n > 0.001?
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02 Oct 2011, 11:05



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Re: What is the largest integer n such that 1/2^n > 0.001?
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Re: What is the largest integer n such that 1/2^n > 0.001?
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