What is the largest integer n such that 1/2^n 0 ?

Question

What is the largest integer n such that  \frac{1}{2^n}> 0.01  ?

(A) 5
(B) 6
(C) 7
(D) 10
(E) 51

Bunuel · Accepted Answer

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

\(\frac{1}{2^n}> 0.01\);

\(\frac{1}{2^n}> \frac{1}{100}\);

\(2^n<100\);

\(2^6=64<100\) and \(2^7=128>100\), therefore, 6 is the largest integer such that \(\frac{1}{2^n}> 0.01\).

Answer: B.

liranmaymoni · Answer

0.01=1/100

So the question is  2^n<100

We should know that:
2^5=32
2^6=64
2^7=128

so the ans is 6. B

Donnie84 · Answer

1/(2^n) > 1/100

Since, 2^6 = 64, 1/64 > 1/100. Answer (B).

It cannot be 7 because 2^7 = 128 and 1/128 < 1/100.

amz14 · Answer

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

PareshGmat · Answer

\frac{1}{2^n} > \frac{1}{100}

With numerators of both sides same, for LHS > RHS, denominator of LHS should be less than that of RHS

2^6 = 64

2^7 = 128

Answer = 6 = B

GGMAT760 · Answer

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

Bunuel · Answer

Cross-multiply  \frac{1}{2^n}> \frac{1}{100}  to get  2^n<100 .

JeffTargetTestPrep · Answer

Converting 0.01 to a fraction we have:

1/(2^n) > 1/100

We can reciprocate the inequality above (remember to switch the inequality sign) to have:

2^n < 100

The largest integer such that 2^n < 100 is 6 since 2^6 = 64 and 2^7 = 128. So n must be 6.

Answer: B

EMPOWERgmatRichC · Answer

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

So the BIGGEST that N could be is 6.

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

dave13 · Answer

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 ?

abhimahna · Answer

Hey dave13 , We are not changing the sign of the inequality. If I am saying 2 < 3 , can't I also say 3 > 2 ? This is exactly the same Bunuel did. \frac{1}{2^n}> \frac{1}{100} is equal to 100> 2^n , which is equal to 2^n<100 Does that make sense?

adkikani · Answer

gmatbusters   chetan2u   niks18   VeritasPrepKarishma   pushpitkc

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?

Bunuel · Answer

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).

BrentGMATPrepNow · Answer

We want: \frac{1}{2^n}> 0.01 Rewrite as: \frac{1}{2^n}> \frac{1}{100} If 2^n = 100 , then \frac{1}{2^n}= \frac{1}{100} If 2^n > 100 , then \frac{1}{2^n} < \frac{1}{100} If 2^n < 100 , then \frac{1}{2^n} > \frac{1}{100} So, for this question, we need the largest value of n so that 2^n < 100 2^6 = 64 and 2^7 = 64 So 6 is the largest integer value of n so that 2^n < 100 Answer: B Cheers, Brent

RAAVAN2003 · Answer

Why not 5 because 1 divided by 32 is 0.03125 which is greater that 0.01 ?

Krunaal · Answer

Please read the question carefully, we want the  largest integer  n

What is the largest integer n such that 1/2^n > 0 ?

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