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

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.

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

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

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

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 \(2^n<100\).

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

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:

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

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 ?

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?

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?

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).

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

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