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.
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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.

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

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

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