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

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What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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11 Feb 2014, 00:45
5
32
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Difficulty:

25% (medium)

Question Stats:

70% (01:17) correct 30% (01:25) wrong based on 2011 sessions

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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

Problem Solving
Question: 86
Category: Arithmetic Exponents; Operations with rational numbers
Page: 73
Difficulty: 600

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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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11 Feb 2014, 00:45
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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 $$\frac{1}{2^n}> 0.01$$.

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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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11 Feb 2014, 01:45
2
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.
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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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11 Feb 2014, 01:38
1
1
1
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
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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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11 Feb 2014, 14:24
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
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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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26 Jun 2014, 02:24
1
$$\frac{1}{2^n} > \frac{1}{100}$$

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

$$2^6 = 64$$

$$2^7 = 128$$

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What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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23 Mar 2018, 05:39
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 ?

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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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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24 Jun 2018, 06:16
1
gmatbusters chetan2u niks18 VeritasPrepKarishma pushpitkc

Quote:
$$\frac{1}{2^n}> \frac{1}{100}$$;

$$2^n<100$$;

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?

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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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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10 Jul 2014, 11:40
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 $$\frac{1}{2^n}> 0.01$$.

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
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Joined: 02 Sep 2009
Posts: 59147
Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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10 Jul 2014, 12:01
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 $$\frac{1}{2^n}> 0.01$$.

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 $$2^n<100$$.
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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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22 Feb 2018, 17:49
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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

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.

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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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28 Feb 2018, 22:49
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

So the BIGGEST that N could be is 6.

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Re: What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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23 Mar 2018, 05:29
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 $$\frac{1}{2^n}> 0.01$$.

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 ?
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What is the largest integer n such that 1/2^n > 0 ?  [#permalink]

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24 Jun 2018, 06:00
1
gmatbusters chetan2u niks18 VeritasPrepKarishma pushpitkc

Quote:
$$\frac{1}{2^n}> \frac{1}{100}$$;

$$2^n<100$$;

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?

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