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Re: What is the least integer n such that [#permalink]
15 Oct 2017, 15:07

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Expert's post

clausen8657 wrote:

How are we supposed to know that 2^10 = 1024??

Is there a shortcut? Thanks!

It's not a bad idea to memorize powers of 2 up to 2^7 From there, you can keep multiplying by 2 to get bigger powers.

2^7 = 128 2^8 = 256 2^9 = 512 2^10 = 1024

Cheers, Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com If you enjoy my solutions, you'll like my GRE prep course. Sign up for GRE Question of the Dayemails

Re: GRE Math Challenge #80-value of n such that (1/2^n)<0.001 [#permalink]
12 May 2018, 07:46

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sandy wrote:

What is the least integer value of n such that \((1/2^n)<0.001\)?

(A) 10 (B) 11 (C) 500 (D) 501 (E) there is no such least value.

We want: 1/(2^n) < 0.001 In other words, we want: 1/(2^n) < 1/1000

Let's start with answer choice A (n = 10) We get: 1/(2^n) = 1/(2^10) = 1/1024 Is 1/1024 < 1/1000? Yes!

Since the question asks us to find the least integer value of n, and since 10 is the smallest answer choice, the correct answer must be A

Cheers, Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com If you enjoy my solutions, you'll like my GRE prep course. Sign up for GRE Question of the Dayemails

Re: What is the least integer n such that [#permalink]
20 Jun 2018, 04:56

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Ronaksingh wrote:

how did you cross multiply? please explain

Let's do this in steps...

We have the inequality: 1/(2^n) < 1/1000 Since 2^n is always POSITIVE, we can multiply both sides by 2^n to get: 1 < (2^n)/1000 Also, since 1000 is POSITIVE, we can multiply both sides by 1000 to get: 1000 < 2^n

Here's a video on dealing with inequalities like this:

_________________

Brent Hanneson – Creator of greenlighttestprep.com If you enjoy my solutions, you'll like my GRE prep course. Sign up for GRE Question of the Dayemails

Re: What is the least integer n such that [#permalink]
20 Nov 2018, 15:30

GreenlightTestPrep wrote:

Ronaksingh wrote:

how did you cross multiply? please explain

Let's do this in steps...

We have the inequality: 1/(2^n) < 1/1000 Since 2^n is always POSITIVE, we can multiply both sides by 2^n to get: 1 < (2^n)/1000 Also, since 1000 is POSITIVE, we can multiply both sides by 1000 to get: 1000 < 2^n

Here's a video on dealing with inequalities like this:

Re: What is the least integer n such that [#permalink]
20 Nov 2018, 16:48

1

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Dear An,

the answer is A. Sandy pointed out that n=10 which is \(2^{10}\) still holds, simply because considering that the stem tells us the least AND that does not exist a number as a result of \(2^n = 1000\), we must have that \(2^{10} = 1024\), we could say that \(n > = 10\)

But these are nuances that you do know and acquire with practice. You develop a sort of instinct.

Be flexible in your approach. You can gain only benefits.

Re: What is the least integer n such that [#permalink]
21 Nov 2018, 16:52

Carcass wrote:

Dear An,

the answer is A. Sandy pointed out that n=10 which is \(2^{10}\) still holds, simply because considering that the stem tells us the least AND that does not exist a number as a result of \(2^n = 1000\), we must have that \(2^{10} = 1024\), we could say that \(n > = 10\)

But these are nuances that you do know and acquire with practice. You develop a sort of instinct.

Be flexible in your approach. You can gain only benefits.

Regards

I got it now. I was confused with this basic thing.

Re: What is the least integer n such that [#permalink]
03 Dec 2018, 05:07

Alternative solution. I multiplied by 1000 in the beginning so:

1000 / 2^n < 1 8*125 / 2^n < 1 ( factoring the 2s out of 1000 ) => 2^n > 8*125 ( the denomenator should be > numerator ) then divide all by 8=2^3 2^(n-3) > 125 => 2^7 > 128 => n-3 = 7 => n = 10