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What is the largest integer n such that 3^n < 0.001?

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What is the largest integer n such that 3^n < 0.001? [#permalink]

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New post 30 Nov 2016, 03:02
2
7
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A
B
C
D
E

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  25% (medium)

Question Stats:

70% (01:09) correct 30% (00:53) wrong based on 366 sessions

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What is the largest integer n such that 3^n < 0.001? [#permalink]

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New post 26 Feb 2017, 01:05
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Bunuel wrote:
What is the largest integer n such that 3^n < 0.001?

A. −13
B. −9
C. −7
D. 0
E. 3


Consider this pattern:

\(3^-1 = \frac{1}{3} = 0.3..\)
\(3^-2 = \frac{1}{9} = 0.1..\)
\(3^-3 = \frac{1}{27} = 0.03..\)
\(3^-4 = \frac{1}{81} = 0.01..\)

With this pattern:
\(3^-5 ~ 0.003\)
\(3^-6 ~ 0.001\)

\(3^-7 < 0.001\)
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Re: What is the largest integer n such that 3^n < 0.001? [#permalink]

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New post 30 Nov 2016, 04:02
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1
Bunuel wrote:
What is the largest integer n such that 3^n < 0.001?

A. −13
B. −9
C. −7
D. 0
E. 3


It's clearly that \(n<0\)
\(3^n<10^{-3} \implies \frac{1}{3^{|n|}}<\frac{1}{10^3} \implies 3^{|n|}>10^3 >9^3 = 3^6 \implies |n| \geq 7 \implies n \leq -7\)

The answer is C
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Re: What is the largest integer n such that 3^n < 0.001? [#permalink]

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New post 30 Nov 2016, 04:45
My answer is C.
Multiplying both side by 1000.
3^n *1000 < 1.
By looking at answer choice only we can eliminate two options i.e. 0 and 3.
Taking value n=-7 itself value just less than 0.001. i.e. 0.0004. As the value goes high i.e. 9 and 13 it will be far less than 0.001.
Hope my answer is right.
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Re: What is the largest integer n such that 3^n < 0.001? [#permalink]

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New post 24 Apr 2017, 06:08
1
Bunuel wrote:
What is the largest integer n such that 3^n < 0.001?

A. −13
B. −9
C. −7
D. 0
E. 3

Substitution of options:

(D) when n = 0,

\(3^n = 3^0 = 1 > 0.001\) X

(C) when n = -7,
3^(-7) = \((1/3)^7\) = (1/9)*(1/9)*(1/9)*(1/3) = (1/243)*(1/9) = (1/2187)

1/2187 < 1/1000 Answer
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What is the largest integer n such that 3^n < 0.001? [#permalink]

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New post 18 Mar 2018, 05:58
2
Bunuel wrote:
What is the largest integer n such that 3^n < 0.001?

A. −13
B. −9
C. −7
D. 0
E. 3


Given \(3^n<0.001\)
=> Multiplying & dividing RHS by 1000 we have
=> \(3^n<\frac{1}{1000}\)
=> OR \(1000<\frac{1}{3^n}\)
=> OR \(1000<3^{-n}\)

Therefore 'n' should be -ve so that the power of \(3^{-n}\) becomes +ve
Thus
=> if \(n=-6\) then \(3^{-(-6)}=3^6=729\)
=> if \(n=-7\) then \(3^{-(-7)}=3^7=2187\) SUFFICIENT
=> if \(n=-8\) then \(3^{-(-8)}=3^8=6561\)

Since 'n' is -ve largest integer value of 'n' that satisfy \(1000<3^{-n}\) is \(n=-7\)

Option "C"

Thanks
Dinesh
What is the largest integer n such that 3^n < 0.001?   [#permalink] 18 Mar 2018, 05:58
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