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If n = (-72)^(1/3) then the value of n is

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If n = (-72)^(1/3) then the value of n is [#permalink]

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Re: If n = (-72)^(1/3) then the value of n is [#permalink]

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New post 18 Jun 2017, 04:05
Bunuel wrote:
If \(n = (-72)^{(\frac{1}{3})}\) then the value of n is

A. -9 < n < -8
B. -8 < n < -7
C. -7 < n < -6
D. -6 < n < -5
E. -5 < n < -4


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https://gmatclub.com/forum/the-value-of ... 86290.html
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Re: If n = (-72)^(1/3) then the value of n is [#permalink]

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Let m= (-72)^1/3
Hence m^3 = -72

Lets check

-1^3 => -1
-2^3 => -8
-3^3 => -27
-4^3 => -64
-5^3 => -125

Hence m must be between (-5,-4)
Hence E.
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Re: If n = (-72)^(1/3) then the value of n is [#permalink]

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New post 18 Jun 2017, 05:21
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Bunuel wrote:
If \(n = (-72)^{(\frac{1}{3})}\) then the value of n is

A. -9 < n < -8
B. -8 < n < -7
C. -7 < n < -6
D. -6 < n < -5
E. -5 < n < -4


\(n = (-72)^{(\frac{1}{3})}\)

Cubing both sides, we get;

\(n^3 = (-72)\)

\(-4^3 = -64\)

\(-5^3 = -125\)

Therefore value of n has to be between, -5 and -4.

\(-5 < n < -4\). Answer (E) ...

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Re: If n = (-72)^(1/3) then the value of n is [#permalink]

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New post 18 Jun 2017, 06:04
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Bunuel wrote:
If \(n = (-72)^{(\frac{1}{3})}\) then the value of n is

A. -9 < n < -8
B. -8 < n < -7
C. -7 < n < -6
D. -6 < n < -5
E. -5 < n < -4


Find some "nice" third powers (aka cube roots)

\((-1)^{(\frac{1}{3})} = -1\)
\((-8)^{(\frac{1}{3})} = -2\)
\((-27)^{(\frac{1}{3})} = -3\)
\((-64)^{(\frac{1}{3})} = -4\)
\((-125)^{(\frac{1}{3})} = -5\)

Since -72 is BETWEEN -64 and -125, \((-72)^{(\frac{1}{3})}\) will be BETWEEN -4 and -5

Answer:
[Reveal] Spoiler:
E


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Re: If n = (-72)^(1/3) then the value of n is [#permalink]

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New post 18 Jun 2017, 06:04
Since \(n = (-72)^{(\frac{1}{3})}\) ,n is definitely negative.

72, when prime-factorized will give \(2^3 * 3^2\)
Cube root of \(2^3\)(8)is 2
Cube root of \(3^2\)(9) is a little more than 2

Hence \((-72)^{(\frac{1}{3})}\) is a little more than -4
This comes in the range -5 < n < -4(Option E)
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Re: If n = (-72)^(1/3) then the value of n is [#permalink]

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New post 18 Jun 2017, 10:20
IMO, E.

Rewritten the question becomes what is the cube root of 72?

4^3 = 64, so N is greater than 4

5^3 = 125, so N is less than 5

E) N is between -4 and -5

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Re: If n = (-72)^(1/3) then the value of n is [#permalink]

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Hi All,

While you will almost certainly have to deal with a square-root a couple of times on Test Day (in both Arithmetic and Geometry questions), cube-roots are relatively rare (you probably won't see them at all on Test Day). That having been said, cube-roots are based on the same general math concepts as square-roots.

For example...
The square-root of 4 is +2 ... since (2)(2) = 4
The cube-root of 27 is +3 ... since (3)(3)(3) = 27
The cube-root of -27 is -3 ... since (-3)(-3)(-3) = -27

In this question, we're asked for the cube-root of -72, but the solution will NOT be an integer (notice how the answer choices are all RANGES - that's a hint that N is not an integer AND that we don't actually have to find the exact value of N; we just have to figure out what numbers it is between). Thus, we have to look for cubes of negative integers:

(-3)^3 = -27
(-4)^3 = -64
(-5)^3 = -125

Since -72 is between -64 and -125, the cube-root of -72 has to be between -4 and -5

Final Answer:
[Reveal] Spoiler:
E


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Re: If n = (-72)^(1/3) then the value of n is [#permalink]

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New post 11 Jan 2018, 13:51
Bunuel wrote:
If \(n = (-72)^{(\frac{1}{3})}\) then the value of n is

A. -9 < n < -8
B. -8 < n < -7
C. -7 < n < -6
D. -6 < n < -5
E. -5 < n < -4


We need to find two integers, one of which,, when raised to the 3rd power, is less than -72, and the other of which, when raised to the 3rd power, is greater than -72.

Since -4^3 = -64 and -5^3 = -125, and since -72 is between those two numbers, n must be between -4 and -5.

Answer: E
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Re: If n = (-72)^(1/3) then the value of n is   [#permalink] 11 Jan 2018, 13:51
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