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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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18 Jun 2017, 03:57
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82% (01:21) correct 18% (01:44) wrong based on 267 sessions

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

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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

Similar questions:
https://gmatclub.com/forum/new-tough-an ... l#p1029224
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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18 Jun 2017, 04:30
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1
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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18 Jun 2017, 05:21
1
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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18 Jun 2017, 06:04
1
Top Contributor
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

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

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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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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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04 Dec 2017, 12:50
1
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

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

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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.

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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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