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If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15 (B) -15 and -10 (C) -10 and -5 (D) -5 and 0 (E) 0 and 5

\(m^3-n^2=-300\) --> \(m=\sqrt[3]{n^2-300}\). To minimize m we should minimize n^2. The lowest value of n^2 is 0, thus the lowest value of m is \(m_{min}=\sqrt[3]{-300}\).

m is less than -5 (since (-5)^3=-125) and more than -10 (since (-10)^3=-1000).

Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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01 Apr 2013, 06:40

2

This post received KUDOS

guerrero25 wrote:

If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15 (B) -15 and -10 (C) -10 and -5 (D) -5 and 0 (E) 0 and 5

\(m^3-n^2=-300\) So, \(m^3 = n^2 - 300\) For \(m^3\) to be minimum, \((n^2 - 300)\) must be minimum For \((n^2 - 300)\) to be minimum, \(n^2\) must be minimum, so \(n^2\) = 0 So \(m^3\) = -300 So m = -6. ..... So m lies between -10 and -5

Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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01 Apr 2013, 02:02

Bunuel wrote:

guerrero25 wrote:

If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15 (B) -15 and -10 (C) -10 and -5 (D) -5 and 0 (E) 0 and 5

\(m^3-n^2=-300\) --> \(m=\sqrt[3]{n^2-300}\). To minimize m we should minimize n^2. The lowest value of n^2 is 0, thus the lowest value of m is \(m_{min}=\sqrt[3]{-300}\).

m is less than -5 (since (-5)^3=-125) and more than -10 (since (-10)^3=-1000).

Answer: C.

Such an Easy approach , Bunuel . I succumbed to the time pressure . I wish I could think like you

Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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25 Apr 2013, 05:57

Bunuel, pls help. how are we considering m as minimum with cube root of -300, is that not an unreal number (negative root)? I was considering the cube root for the lowest positive value of m
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"When the going gets tough, the tough gets going!"

Bunuel, pls help. how are we considering m as minimum with cube root of -300, is that not an unreal number (negative root)? I was considering the cube root for the lowest positive value of m

Even roots from negative number is undefined on the GMAT: \(\sqrt[{even}]{negative}=undefined\), for example \(\sqrt{-25}=undefined\).

Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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25 Apr 2013, 07:53

sdas wrote:

Bunuel, pls help. how are we considering m as minimum with cube root of -300, is that not an unreal number (negative root)? I was considering the cube root for the lowest positive value of m

To add to what Bunnuel said. Try to think in reverse. You can always multiply a negative number 3 times to get an odd number, but you cannot multiply a negative number 2 times to get a negative number
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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10 Jul 2014, 08:06

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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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10 Jul 2014, 13:07

The way I intepreted this problem, is m can be any negative number. M does not have to be an integer. I immediately chose the greatest negative range as the answer because I figured I could offset it with some (n^2) to equal -300. For example if I chose m to be -20 than (m^(3)) would be -8000. and I would find a number that for (n2) that is equal to 7970.

Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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11 Jul 2014, 04:09

bankerboy30 wrote:

The way I intepreted this problem, is m can be any negative number. M does not have to be an integer. I immediately chose the greatest negative range as the answer because I figured I could offset it with some (n^2) to equal -300. For example if I chose m to be -20 than (m^(3)) would be -8000. and I would find a number that for (n2) that is equal to 7970.

Bunuel whats wrong with this logic?

Hi Bankerboy30,

In your case, you would need to find n such that square of n would equal -7700 (300-8000). Now, we know that square of a real number cannot be negative and we don't deal with imaginary numbers in GMAT.

So, you need to go by a limitation that square of n can be minimum ZERO, not less than that. If you use that, you will get the answer as Bunuel got.

Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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08 Apr 2016, 18:23

guerrero25 wrote:

If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15 (B) -15 and -10 (C) -10 and -5 (D) -5 and 0 (E) 0 and 5

m will be minimum when n=0, otherwise by deducting a positive number, the negative will get even bigger. m^3 = 300 ok... -5x-5x-5=-125..so clearly can be lower than -5. D and E are out. -10x-10x-10=-1000 clearly not lower than -10. A and B out. C remains.

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