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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]
01 Apr 2013, 00:51

5

This post received KUDOS

Expert's post

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).

Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]
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]
01 Apr 2013, 06:40

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]
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 _________________

"When the going gets tough, the tough gets going!"

Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]
25 Apr 2013, 07:34

Expert's post

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

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. _________________

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

The question is not can you rise up to iconic! The real question is will you ?

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

Does it help?

AEL

gmatclubot

Re: If m^3-n^2=-300, then the lowest possible value of m is
[#permalink]
11 Jul 2014, 04:09