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

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If m^3-n^2=-300, then the lowest possible value of m is [#permalink]  01 Apr 2013, 00:22
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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
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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
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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).

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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).

Such an Easy approach , Bunuel . I succumbed to the time pressure . I wish I could think like you
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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
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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
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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 $$\sqrt[3]{-64} =-4$$.
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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
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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.

Bunuel whats wrong with this logic?
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]  11 Jul 2014, 01:07
$$m^3 - n^2 = -300$$

$$n^2 = 300 + m^3$$

$$5^3 = 125; & 10^3 > 300$$

So least value of m should be between -5 & -10

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