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

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guerrero25
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If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   01 Apr 2013, 01:22
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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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Bunuel
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   01 Apr 2013, 01:51
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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).

Answer: C.
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   01 Apr 2013, 03:02
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 :wink:
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   01 Apr 2013, 07:40
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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


\(m^3-n^2=-300\)
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] New post   25 Apr 2013, 06:57
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] New post   25 Apr 2013, 08:34
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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] New post   25 Apr 2013, 08:53
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] New post   10 Jul 2014, 14:07
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] New post   11 Jul 2014, 02:07
\(m^3 - n^2 = -300\)

Adjusting the -ve signs

\(n^2 = 300 + m^3\)

\(5^3 = 125; & 10^3 > 300\)

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

Answer = C
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   11 Jul 2014, 05:09
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
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   05 Aug 2015, 20:56
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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
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   08 Apr 2016, 19:23
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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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   15 Apr 2016, 02:39
Here the value of M must lie between -6 and -7
Go ahead push that C :)
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   18 Jul 2018, 10:39
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.




Bunuel hello there :) how are you ? :-) can you please explain how after this \(m^3-n^2=-300\) you get this \(m=\sqrt[3]{n^2-300}\). Exponent 3 is outside of darical sign and exponent 2 is inside radical sign :?
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   18 Jul 2018, 10:42
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dave13 wrote:
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.




Bunuel hello there :) how are you ? :-) can you please explain how after this \(m^3-n^2=-300\) you get this \(m=\sqrt[3]{n^2-300}\). Exponent 3 is outside of darical sign and exponent 2 is inside radical sign :?


\(m^3-n^2=-300\);

\(m^3=n^2-300\);

\(m=\sqrt[3]{n^2-300}\).
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If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   18 Jul 2018, 11:03
Bunuel from here \(m^3=n^2-300\) how do you get ths \(m=\sqrt[3]{n^2-300}\) :? what are you doing such that exponent 3 goes to the right :) please help :-)
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   18 Jul 2018, 11:05
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dave13 wrote:
Bunuel from here \(m^3=n^2-300\) how do you get ths \(m=\sqrt[3]{n^2-300}\) :? what are you doing such that exponent 3 goes to the right :) please help :-)


Take the cube root. The same way we get \(x=\sqrt[3]{y}\) from \(x^3 = y\).
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   18 Jul 2018, 12:12
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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


\(m^3-n^2=-300\)
\(m^3 + 300 = n^2\)

To minimize the value of \(m^3\), we must minimize the right side of the equation in blue.
Since the square of a value cannot be negative, the least possible option for the right side is 0:
\(m^3 + 300 = 0\)
\(m^3 = -300\)

Since \(-5^3 = -125\) and \(-10^3 = -1000\), m must be between -10 and -5.

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C
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   19 Jul 2018, 00:46
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



\(n^2\) =\(m^3\)+300
since L.H.S (+), R.H.S should be positive
so option (a), (b) is wrong
from (c) (d) and (e) lowest possible value of m
we can get from option c
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If m^3-n^2=-300, then the lowest possible value of m is [#permalink] New post   24 Jul 2018, 02:33
When we look at the answer choices, we can see that A and B is too high to get -300 even if we take n=0. Thus eliminate A and B. Since we need the lowest value of m assume that n=0 (it will help minimize m as the higher absolute value of a negative number the lower that number, thus the case, in which m^3= -300 gives the possible lowest value for m ). We get m = -6, smth, which is in the range of -10 and -5. Hence C.




Hope it helps!
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