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# M28-14

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Intern
Joined: 04 Jun 2018
Posts: 20

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21 Sep 2018, 11:55
Bunuel wrote:
Official Solution:

If $$m$$ is a negative integer and $$m^3 + 380 = 381m$$, then what is the value of $$m$$?

A. $$-21$$
B. $$-20$$
C. $$-19$$
D. $$-1$$
E. $$None \ of \ the \ above$$

Given $$m^3 + 380 = 380m+m$$.

Re-arrange: $$m^3-m= 380m-380$$.

$$m(m+1)(m-1)=380(m-1)$$. Since $$m$$ is a negative integer, then $$m-1\neq{0}$$ and we can safely reduce by $$m-1$$ to get $$m(m+1)=380$$.

So, we have that 380 is the product of two consecutive negative integers: $$380=-20*(-19)$$, hence $$m=-20$$.

Hi Bunuel
I have a small conceptual doubt regarding this option.
When you say that we can safely eliminate x-1 on both sides, What does it exactly mean?
For example:
X^2=X. Here if we eliminate X on both sides, we say that X=0 is also one of the possible solutions.
Similarly, if we eliminate x-1 from this option, can we say that x-1=0 was one of the solutions of x but since x is a negative integer, X can't be negative.

I hope I have been able to convey my doubt.
Realllllu hoping for your reply on this concept. And if you can suggest some questions that test this concept, I will be really glad.

Regards
Nitesh
Math Expert
Joined: 02 Sep 2009
Posts: 50007

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22 Sep 2018, 02:45
nitesh50 wrote:
Bunuel wrote:
Official Solution:

If $$m$$ is a negative integer and $$m^3 + 380 = 381m$$, then what is the value of $$m$$?

A. $$-21$$
B. $$-20$$
C. $$-19$$
D. $$-1$$
E. $$None \ of \ the \ above$$

Given $$m^3 + 380 = 380m+m$$.

Re-arrange: $$m^3-m= 380m-380$$.

$$m(m+1)(m-1)=380(m-1)$$. Since $$m$$ is a negative integer, then $$m-1\neq{0}$$ and we can safely reduce by $$m-1$$ to get $$m(m+1)=380$$.

So, we have that 380 is the product of two consecutive negative integers: $$380=-20*(-19)$$, hence $$m=-20$$.

Hi Bunuel
I have a small conceptual doubt regarding this option.
When you say that we can safely eliminate x-1 on both sides, What does it exactly mean?
For example:
X^2=X. Here if we eliminate X on both sides, we say that X=0 is also one of the possible solutions.
Similarly, if we eliminate x-1 from this option, can we say that x-1=0 was one of the solutions of x but since x is a negative integer, X can't be negative.

I hope I have been able to convey my doubt.
Realllllu hoping for your reply on this concept. And if you can suggest some questions that test this concept, I will be really glad.

Regards
Nitesh

I think it's explained in the solution:

$$m(m+1)(m-1)=380(m-1)$$. Since $$m$$ is a negative integer, then $$m-1\neq{0}$$ and we can safely reduce by $$m-1$$ to get $$m(m+1)=380$$..

Another example: (x + 1)x = x + 1. The solution of this is: (x + 1)(x - 1) = 0 --> x = -1 or x = 1. So, we cannot reduce (x + 1)x = x + 1 by x + 1 because x + 1 can be 0 and we cannot divide by 0. If we do we'll loose the root, in this case x = -1. BUT if we were told there that x is positive, or say that x is not -1, then we'd have that x + 1 is NOT 0, which would allow us to reduce by x + 1 and get x = 1.

Hope it's clear.
_________________
Intern
Joined: 08 Aug 2018
Posts: 5
GMAT 1: 210 Q1 V1

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23 Sep 2018, 18:26
I think this is a high-quality question. Great question
Intern
Joined: 27 Nov 2017
Posts: 13

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15 Oct 2018, 20:40
I think this is a poor-quality question. Question is wrong
Math Expert
Joined: 02 Sep 2009
Posts: 50007

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15 Oct 2018, 21:03
Samakshkapoor92 wrote:
I think this is a poor-quality question. Question is wrong

Please elaborate your claim. The question is mathematically correct and to the highest GMAT standards.
_________________
Re: M28-14 &nbs [#permalink] 15 Oct 2018, 21:03

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# M28-14

Moderators: chetan2u, Bunuel

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