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M28-14

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Re: M28-14  [#permalink]

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New post 21 Sep 2018, 10:55
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 \(m=-20\).


Answer: B





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
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Joined: 02 Sep 2009
Posts: 52231
Re: M28-14  [#permalink]

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New post 22 Sep 2018, 01:45
1
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 \(m=-20\).


Answer: B





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.
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Re M28-14  [#permalink]

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New post 23 Sep 2018, 17:26
I think this is a high-quality question. Great question
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Re M28-14  [#permalink]

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New post 15 Oct 2018, 19:40
I think this is a poor-quality question. Question is wrong
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Re: M28-14  [#permalink]

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New post 15 Oct 2018, 20:03
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Re: M28-14  [#permalink]

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New post 26 Oct 2018, 05:16
siddhanthsivaraman wrote:
Hi Bunuel,
if this is the product of two consecutive negative integers,then isnt m=-19 instead of -20?(m+1=-20)




Yes.I think plugging numbers is the best approach here.
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Re: M28-14  [#permalink]

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New post 26 Oct 2018, 09:11
Plug-in the choices will be easy and we'll get to the answer fast.
and as aserghe1 mentioned last digit comparison of the LHS and RHS would be quite handy
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Re: M28-14  [#permalink]

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New post 11 Dec 2018, 21:01
TRIAL & ERROR

Plugging numbers-

m^3+380-381m=0

plug value m=-20
so,

(-20)^3+ 380-381(-20)=0
-8000+380+7620=0
-8000+8000= 0

so, m=-20
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Re: M28-14 &nbs [#permalink] 11 Dec 2018, 21:01

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