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M28-14

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Bunuel
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M28-14 [#permalink] New post   16 Sep 2014, 01:28
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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. \(None \ of \ the \ above\)
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B
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Bunuel
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Re M28-14 [#permalink] New post   16 Sep 2014, 01:28
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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. \(None \ of \ the \ above\)


To simplify the equation \(m^3 + 380 = 380m + m\), we can rearrange the terms such that the ones involving 380 are on one side.

\(m^3-m= 380m-380\).

Then, we can factor it:

\(m(m+1)(m-1)=380(m-1)\).

Since \(m\) is a negative integer, we know that \(m-1\neq{0}\), so we can safely divide both sides of the equation by it to get \(m(m+1)=380\).

Hence, we have that 380 is the product of two consecutive negative integers \(m\) and \(m+1\). By trial and error it's easy to find that \(380=-20*(-19)\), therefore, \(m=-20\).


Answer: B
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shasadou
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Re: M28-14 [#permalink] New post   25 Dec 2015, 01:41
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not having that algebraic gift i did it the following way: the stem simplifies to m(m^2-381)= -380, where we know that m is -ve therefore (m^2-381) should be +ve for the product to be a -ve number.

-1 and -19 rule out because they would leave a -ve results int he brackets and hence give +ve product which cannot be true. therefore we need to brute force options -20 and -21. start with -20 and you get the desired result
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BrentGMATPrepNow
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M28-14 [#permalink] New post   30 Mar 2018, 08:06
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Bunuel wrote:
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. \(None \ of \ the \ above\)


This can be solved with some TRICKY factoring.

Given: m³ - 381m + 380 = 0
Rewrite -381m as -m - 380m to get: m³ - m - 380m + 380 = 0
Factor IN PARTS to get: m(m² - 1) - 380(m - 1) = 0
Factor m² - 1 to get: m(m+1)(m-1) - 380(m-1) = 0
Collect "like" terms to get: (m-1)[ m(m+1) - 380] = 0
Simplify: (m-1)(m² + m - 380) = 0
Factor to get: (m-1)(m+20)(m-19) = 0

So, the possible m-values are 1, -20 and 19
The question tells us that m is NEGATIVE

So, m must equal -20

Answer: B

Cheers,
Brent
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Re: M28-14 [#permalink] New post   23 Sep 2018, 18:26
I think this is a high-quality question. Great question
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Bunuel
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Re: M28-14 [#permalink] New post   08 Mar 2023, 06:40
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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joe123x
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M28-14 [#permalink] New post   16 Mar 2023, 04:34
can we solve it in an alternative way?

such as: if we take -20^3 that gives us a negative number with zeros, and we add 380 and that leaves us with an ending of 20.
and since 381 ends with 1, when multiplied by -20 also gives us a number that ends with 20.

can I apply this logic?

wrong way of thinking. is there an alternative solution ?
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Bunuel
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Re: M28-14 [#permalink] New post   16 Mar 2023, 04:54
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joe123x wrote:
can we solve it in an alternative way?

such as: if we take -20^3 that gives us a negative number with zeros, and we add 380 and that leaves us with an ending of 20.
and since 381 ends with 1, when multiplied by -20 also gives us a number that ends with 20.

can I apply this logic?

wrong way of thinking. is there an alternative solution ?


You can check slightly different solution here or search for it in this thread.
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Re M28-14 [#permalink] New post   12 Aug 2023, 10:48
I think this is a high-quality question and I agree with explanation.
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Re: M28-14 [#permalink] New post   14 Oct 2023, 00:57
I have an alternative way of approaching this.
m^3 + 380 - 381m = 0
m(m^2 - 381) = -380
Since m is -ve, (m^2 - 381) must be positive to ensure RHS is -ve. 19^2 = 361 so we know m needs to be >19 for (m^2 - 381) to be positive.
And since m is an integer, we also know m^2 - 381 needs to be an integer and they are both in fact factors of 380. Most obvious choice is B. Then plug in the value to ensure the relationship holds!
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Re: M28-14 [#permalink] New post   15 Feb 2024, 01:07
­Bunuel, this is such a beautiful question!

I took away many learning point gems from it. I have a question.

Other than having the instincts of a mathematical genius like you, how did you see, or decide to look at the equation and decide, "Okay, I am going to break 381m into 380m + m, and then I will have the factor of (m-1) which I will be able to cancel, so I am left with consecutive integers"?

Thats like 2-3 steps ahead like chess, and I would also like to learn how to think like this, or if there were any hints or triggers to pick this lock. Did the 380 and 381 tip you off? Just for future questions like this.

Thanks!

 
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Bunuel
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Re: M28-14 [#permalink] New post   15 Feb 2024, 01:22
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CoffeeNCream wrote:
­Bunuel, this is such a beautiful question!

I took away many learning point gems from it. I have a question.

Other than having the instincts of a mathematical genius like you, how did you see, or decide to look at the equation and decide, "Okay, I am going to break 381m into 380m + m, and then I will have the factor of (m-1) which I will be able to cancel, so I am left with consecutive integers"?

Thats like 2-3 steps ahead like chess, and I would also like to learn how to think like this, or if there were any hints or triggers to pick this lock. Did the 380 and 381 tip you off? Just for future questions like this.

Thanks!


 

­When I saw the equation with 380 and 381m, I had a hunch that factoring might help. The cubic term and the number 380 seemed like clues. So, I thought about breaking down 381m into 380m and m. Then, noticing the difference of cubes pattern, I figured I could cancel out the common factor of (m-1). It's like solving a puzzle - you get better at it with practice and familiarity with algebraic techniques. So, recognizing patterns and using logical deduction guided me to my solution approach.
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Re: M28-14 [#permalink]
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