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Re M2814 [#permalink]
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16 Sep 2014, 01:28



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Re: M2814 [#permalink]
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20 Oct 2014, 22:44
If the question statement is "m is a negative integer and m^3 + 380 = 381m", how did you derive that "given m^3 + 380 = 380m+m." Is this a typo and there is a +m missing?



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21 Oct 2014, 01:10



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Re: M2814 [#permalink]
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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^2381)= 380, where we know that m is ve therefore (m^2381) 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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Re: M2814 [#permalink]
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29 Aug 2016, 05:14
Hi Bunuel, if this is the product of two consecutive negative integers,then isnt m=19 instead of 20?(m+1=20)



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29 Aug 2016, 05:26



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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 \(m^3 + 380 = 381m\). Rearrange: \(m^3m= 380m380\). \(m(m+1)(m1)=380(m1)\). Since \(m\) is a negative integer, then \(m1\neq{0}\) and we can safely reduce by \(m1\) 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 Great explanation as always. However, I approached it differently. \(m^3 + 380 = 381m\) \(m(m^2381)=380\) then \(m^2380=\frac{380}{m}\) next \(m^2=381\frac{380}{m}\), and now we plug in the answer choices and only (C) 19 gives us the right answer What do you think Bunuel?



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Re: M2814 [#permalink]
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17 Jun 2017, 10:50
MikeMighty 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 \(m^3 + 380 = 381m\). Rearrange: \(m^3m= 380m380\). \(m(m+1)(m1)=380(m1)\). Since \(m\) is a negative integer, then \(m1\neq{0}\) and we can safely reduce by \(m1\) 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 Great explanation as always. However, I approached it differently. \(m^3 + 380 = 381m\) \(m(m^2381)=380\) then \(m^2380=\frac{380}{m}\) next \(m^2=381\frac{380}{m}\), and now we plug in the answer choices and only (C) 19 gives us the right answer What do you think Bunuel? Yes, you can use plugin method to solve this question.
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Re: M2814 [#permalink]
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18 Jun 2017, 06:56
MikeMighty wrote: Great explanation as always. However, I approached it differently. \(m^3 + 380 = 381m\) \(m(m^2381)=380\) then \(m^2380=\frac{380}{m}\) next \(m^2=381\frac{380}{m}\), and now we plug in the answer choices and only (C) 19 gives us the right answerWhat do you think Bunuel? Hi Mike, I would like to highlight that the correct answer is B not C. The right answer is 20. I did the same like you till a point and used other reasoning. Usign sense of numbers and memorizing some square of integers a re perfect. I will elaborate. \(m(m^2381)=380\).........I need two numbers that gives 380 m^2 can't me less than or equal to 381 otherwise, it will be as follows: Case 1:Negative ( Positive less 381  381) = positive value .......Incorrect Case 2:Negative ( Positive equal to 381  381) = 0 .......Incorrect D) 1 This will give you case 1......Eliminate D C ) 19 m^2 = 381...So it is case 2.....Eliminate C B) 20 20 ( 400 381)= 20 * 19 = 380.........Correct Answer B



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Re: M2814 [#permalink]
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03 Jul 2017, 22:55
hi brunel could you pls clear how to elaborate:
m3−m = m(m+1)(m−1)



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Re: M2814 [#permalink]
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03 Jul 2017, 23:11



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Re: M2814 [#permalink]
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28 Sep 2017, 08:04
Is this approach correct?
Divide both side by M m^2+ 380/m=381
Then plug in numbers and check.



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Re: M2814 [#permalink]
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28 Sep 2017, 08:06



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I did this a bit differently though not sure if that was fluke or not 
On the Right side we have 
381m = which would clearly be a multiple of m
On the left we have \(m^3\) + 380. Now \(m^3\) is clearly a multiple of m too so 380 must be a multiple of m too.
Now test for are 20, 19 and 1. 1 and 19 dont work when used in the equation. 20 works. Hence B.
Is this the right approach?



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Re: M2814 [#permalink]
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18 Mar 2018, 03:11
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. \(None \ of \ the \ above\) HI Bunuel, \(m^3 + 380 = 381m\) ==> \(m^3381m=380\) \(m(m^2381)=380\)==> m=380 or \((m^2381)=380\) How to proceed further?
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NandishSS wrote: 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. \(None \ of \ the \ above\) HI Bunuel, \(m^3 + 380 = 381m\) ==> \(m^3381m=380\) \(m(m^2381)=380\)==> m=380 or \((m^2381)=380\) How to proceed further? HI GMATPrepNow / Brent Can you pls help me with above query?
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Re: M2814 [#permalink]
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30 Mar 2018, 08:06
NandishSS wrote: NandishSS wrote: 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. \(None \ of \ the \ above\) HI Bunuel, \(m^3 + 380 = 381m\) ==> \(m^3381m=380\) \(m(m^2381)=380\)==> m=380 or \((m^2381)=380\) How to proceed further? HI GMATPrepNow / Brent Can you pls help me with above query? 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)( m1)  380( m1) = 0 Collect "like" terms to get: (m1)[ m(m+1)  380] = 0 Simplify: (m1)(m² + m  380) = 0 Factor to get: (m1)(m+20)(m19) = 0 So, the possible mvalues 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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