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Re M2004
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16 Sep 2014, 00:07
Official Solution:If \(A\), \(B\), \(C\), and \(D\) are integers such that \(A  C + B\) is even and \(D + B  A\) is odd, which of the following expressions is always odd? A. \(A + D\) B. \(B + D\) C. \(C + D\) D. \(A + B\) E. \(A + C\) Rewrite \(A  C + B\) as \((A + B)  C\) and \(D + B  A\) as \(D + (B  A)\). If \(A + B\) is even, \(B  A\) is even too. If \(A + B\) is odd, \(B  A\) is odd too. Now it follows from the stem that:  if \(A + B\) is even, \(C\) is even and \(D\) is odd
 if \(A + B\) is odd, \(C\) is odd and \(D\) is even
Thus, \(C + D\) is always odd. Answer: C
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Re: M2004
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20 Dec 2014, 18:54
I think I may have found a much more simple methodology to answer this question:
(1) A  C + B = even; thus, A  C + B = 2k (where k is an integer)
(2) D + B  A = odd; thus, D + B  A = 2k + 1 (where k is an integer)
rearrange (2): D + B  A  1 = 2k
Sub: rearranged (2) into (1): A  C + B = D + B  A  1 > 2A + 1 = C + D
thus, per my method above, if C + D = 2A + 1 (where A is an integer), C + D is ALWAYS ODD



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Re M2004
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08 Jan 2015, 09:41
I think this question is good and helpful.



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Re: M2004
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15 Apr 2015, 08:06
bartone89 I believe your method is wrong because it assumes the values of the two equations are consecutive. ie 2k and 2k+1. I hope I havent misunderstood you solution. Anyway, I think I may have found a better method. The two equations are AC+B and D+BA, even and odd respectively. On subtracting the two, you get, AC+BDB+A = 2ACD (which is odd, since even  odd = odd) 2A(C+D) is odd. Which means C+D is odd since we know that 2A has to be even. C+D is one of the options. Hope this makes sense.



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Re: M2004
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19 Jan 2016, 05:25
There is an easier way this question can be solved. Lets say AC+B = 2n (even) and after rearranging the second equation, A+D+B=2m+1 (odd) Adding the two equations we get, 2B+DC=2(m+n)+1 From this, 2B is even and 2(m+n) is even, hence we conclude, DC=odd When difference of two integers is odd, there sum will also be odd; Hence D+C = odd
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Re: M2004
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20 Feb 2016, 12:58
I solved it in a different way: add both expressions AC+B+D+BA=odd. ok, rewrite everything as: 2B+DC=odd 2B is always even, it must bean only 1 thing: DC is always odd. if DC is odd, then D+C is always odd.



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Re: M2004
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26 Apr 2016, 17:25
mvictor wrote: I solved it in a different way: add both expressions AC+B+D+BA=odd. ok, rewrite everything as: 2B+DC=odd 2B is always even, it must bean only 1 thing: DC is always odd. if DC is odd, then D+C is always odd. Very good solution mvictor!



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Re: M2004 I solved it with a different approach! Just Sharing
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27 Nov 2016, 05:40
Add or Substract the two equations
AC+B=Even D+BA=Odd =2A+CD (EvenOdd is always Odd) and we know 2A is always even
Therefore CD should be Odd and thus also C+D



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Re: M2004
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06 May 2017, 13:56
y4nkee wrote: mvictor wrote: I solved it in a different way: add both expressions AC+B+D+BA=odd. ok, rewrite everything as: 2B+DC=odd 2B is always even, it must bean only 1 thing: DC is always odd. if DC is odd, then D+C is always odd. Very good solution mvictor! I agree very good solution. But it still requires timing luck or incredible number sense. I solved it through plugging in values. My method take no number sense needed and does not require too much logic. AC+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 12+3 = 2 which is even. D+BA is odd. Reuse the same variables from before. D + 3  1 is D + 2. Set D to an easy odd number which is 5. So D = 5. Now you have values for A through D or (A,B,C,D) = (1,3,2,5). Work through the answers. A. A+D = 1 + 5 = 6 B. B+D = 3 + 5 = 8 C. C+D = 2 + 5 = 7. This is odd. No need to do the rest, the answer is C. D. A+B E. A+C



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Re: M2004
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08 Jun 2017, 07:06
Given AC+B =Even and D+BA=Odd Subtract 1) and 2) 2ACD=Evenodd=Odd=> 2A=Odd (C+D) Left side of the equation is even hence C+D has to be Odd.



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Re: M2004
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25 Jun 2017, 03:01
Given AC+B=Even1 & D+BA = Odd2 Add 1 & 2 even + odd = odd hence CD= 2B  Odd= Odd if CD is odd, then C+D is also Odd. (eg 43=1, 4+3=7)



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Re: M2004
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18 Sep 2017, 18:15
+Ac+B= even D+BA = odd Now equation 1  equation2 which gives 2A(D+C)=odd 2A is always even so (D+C) is always odd Hence C



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Re: M2004
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20 Nov 2017, 15:41
joondez wrote: y4nkee wrote: mvictor wrote: I solved it in a different way: add both expressions AC+B+D+BA=odd. ok, rewrite everything as: 2B+DC=odd 2B is always even, it must bean only 1 thing: DC is always odd. if DC is odd, then D+C is always odd. Very good solution mvictor! I agree very good solution. But it still requires timing luck or incredible number sense. I solved it through plugging in values. My method take no number sense needed and does not require too much logic. AC+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 12+3 = 2 which is even. D+BA is odd. Reuse the same variables from before. D + 3  1 is D + 2. Set D to an easy odd number which is 5. So D = 5. Now you have values for A through D or (A,B,C,D) = (1,3,2,5). Work through the answers. A. A+D = 1 + 5 = 6 B. B+D = 3 + 5 = 8 C. C+D = 2 + 5 = 7. This is odd. No need to do the rest, the answer is C. D. A+B E. A+C the thing is  when presented with such expressions, you have to either add or subtract one from another. the thing is, you have 3 variables in each expression, therefore, you might lose some extra time to test all the possible options... my approach is very direct and fast though...and I always have the same approach to solving similar questions.



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Re: M2004
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31 May 2018, 07:29
The basic Rule to solve this question is if A+B = even then AB will also be even and vice versa Same way if A+B is odd the AB will also be odd and vice versa Now AC+B = Even >1 D+BA = Odd >2 adding 1 & 2 we have DC +2B = Odd DC = Odd 2B DC = Odd
hence D+C is odd.



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Re M2004
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09 Jun 2018, 00:02
How to approach question in this way. There can be multiple other combinations to start with.



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Re: M2004
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11 Jun 2018, 06:53
A  C + B = even A+D +B = ODD  D  C + 2B = ODD > 2B is always odd.
so DC = ODD Therefore, D+C = ODD
Please let me know if the approach is correct and can be followed
Thanks & Warm Regards, Richin



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Re: M2004
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23 Oct 2018, 18:48
Bunuel is there a way i can find similar problems? i tried searching for similar prblms but all i could find are gmatclub test problems.... please suggest a way... and i request a way to find similar topics ...youre too kind to provide the links but i cannot trouble every single time...



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Re: M2004
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23 Oct 2018, 19:38










