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If A, B, C, and D are integers such that A  C + B is even [#permalink]
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20 Sep 2011, 03:29
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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
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Last edited by Bunuel on 05 Mar 2013, 03:02, edited 1 time in total.
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Re: Even/Odd [#permalink]
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20 Sep 2011, 03:57
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jamifahad wrote: 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?
\(A + D\) \(B + D\) \(C + D\) \(A + B\) \(A + C\)
I am wondering if there is any matrix method to do these sort of Even/Odd question. If anybody knows about any such method, kindly post. ***THIS SOLUTION HAS FLAWS AS POINTED OUT BY TomB***Sorry, didn't use the matrix for this one. AC+B=even; means Either one of these Or two of these Or All three of them ARE EVEN. But, NOT all of them are ODD. D + B  A=odd Means, ALL of them are ODD That confirms; A, B, D are ODD; and from before, C must be EVEN. Ans: "C"
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Last edited by fluke on 21 Sep 2011, 09:46, edited 1 time in total.
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Re: Even/Odd [#permalink]
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21 Sep 2011, 01:33
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Hi Jamifahad,
Even matrix method takes more than two mins in this problem since you have to jot down the even and odd values.
Try subtracting the equations (A  C + B)  (D + B  A) = 2A(C+D)
Even  Odd = Odd
So 2A(C+D) is odd
2A is always even , so the subtrahend should always be odd, so C+D is odd. Hope I am clear.



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Re: Even/Odd [#permalink]
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21 Sep 2011, 09:17
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Hai fluke
How did you get that D,B,A are odd. My reasoning is as follows:
D + B A=odd D=odd and BA could be even . odd+even =odd.
BA= oddodd=even;eveneven=even. Then D is odd but A,B could be even or odd.
another possibility: D + B A=odd. so D= even and BA=odd, then eithe B or A could be either even or odd.
Therefore D=even, B and A are either even or odd.
please correct me if i am wrong



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Re: Even/Odd [#permalink]
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21 Sep 2011, 09:24
hai @abhi6001
can you explain 2A(C+D)=odd. how C+D is odd.
Thanks



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Re: Even/Odd [#permalink]
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21 Sep 2011, 09:45
TomB wrote: Hai fluke
How did you get that D,B,A are odd. My reasoning is as follows:
D + B A=odd D=odd and BA could be even . odd+even =odd.
BA= oddodd=even;eveneven=even. Then D is odd but A,B could be even or odd.
another possibility: D + B A=odd. so D= even and BA=odd, then eithe B or A could be either even or odd.
Therefore D=even, B and A are either even or odd.
please correct me if i am wrong My earlier post is wrong. thanks for pointing out. I'll have to use the matrix after all.
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Re: Even/Odd [#permalink]
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21 Sep 2011, 10:04
jamifahad wrote: 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?
\(A + D\) \(B + D\) \(C + D\) \(A + B\) \(A + C\)
Start with all possible combinations for A and B: ODD: O EVEN: E Now fill in the column 1, i.e. C's type with either odd or even; based on statement 1 Now fill in the column 4, i.e. D's type with either odd or even; based on statement 2 By POE, for each of the rows above: "C+D=ODD" Ans: "C"
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Re: Even/Odd [#permalink]
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21 Sep 2011, 10:31
C.
Even  Odd = Odd Therefore: AC+B  (D+BA) = ODD .. 2ADC = ODD ( now we know anything multiplied by 2 makes it Even.. so 2A cannot be the factor to cause this to be odd.. ) so we can simplfy it : (D+C) = ODD
Odd +/1 is ODD.. so D+C is odd!



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Re: Even/Odd [#permalink]
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21 Sep 2011, 14:16
We know
Even + Odd = Odd
So,
A  C + B = Even E  E + E = Even  1 O  O + O = Even  2
D + B  A = ODD 0 + E  E = ODD  1 E + O  O = ODD  2
So considering 1 and 2
If C is Even then D is odd & if C is Odd then D is Even
So C + D should always be odd



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Re: Even/Odd [#permalink]
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21 Sep 2011, 18:17
AC+B = Even D+BA = Odd is equivalent to A+B+C = Even A+B+D = Odd We are only interested in C and D now, and need only to consider two scenarios: A+B is odd or even.
A+B  C  D  Odd  Odd  Even  Even  Even  Odd 
C+D is always odd. c)



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Re: Even/Odd [#permalink]
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21 Sep 2011, 21:47
jamifahad wrote: 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?
\(A + D\) \(B + D\) \(C + D\) \(A + B\) \(A + C\)
I am wondering if there is any matrix method to do these sort of Even/Odd question. If anybody knows about any such method, kindly post. Looks good question.... We always need simple, easy and quick/fast approach: AC+B = EVEN ............I D+BA = ODD..............II Add I and II: AC+B+D+BA = EVEN+ODD 2BC+D = ODD No matter whether B is even or odd, 2B has to be an even number. If so, then remaining number, (DC), has to be odd. If (DC) is odd, (D+C) has also to be always odd. Therefore C.



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Re: Even/Odd [#permalink]
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21 Sep 2011, 23:50
subtracting and adding the two terms 2A (C+D) = odd meaning C+D = odd 2B  C+D = odd meaning (CD) = odd. not provided in the options. hence C it is.
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Re: Even/Odd [#permalink]
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09 May 2012, 14:32
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jamifahad wrote: 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?
\(A + D\) \(B + D\) \(C + D\) \(A + B\) \(A + C\)
I am wondering if there is any matrix method to do these sort of Even/Odd question. If anybody knows about any such method, kindly post. Make it plain and simple: A  C + B = even D + B  A = odd Add both: A  C + B + D + B  A = Even +odd 2B  C + D = odd 2B + (D  C) = odd Since 2B is even, (DC) must be odd. Alternatively, C could be odd or even and same is D but if C is odd, D must be even and vice versa. However it is not necessary to find out whether C is odd or even. If so, either (C+D) or (CD) is odd. So its C that \(C + D\) is always odd.



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Re: If A, B, C, and D are integers such that A  C + B is even [#permalink]
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05 Mar 2013, 02:47
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Bunuel /Karishma/ Experts, This is how I solved: A−C+B is even and D+B−A is odd Add them 2B + DC must be odd since 2B is even , DC must be odd. So C+D must be ODD :D Kudos if this helps anybody
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Re: If A, B, C, and D are integers such that A  C + B is even [#permalink]
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Re: If A, B, C, and D are integers such that A  C + B is even [#permalink]
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05 Mar 2013, 06:08
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Alternatively, you could also subtract both equations:
A C + B = EVEN D + B  A = ODD
Subtracting the 1st from the 2nd:
D +C  2A = EVEN + ODD
Or,
D + C = EVEN + EVEN + ODD = ODD



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Re: If A, B, C, and D are integers such that A  C + B is even [#permalink]
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05 Mar 2013, 06:14
caioguima wrote: Alternatively, you could also subtract both equations:
A C + B = EVEN D + B  A = ODD
Subtracting the 1st from the 2nd:
D +C  2A = EVEN + ODD
Or,
D + C = EVEN + EVEN + ODD = ODD yeah you can do that since these are equations. but know that u shud never subtract in case of inequalities.. . Kudos if this helps.
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Re: If A, B, C, and D are integers such that A  C + B is even [#permalink]
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03 Mar 2018, 09:10
jamifahad wrote: 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 Given: A  C + B = EVEN and D + B  A = ODDAdd them to get: ( A  C + B) + ( D + B  A) = EVEN + ODDSimplify: 2B + D  C = ODD Since 2B is ALWAYS even, we get: EVEN + D  C = ODD So: D  C = ODD  EVEN This means: D  C = ODD If D  C is ODD, then we know that one of the numbers is EVEN and the other number is ODD If one of the numbers is EVEN and the other number is ODD, then C + D must also be ODDAnswer: C Cheers, Brent
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Re: If A, B, C, and D are integers such that A  C + B is even
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