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If A, B, C, and D are integers such that A - C + B is even

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If A, B, C, and D are integers such that A - C + B is even [#permalink] New post 20 Sep 2011, 02:29
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 05 Mar 2013, 02:02, edited 1 time in total.
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Re: Even/Odd [#permalink] New post 20 Sep 2011, 02:57
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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.

A-C+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, 08:46, edited 1 time in total.
Added the warning after TomB pointed out error
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Re: Even/Odd [#permalink] New post 21 Sep 2011, 00:33
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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] New post 21 Sep 2011, 00:41
@abhi6001
Awesome buddy!
+1. and welcome.
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Re: Even/Odd [#permalink] New post 21 Sep 2011, 08:17
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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 B-A could be even . odd+even =odd.

B-A= odd-odd=even;even-even=even. Then D is odd but A,B could be even or odd.

another possibility: D + B- A=odd. so D= even and B-A=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] New post 21 Sep 2011, 08:24
hai @abhi6001

can you explain 2A-(C+D)=odd. how C+D is odd.

Thanks
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Re: Even/Odd [#permalink] New post 21 Sep 2011, 08:45
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 B-A could be even . odd+even =odd.

B-A= odd-odd=even;even-even=even. Then D is odd but A,B could be even or odd.

another possibility: D + B- A=odd. so D= even and B-A=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] New post 21 Sep 2011, 09:04
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




CABD
OE
EO
OO
EE


Now fill in the column 1, i.e. C's type with either odd or even; based on statement 1




CABD
OOE
OEO
EOO
EEE


Now fill in the column 4, i.e. D's type with either odd or even; based on statement 2




CABD
OOEE
OEOE
EOOO
EEEO


By POE, for each of the rows above:
"C+D=ODD"

Ans: "C"
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Re: Even/Odd [#permalink] New post 21 Sep 2011, 09:31
C.

Even - Odd = Odd
Therefore:
A-C+B - (D+B-A) = ODD
.. 2A-D-C = 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] New post 21 Sep 2011, 13:16
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] New post 21 Sep 2011, 17:17
A-C+B = Even
D+B-A = 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+BCD
OddOddEven
EvenEvenOdd


C+D is always odd.

c)
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Re: Even/Odd [#permalink] New post 21 Sep 2011, 20:47
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:

A-C+B = EVEN ............I
D+B-A = ODD..............II

Add I and II:

A-C+B+D+B-A = EVEN+ODD
2B-C+D = ODD

No matter whether B is even or odd, 2B has to be an even number. If so, then remaining number, (D-C), has to be odd.

If (D-C) is odd, (D+C) has also to be always odd.

Therefore C.
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Re: Even/Odd [#permalink] New post 21 Sep 2011, 22:50
subtracting and adding the two terms
2A- (C+D) = odd meaning C+D = odd

2B - C+D = odd meaning (C-D) = odd. not provided in the options.

hence C it is.
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Re: Even/Odd [#permalink] New post 09 May 2012, 13:32
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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, (D-C) 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 (C-D) 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] New post 05 Mar 2013, 01:47
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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 + D-C must be odd

since 2B is even , D-C must be odd.

So C+D must be ODD :D :-D :-D :-D :lol: :lol:

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] New post 05 Mar 2013, 02:06
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Sachin9 wrote:
Bunuel /Karishma/ Experts,
This is how I solved:

A−C+B is even and D+B−A is odd

Add them

2B + D-C must be odd

since 2B is even , D-C must be odd.

So C+D must be ODD :D :-D :-D :-D :lol: :lol:

Kudos if this helps anybody


Yes, your approach is correct. +1.
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Re: If A, B, C, and D are integers such that A - C + B is even [#permalink] New post 05 Mar 2013, 05:08
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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] New post 05 Mar 2013, 05:14
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] New post 23 Jul 2014, 21:09
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Re: If A, B, C, and D are integers such that A - C + B is even   [#permalink] 23 Jul 2014, 21:09
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