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Is A + B + C even?

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Is A + B + C even? [#permalink]

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Is A + B + C even?

(1) A - C - B is even

(2) (A - C)/B is odd
[Reveal] Spoiler: OA

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Is A + B + C even? [#permalink]

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Is A + B + C even?

Note that we are not told that a, b and c are integers.


(1) \(a-c-b=even\), if the variables are integers then \(a+b+c\) will be even but if they are not: \(a=3.5\), \(b=1\), \(c=0.5\) --> \(a-c-b=2=even\), but \(a+b+c=5=odd\). Not sufficient.


(2) \(\frac{a-c}{b}=odd\). The same here: if the variables are integers then \(a+b+c\) will be even but if they are not: \(a=3.5\), \(b=1\), \(c=0.5\) --> \(\frac{a-c}{b}=3=odd\), but \(a+b+c=5=odd\). Not sufficient


(1)+(2) \(a+b+c\) may or may not be even (again if variables are integers: YES but if \(a=3.5\), \(b=1\), \(c=0.5\) answer is NO). Not sufficient.


Answer: E.
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Re: Even/Odd [#permalink]

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New post 17 Sep 2011, 14:22
statement 1 ) A - c -b is even

let A = 10.5, B =2 and c =4.5

10.5 - 4.5 -2 = 4 =even, but A+B+C = 17 (NOT EVEN.)

Pick A,B,C AS 2 THEN A- B- C = -2 = even , A+B+C =6 (EVEN )

STATEMENT 1 INSUFFICIENT.

Statement 2 ) (A-C)/b = odd

again, let A = 10.5, B =2 and c =4.5

(10.5-4.5)/2 = 3 =ODD, AND A+B+C = 17 = ODD

BUT Let A= 8, B= 2 And C =2

> (A-C)/B = 3 = ODD, BUT A + B +C = 12 = EVEN

STATEMENT 2 INSUFFICIENT.


the same values as i have listed above will also prove combining both statements dont provide any definitive answer either.

Answer E

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Re: Is A+B+C even? 1) A-C-B is even 2) (A-C)/B is odd [#permalink]

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New post 16 Dec 2011, 21:49
E it is.

I tried to solve algebraically. However, I have a question.

If it were mentioned that all three numbers were integers, how would the answer change? How to figure this out algebraically? (Without having to pick numbers)
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Re: Is A+B+C even? 1) A-C-B is even 2) (A-C)/B is odd [#permalink]

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New post 17 Dec 2011, 11:13
i dont an algebric way to do it but here is a simple permutation on A,B,C "if given is that A, B , C are integers" .

then option 1 A-B-C is even only in below cases
A-C-B
E-E-E
E-O-O (E=Even, O=Odd)
O-O-E
O-E-O
in all cases A+ B +c will be even .

stmt 2: A-B/C is odd

A/C - B/C should be odd.
E O
O E


A/C can be even in both cases E/E or E/O eg: 4/2 or 6/3.
B/C can be even in both cases E/E or E/O eg: 4/2 or 6/3.

in the same way we can do for Odd case .

i.e. this stmt is insuff.
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Is A + B + C even? [#permalink]

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Is A + B + C even?

(1) A - C - B is even

(2) (A - C)/B is odd

Last edited by Bunuel on 29 Jul 2012, 06:01, edited 2 times in total.
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Re: Odd/Even (Ultra hard) from Math Flashcard [#permalink]

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New post 08 Apr 2012, 07:41
I think the OA is incorrect.

1)A-B-C can only be even, if either all are even or two are odd and one is even, in both cases a+b+c is even.

2) (A-C)/B can be odd if all are even or if all are odd, so A+B+C is even or odd, which makes the statement insufficient.

IMO A is correct.

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Re: Is A + B + C even? (1) A - C - B is even (2) (A - C)/B is [#permalink]

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catty2004 wrote:
Is A + B + C even?

(1) A - C - B is even

(2) (A - C)/B is odd


Notice that we are not told that a, b and c are integers.

Q: is \(a+b+c=even\)?

(1) \(a-c-b=even\), if the variables are integers then \(a+b+c\) will be even but if they are not, for example if \(a=3.5\), \(b=1\), \(c=0.5\) then \(a-c-b=2=even\), but \(a+b+c=5=odd\). Not sufficient.

(2) \(\frac{a-c}{b}=odd\). The same here: if the variables are integers then \(a+b+c\) will be even but if they are not for example if \(a=3.5\), \(b=1\), \(c=0.5\) then \(\frac{a-c}{b}=3=odd\), but \(a+b+c=5=odd\). Not sufficient


(1)+(2) \(a+b+c\) may or may not be even (again if variables are integers: YES but if \(a=3.5\), \(b=1\), \(c=0.5\) answer is NO). Not sufficient.

Answer: E.
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Re: Is A + B + C even? (1) A - C - B is even (2) (A - C)/B is [#permalink]

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sanjoo wrote:
Bunuel wrote:
catty2004 wrote:
Is A + B + C even?

(1) A - C - B is even

(2) (A - C)/B is odd


Notice that we are not told that a, b and c are integers.

Q: is \(a+b+c=even\)?

(1) \(a-c-b=even\), if the variables are integers then \(a+b+c\) will be even but if they are not, for example if \(a=3.5\), \(b=1\), \(c=0.5\) then \(a-c-b=2=even\), but \(a+b+c=5=odd\). Not sufficient.

(2) \(\frac{a-c}{b}=odd\). The same here: if the variables are integers then \(a+b+c\) will be even but if they are not for example if \(a=3.5\), \(b=1\), \(c=0.5\) then \(\frac{a-c}{b}=3=odd\), but \(a+b+c=5=odd\). Not sufficient


(1)+(2) \(a+b+c\) may or may not be even (again if variables are integers: YES but if \(a=3.5\), \(b=1\), \(c=0.5\) answer is NO). Not sufficient.

Answer: E.



I was asking abt statement 2.. U said..if it were mentioned that a b and c are intergers than A+b+c=even ..and then this statement wud b sufficient.. But i got this now..i have solved this ..bt i tuk some time while proving this statemt 2 ...



It goes like this:

Q: Is A+B+C= EVEN

S1: A-B-C= Even i.e. We can write A-B-C= 2N
=>A=2N + B+C
Substituting in the main equation
we get

2N+2(B+C)=2[N+B+C]

Now, N is an Integer, but we dont know whether, B+C= Integer or not. Hence N+B+C mayor may not be an Integer. Therefore, S1 is insufficient.

Same case with Statement 2, which on simplification gives, A=(2N+1)B + C
=> A+B+C= 2(B+C+NB)

Again same same scenario, We are not aware whether, the value in the parenthesis is an Integer or not.

Hence, S2 is also Insufficient.

S1 & S2 together also does not resolve the issue of whether B+C is integer or not.

Hence answer is E.

I hope this is clear. Thanks :)

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Re: Is A + B + C even? [#permalink]

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New post 30 Nov 2014, 14:35
Note that we are not told that a, b and c are integers.

Q: is a+b+c=even?

(1) a-c-b=even, if the variables are integers then a+b+c will be even but if they are not: a=3.5, b=1, c=0.5 —> a-c-b=2=even, but a+b+c=5=odd. Not sufficient.

(2) \fraca-cb=odd. The same here: if the variables are integers then a+b+c will be even but if they are not: a=3.5, b=1, c=0.5 —> \fraca-cb=3=odd, but a+b+c=5=odd. Not sufficient

(1)+(2) a+b+c may or may not be even (again if variables are integers: YES but if a=3.5, b=1, c=0.5 answer is NO). Not sufficient.

Answer: E.

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Is A + B + C even? [#permalink]

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New post 15 Apr 2015, 02:43
K is integer for sure.

1. A - (B+C) = 2K
=> A = 2K + B + C
A +B + C = 2K +2B + 2C = 2(K + B + C)

If B: Integer; C = 0.5
=> A + B + C = 2(K+B) + 1

2. (A - C)/B = 2k +1
=> A - C = 2kB + B
=> A = 2kB + B + C
=> A + B + C = 2(kB + B + C)
If B: Integer; C = 0.5

=> A + B + C = 2(kB + B) + 1

Quite tricky.

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Re: Is A + B + C even? [#permalink]

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New post 16 Apr 2015, 05:12
What a great question! I started this one writing out all the cases but then realized we don't know if a,b and c are integers. Got answer E and thought it took me at most 2.5 mins... nope just over 3:50. Although I finally got it, this question has made me realize I really need to improve my speed.
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Re: Is A + B + C even? [#permalink]

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New post 09 Dec 2017, 23:15
catty2004 wrote:
Is A + B + C even?

(1) A - C - B is even

(2) (A - C)/B is odd


Please find the solution as attached.

ANswer: Option E
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Re: Is A + B + C even?   [#permalink] 09 Dec 2017, 23:15
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