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# M22-22

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Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128564 [0], given: 12180

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16 Sep 2014, 01:17
Expert's post
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Difficulty:

95% (hard)

Question Stats:

30% (01:55) correct 70% (02:04) wrong based on 64 sessions

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Is $$a+b+c$$ even?

(1) $$a-c-b$$ is even.

(2) $$\frac{a-c}{b}$$ is odd
[Reveal] Spoiler: OA

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Math Expert
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16 Sep 2014, 01:17
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Official Solution:

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

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

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

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23 Dec 2014, 16:27
Would the answer be D if it was given that a, b, and c are integers?

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24 Dec 2014, 07:23
amishra1 wrote:
Would the answer be D if it was given that a, b, and c are integers?

Yes, in that case the answer would be D.
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26 Jun 2015, 05:03
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Here's how I would look at Statement A,

a - c - b = even
a + c + b - 2c - 2b = even
a + c + b - 2(c+b) = even

For this statement to be true,
a + c + b = even, since only "even - even = even"

Am I missing something?

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26 Jun 2015, 06:06
mukulc wrote:
Here's how I would look at Statement A,

a - c - b = even
a + c + b - 2c - 2b = even
a + c + b - 2(c+b) = even

For this statement to be true,
a + c + b = even, since only "even - even = even"

Am I missing something?

Yes.

I guess you haven't read the solution. Very first sentence there says: "note that we are not told that a, b and c are integers".
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12 Mar 2016, 23:50
For statement 1, is it correct to say that if A+B+C=Even and A,B,C are integers, then A-B-C is also even under any order of A,B,C (i.e., also even C-A-B or B-A-C)?
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Math Expert
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13 Mar 2016, 08:22
happyface101 wrote:
For statement 1, is it correct to say that if A+B+C=Even and A,B,C are integers, then A-B-C is also even under any order of A,B,C (i.e., also even C-A-B or B-A-C)?

Yes, this would be correct.
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01 Aug 2016, 12:02
I think this is a high-quality question and I agree with explanation. Good one!

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Senior Manager
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Location: India
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22 Aug 2016, 05:36
I think this is a high-quality question and I agree with explanation. Bombastic Question Lol great trap! You people never fail to surprise me!

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Manager
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01 Dec 2016, 07:33
Sigh, how many more times am I going to forget to check if I'm dealing with integers. Quickest and proudest answer D, thinking I'm all that smart, just to be blown by the "might not be integers" detail.

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Intern
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30 Sep 2017, 12:30
Here is a method I used to solve this question in the most efficient manner that I haven't seen discussed here.

Stem:$$A+B+C= Even$$
(1)$$A-B-C=E$$
This becomes $$A=E+B+C$$
Important property: In a term of the form (Even number) + X, the (Even number) plays no role in the Even-Odd nature of the term
In turn, this becomes A=B+C,
Plugging into the equation we get:$$B+C+B+C=E?$$this becomes $$2B+2C=E$$
If A & B are integers, this is sufficient, but A & B can both be .1

(2)$$\frac{A-C}{B}=O$$
Or $$O*B=A-C$$
Important property: In a term of the form (Odd number)*(X), the (Odd number) plays no role in the Even-Odd nature of the term
$$B=A-C$$
Plugging into the equation we are left with $$2A=E?$$
If A is an integer, this is sufficient, but A can be .1

Therefore E

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M22-22   [#permalink] 30 Sep 2017, 12:30
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