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# M27-13

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:27
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Question Stats:

53% (01:05) correct 48% (01:29) wrong based on 40 sessions

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If $$a$$, $$b$$ and $$c$$ are integers, is $$abc$$ an even integer?

(1) $$b$$ is halfway between $$a$$ and $$c$$

(2) $$a = b - c$$
[Reveal] Spoiler: OA

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16 Sep 2014, 00:27
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Official Solution:

In order for the product of the integers to be even, at least one of them must be even.

(1) $$b$$ is halfway between $$a$$ and $$c$$. On the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\frac{a+c}{2}$$. Now, does that mean that at least one of them is even? Not necessarily: $$a=1$$, $$b=3$$ and $$c=5$$, of course it's also possible that for example $$b= 4 = even$$, for $$a=1$$ and $$c=7$$. Not sufficient.

(2) $$a = b - c$$. Re-arrange: $$a+c=b$$. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.

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22 Sep 2016, 08:26
I think this is a high-quality question and I agree with explanation.
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Joined: 26 Jul 2016
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Location: India
Concentration: Operations, Entrepreneurship
Schools: Tepper '19
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31 Oct 2016, 00:44
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Bunuel wrote:
Official Solution:

In order for the product of the integers to be even, at least one of them must be even.

(1) $$b$$ is halfway between $$a$$ and $$c$$. On the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\frac{a+c}{2}$$. Now, does that mean that at least one of them is even? Not necessarily: $$a=1$$, $$b=3$$ and $$c=5$$, of course it's also possible that for example $$b= 4 = even$$, for $$a=1$$ and $$c=7$$. Not sufficient.

(2) $$a = b - c$$. Re-arrange: $$a+c=b$$. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.

Bunuel please try putting in number 121 !! 1=2-1 but 121 is an odd number
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31 Oct 2016, 01:19
yashrakhiani wrote:
Bunuel wrote:
Official Solution:

In order for the product of the integers to be even, at least one of them must be even.

(1) $$b$$ is halfway between $$a$$ and $$c$$. On the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\frac{a+c}{2}$$. Now, does that mean that at least one of them is even? Not necessarily: $$a=1$$, $$b=3$$ and $$c=5$$, of course it's also possible that for example $$b= 4 = even$$, for $$a=1$$ and $$c=7$$. Not sufficient.

(2) $$a = b - c$$. Re-arrange: $$a+c=b$$. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.

Bunuel please try putting in number 121 !! 1=2-1 but 121 is an odd number

I tried to decipher your post but could not. Please elaborate. Thank you.
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31 Oct 2016, 02:11
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Bunuel wrote:
yashrakhiani wrote:
Bunuel wrote:
Official Solution:

In order for the product of the integers to be even, at least one of them must be even.

(1) $$b$$ is halfway between $$a$$ and $$c$$. On the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\frac{a+c}{2}$$. Now, does that mean that at least one of them is even? Not necessarily: $$a=1$$, $$b=3$$ and $$c=5$$, of course it's also possible that for example $$b= 4 = even$$, for $$a=1$$ and $$c=7$$. Not sufficient.

(2) $$a = b - c$$. Re-arrange: $$a+c=b$$. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.

Bunuel please try putting in number 121 !! 1=2-1 but 121 is an odd number

I tried to decipher your post but could not. Please elaborate. Thank you.

Bunuel ' 'abc' an even integer this means that abc is a 3 digit even integer , statement 2 says a=b-c , 1=2-1 & 121 is an odd number . Bunuel Please check if the question is indeed asking for condition if abc is an even number or if the question intends to ask the condition for A * B * C is an even number.

According to the provided answer clearly the question is asking for value of abc and not a*b*c .
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31 Oct 2016, 02:25
yashrakhiani wrote:
Bunuel ' 'abc' an even integer this means that abc is a 3 digit even integer , statement 2 says a=b-c , 1=2-1 & 121 is an odd number . Bunuel Please check if the question is indeed asking for condition if abc is an even number or if the question intends to ask the condition for A * B * C is an even number.

According to the provided answer clearly the question is asking for value of abc and not a*b*c .

abc is a*b*c. If it were 3-digit number it would have been explicitly mentioned.
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21 Nov 2016, 06:47
Brilliant Question this one.
Here we need to check if abc is even or not
As all of a,b,c are integers (very important piece of information )=> If anyone of a or b or c is even, we can say with assurance that abc will be even
Lets look at statements now
Statement 1
b=a+c/2
2,3,4 => abc=even
7,9,11=> abc is odd
hence not sufficient

Statement 2
Here a=b-c
making cases for b,c=>
even,even
even,odd
odd,even
odd,odd

So corresponding to the above cases a will be=>
even
odd
odd
even

Hence clearly one out of a,b,c is always even
Conversely => as a=b-c and is b and c are both odd => a will be even
hence abc is always even
Hence B
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24 Nov 2017, 07:23
Bunuel wrote:
Official Solution:

In order for the product of the integers to be even, at least one of them must be even.

(1) $$b$$ is halfway between $$a$$ and $$c$$. On the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\frac{a+c}{2}$$. Now, does that mean that at least one of them is even? Not necessarily: $$a=1$$, $$b=3$$ and $$c=5$$, of course it's also possible that for example $$b= 4 = even$$, for $$a=1$$ and $$c=7$$. Not sufficient.

(2) $$a = b - c$$. Re-arrange: $$a+c=b$$. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.

Hi Bunuel - isn't it possible that one of the integers a,b,c is zero? The question doesn't state that the integers are either positive or distinct numbers. In the case that they aren't and a=0 while b=c, then wouldn't we have the case that a*b*c = 0 and thus both statements are insufficient? This is the answer I got, so if you could please explain I'd be really grateful.
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24 Nov 2017, 07:25
Udsey1234 wrote:
Bunuel wrote:
Official Solution:

In order for the product of the integers to be even, at least one of them must be even.

(1) $$b$$ is halfway between $$a$$ and $$c$$. On the GMAT we often see such statement and it can ALWAYS be expressed algebraically as $$b=\frac{a+c}{2}$$. Now, does that mean that at least one of them is even? Not necessarily: $$a=1$$, $$b=3$$ and $$c=5$$, of course it's also possible that for example $$b= 4 = even$$, for $$a=1$$ and $$c=7$$. Not sufficient.

(2) $$a = b - c$$. Re-arrange: $$a+c=b$$. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.

Hi Bunuel - isn't it possible that one of the integers a,b,c is zero? The question doesn't state that the integers are either positive or distinct numbers. In the case that they aren't and a=0 while b=c, then wouldn't we have the case that a*b*c = 0 and thus both statements are insufficient? This is the answer I got, so if you could please explain I'd be really grateful.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check below for more:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Re: M27-13   [#permalink] 24 Nov 2017, 07:25
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# M27-13

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