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Bunuel
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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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Bunuel
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.


Answer: B

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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Bunuel
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.


Answer: B

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. Zero is an INTEGER.

2. Zero 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. Zero is neither positive nor negative (the only one of this kind).

4. Zero is divisible by EVERY integer except 0 itself (\(\frac{x}{0} = 0\), so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x).

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))

9. \(0^0\) case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), \(0^n = 0\).

11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.

12. \(0! = 1! = 1\).



Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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it should be clarified that the question is asking about the product of abc not abc as an integer. If not, kindly how would I understand it, with the given info .
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nuruddinies
it should be clarified that the question is asking about the product of abc not abc as an integer. If not, kindly how would I understand it, with the given info .

If abc were a three-digit number it would have been mentioned explicitly. Without that, abc can only be a*b*c, since only multiplication sign (*) is usually omitted.
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Step 1: Analyse Question Stem

The values, a, b and c are integers. Therefore, each of them can be odd or even.
We have to find out if the product abc is even. The product abc will be even if at least one of a, b or c is even.

Therefore, the question to be answered is “Is at least one of a, b or c, even?”

Step 2: Analyse Statements Independently (And eliminate options) – AD / BCE

Statement 1: b is halfway between a and c

If a =1 and c = 5, b = 3. Is at least one of the numbers even? NO.

If a = 2 and c = 6, b = 4. Is at least one of the numbers even? YES.

The data in statement 1 is insufficient to answer the question with a definite YES or NO.
Statement 1 alone is insufficient. Answer options A and D can be eliminated.

Statement 2: a = b−c

Rearranging the equation, we have,

a + c = b.

Since b is an integer, it can be odd or even.

If b = odd, a + c = odd; this means one of a or c must be even. Is at least one of the numbers even? YES

If b = even, we have already proved that the product abc is even.

The data in statement 2 is sufficient to answer the question with a definite YES.
Statement 2 alone is sufficient. Answer options C and E can be eliminated.

The correct answer option is B.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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What if a, b and c are 0?
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Ontherox
What if a, b and c are 0?

Have you checked this post: https://gmatclub.com/forum/m27-184489.html#p1967712

If a = b = c = 0, then abc = 0, which is even.
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