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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.

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

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

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

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.
_________________

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

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 .

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.
_________________

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
_________________

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.

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. 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.