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Re M2713
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16 Sep 2014, 00:27
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) \(a = b  c\). Rearrange: \(a+c=b\). 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
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Re M2713
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22 Sep 2016, 08:26
I think this is a highquality question and I agree with explanation.



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Re: M2713
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31 Oct 2016, 00:44
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) \(a = b  c\). Rearrange: \(a+c=b\). 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=21 but 121 is an odd number



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Re: M2713
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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) \(a = b  c\). Rearrange: \(a+c=b\). 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=21 but 121 is an odd number I tried to decipher your post but could not. Please elaborate. Thank you.
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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) \(a = b  c\). Rearrange: \(a+c=b\). 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=21 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=bc , 1=21 & 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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Re: M2713
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Re: M2713
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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=bc 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=bc and is b and c are both odd => a will be even hence abc is always even Hence B
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Re: M2713
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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) \(a = b  c\). Rearrange: \(a+c=b\). 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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Re: M2713
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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) \(a = b  c\). Rearrange: \(a+c=b\). 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. Check below for more: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Bunuel wrote: If \(a\), \(b\) and \(c\) are integers, is \(abc\) an even integer?
(1) \(b\) is halfway between \(a\) and \(c\)
(2) \(a = b  c\) (1) \(b\) is halfway between \(a\) and \(c\) For b to be midway between a and c there are many possible number choices. 1,2,3 → abc = even 2,3,4 → abc = even 1,3,5 → abc = odd 2,4,6 → abc = even
Thus, INSUFFICIENT(2) \(a = b  c\) a = b  c → b = a + c If a and c are odd, b is even → abc = even If a and c are even, b is even → abc = even If a is odd and b is even or vice versa, b is odd → abc = even
Thus, SUFFICIENT
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