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odd and even question

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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Thu Mar 03, 2016 4:29 pm
If a and b are positive integers such that a < b, is b even?

(1) b/2 - a/2 is an integer.

(2) 3b/4 - a/2 is an integer.
Target question: Is b even?

Statement 1: b/2 - a/2 is an integer
We can combine the fractions to get (b-a)/2 is an integer.
If (b-a)/2 is an integer, then b-a must be even.
So, statement 1 is really just telling us that b-a is even. There are several pairs of values that satisfy this condition. Here are two:
case a: a=3 and b=5, in which case b is not even
case b: a=2 and b=6, in which case b is even
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 3b/4 - a/2 is an integer
We can combine the fractions to get (3b-2a)/4 is an integer.
If (3b-2a)/4 is an integer, then 3b-2a must be divisible by 4.
If 3b-2a is divisible by 4, then 3b-2a must be even
Well, we know that 2a will be even for all integer values of a
So, if 3b - 2a = even, then 3b must be even.
If 3b is even, then b must be even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

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by DavidG@VeritasPrep » Thu Mar 03, 2016 4:35 pm
didieravoaka wrote:Does someone can show a quick and easy solution. I got one that was too long.

Thanks.

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Statement 1: (b-a)/2 = Integer; So b-a = 2*integer; Now we know that b-a = even. It's possible that both b and a are ODD or that both are EVEN. (You can also test numbers:

Case1: b =3 and a = 1 (NO, b is not even)

Case2: b = 4 and a = 2 (YES, b is even)

So statement 1 is not sufficient.

Statement 2: 3b/4 - a/2 = integer
or 3b/4 - 2a/4 = integer
(3b - 2a)/4 = integer
3b - 2a = 4*integer; Because 4* integer must be an even number, we know that 3b - 2a = Even. Moreover, we know that 2a is even. So 3b - Even = Even
3b = Even + Even
3b = Even
If 3b is Even, b must be Even. So statement 2 alone is sufficient. Answer is B.
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by didieravoaka » Thu Mar 03, 2016 4:42 pm
Thanks to both Brent and David, I really like both of your solutions. I appreciate your help.

Marc.
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by Matt@VeritasPrep » Fri Mar 04, 2016 4:24 pm
I'm a little late to the party, but:

S1::

(b - a)/2 = an integer

(b - a) = 2 * an integer

(b - a) = even

We get an even with (Odd - Odd) and with (Even - Even), so this is no good.

S2::

(3b - 2a)/4 = an integer

3b - 2a = 4 * an integer

3b - 2a = even

3b = even + 2a

3b = even + even

3b = even

If b were odd, we'd have 3b = 3 * odd, but that can't be even! So we must have b = even, and we're set.
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