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If a and b are positive integers such that a – b and a/b

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If a and b are positive integers such that a – b and a/b

by mitzwillrockgmat » Thu Jun 10, 2010 12:26 pm
If a and b are positive integers such that a - b and a/b are both even integers, which of the following must be an odd integer?

A. a/2
B. b/2
C. (a+b)/2
D. (a+2)/2
E. (b+2)/2

Hello, can some explain to me what I'm doing wrong here! This is what I've done so far....

if a - b is equal to an even integer then either both are even or both are odd (that's some 'parallelism' right there!)

if a/b is equal to an even integer then both a & b are even or one is even and the other is odd.

Combined one can see that both a & b are even.

A. a/2 - can be even or odd : 4/2 = 2 & 6/2=3

B. b/2 - same as A, can be even or odd : 4/2 = 2 & 6/2=3

C. (a+b)/2 - a=2, b=4 ... 6/2= 3 or if a=4, b= 8 ...12/2=6

D. (a+2)/2 - a=2 so a+2= 4, 4/2=2 or a=4 so a+2= 6 , 6/2=3

E. (b+2)/2 - it is the same as D.

Can someone please explain what I'm doing wrong??? By the way ans is D.
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by Patrick_GMATFix » Thu Jun 10, 2010 12:49 pm
Hi mitzwillrockgmat,

In analyzing D, you wrote: D. (a+2)/2 - a=2 so a+2= 4, 4/2=2 or a=4 so a+2= 6 , 6/2=3 . a cannot equal 2, because in that case there would be no even value of b that satisfies a/b is even. a=2 contradicts the prompt data.

Your analysis prior to looking at the answers is good, except when you write:"if a/b is equal to an even integer then both a & b are even or one is even and the other is odd. ". This part is not accurate a/b = even means that a must be even and b could be either.

Anyway, your conclusion that a and b are both even is correct; however, there is another important point you are missing: Since a/b is even, a must be a multiple of 4. This is because when a is divided by another even number (b), after everything cancels out, the result is even. Thus a must have been a multiple of 2*2. For example, a/b could be 12/2 or 24/6, because 12 and 24 are multiples of 4. a/b could not equal 6/2 or 18/2.


When looking at D, remember that a is a multiple of 4.

D. (a+2)/2=(a/2)+(2/2) = (a/2)+1. a/2 must be even, so (a/2)+1 must be odd

Hope that makes sense,
-Patrick
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by mitzwillrockgmat » Thu Jun 10, 2010 1:08 pm
Patrick_GMATFix wrote:Hi mitzwillrockgmat,

In analyzing D, you wrote: D. (a+2)/2 - a=2 so a+2= 4, 4/2=2 or a=4 so a+2= 6 , 6/2=3 . a cannot equal 2, because in that case there would be no even value of b that satisfies a/b is even. a=2 contradicts the prompt data.

Your analysis prior to looking at the answers is good, except when you write:"if a/b is equal to an even integer then both a & b are even or one is even and the other is odd. ". This part is not accurate a/b = even means that a must be even and b could be either.

Anyway, your conclusion that a and b are both even is correct; however, there is another important point you are missing: Since a/b is even, a must be a multiple of 4. This is because when a is divided by another even number (b), after everything cancels out, the result is even. Thus a must have been a multiple of 2*2. For example, a/b could be 12/2 or 24/6, because 12 and 24 are multiples of 4. a/b could not equal 6/2 or 18/2.


When looking at D, remember that a is a multiple of 4.

D. (a+2)/2=(a/2)+(2/2) = (a/2)+1. a/2 must be even, so (a/2)+1 must be odd

Hope that makes sense,
-Patrick

Whew! got it now, thanks!! By the way, at what level would you rate this question, please don't say 400!
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by mitzwillrockgmat » Thu Jun 10, 2010 1:47 pm
mitzwillrockgmat wrote:
Patrick_GMATFix wrote:Hi mitzwillrockgmat,

In analyzing D, you wrote: D. (a+2)/2 - a=2 so a+2= 4, 4/2=2 or a=4 so a+2= 6 , 6/2=3 . a cannot equal 2, because in that case there would be no even value of b that satisfies a/b is even. a=2 contradicts the prompt data.

Your analysis prior to looking at the answers is good, except when you write:"if a/b is equal to an even integer then both a & b are even or one is even and the other is odd. ". This part is not accurate a/b = even means that a must be even and b could be either.

Anyway, your conclusion that a and b are both even is correct; however, there is another important point you are missing: Since a/b is even, a must be a multiple of 4. This is because when a is divided by another even number (b), after everything cancels out, the result is even. Thus a must have been a multiple of 2*2. For example, a/b could be 12/2 or 24/6, because 12 and 24 are multiples of 4. a/b could not equal 6/2 or 18/2.


When looking at D, remember that a is a multiple of 4.

D. (a+2)/2=(a/2)+(2/2) = (a/2)+1. a/2 must be even, so (a/2)+1 must be odd

Hope that makes sense,
-Patrick
Whew! got it now, thanks!! By the way, at what level would you rate this question, please don't say 400! :s

So, based on your corrections, the answer should be as follows, correct?

if a - b is equal to an even integer then either both are even or both are odd

if a/b is equal to an even integer then a is even & b is even or odd.

Combined one can see that both a & b are even.

Furthermore, a cannot = 2 because 2 is a prime so it is only divisible by 2 and 1. As stated above b is even so cannot be 1
& cannot be 2 coz 2/2 =1 which is odd.

Moreover, if a is not 2 it can ONLY be a multiple of 4 because only multiples of 4 satisfy the above 2 statements:

a/b = 4/2 =2, 8/2=4, 8/4=2, 12/6=2 but not 10/2=5, 12/4=3, 14/2=7 & so on...

Thus, we finally know that a is a multiple of 4 & b can be any even no that satisfies a/b=even no.

A. a/2 - is always even -> 4/2=2 , 8/4=2 , 16/2=8

B. b/2 - can be even or odd -> 4/2=2 , 6/2=3

C. (a+b)/2 - can be even or odd -> a=4, b=2 ... 6/2=3 or if a=4, b= 8 ... 12/2=6

D. (a+2)/2 - always odd! -> a=4 ... a+2=6 ... 6/2=3 or a=8 ... a+2=10 ... 10/2=5 or a=12...a+2=14...14/2=7

E. (b+2)/2 - can be even or odd -> b=4 ... b+2=6 ...6/2=3 or b=2...b+2=4 ... 4/2 =2
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by Patrick_GMATFix » Thu Jun 10, 2010 1:52 pm
Much better solution :-)

I'd rate it at high 6's, low 7's.
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by mitzwillrockgmat » Thu Jun 10, 2010 1:54 pm
Patrick_GMATFix wrote:Much better solution :-)

I'd rate it at high 6's, low 7's.
okay, i feel much better then :) thanks patrick! :)
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