Problem Solving

For a-b and a/b to be positive integer both a and b should be positive

Only one scenario can exists for odd integer result.

Suppose consider any value for a like 8 and b=2


b/2 is odd


So Answer is B

a -b = 2k1 ( given even) -- (1)
a/b = 2k2 (given ) --- (2)
one more condition b neq = 0 for (2) to be defined.
k1, k2 can be -2 ,. -1 ..., 1, 2, ....
from 2
a = 2 (k2*b) ===> even (no matter what is b)
from 1
even - b = even ==> b = even

Hence b/2 is must be an odd (B)

from 2 ==> a/2 = k2*b ===> it will be even .
==> a = even.
suppose a = -b which will be even ==> (a -b)/2 = -2b/2 == -b ==> even.
similarly, option D and E eliminated.

Hence Ans : B

Hi Guys,

Even I calculated B, but the answer is "D". I don't know whether its correct.

@selango[/quote]For a-b and a/b to be positive integer both a and b should be positive

Only one scenario can exists for odd integer result.

Suppose consider any value for a like 8 and b=2


b/2 is odd


So Answer is B

take a = 12 b = 6 ==> a -b == 6 (even satisfy 1)
a/b = 2 --> satisfy (2)
Now see option (a+b)/2 = 18/2 = 9 ==> which is odd

how did you conclude ans B?

Hi Guys,

Even I calculated B, but the answer is "D". I don't know whether its correct.

let's say a = -2
then D == (a +2)/2 = (-2 + 2)/2 = 0
==> is it odd ???

Guys,

I agree wid the reasoning, but if we assume a=12, b=6. In this situation both "a-b" & "a/b" are even. However if we calculate (a+b)/2 it comes out to 9. which is odd. going by this explanation correct answer can be "C". My only doubt is that question says "Which must be odd" and for all options, we can get a combination which is odd. Hence which should be mark.

My previous two post was not correct explanation.
Here is correct one. going by my 1st post
a = 2*(k2b) ===> a is even
from (1) a -b = 2k1 ==> even - b = even ==> b = even [must]

But if b is even that does not mean b/2 is odd i.e if b = 4 ==> b/2 = 2 (even)
Hence option B is ruled out.

now a/2 = k2*b Since b is even hence a/2 is even [must]. ==> option A ruled out.
Hence option D a/2 +1 must be an odd.

Option C: a/2 + b/2 ==> even + (odd or even it depends on value b) ==> even or odd
Hence option C is also ruled out.

Option E : b/2 + 1 ==> again (odd or even) + odd ==> odd or even.

In one of my previous post I ruled out option D on the basis of ( a = -2 )
in fact a will not eq to -2 otherwise a/b will not be even. [ if a = -2 then b = 1 or -1 to become a/b even. then in this case a-b will become odd]
Hence correct Answer should be D.

Thanks Akdayal. Now it looks correct.

akdayal wrote:My previous two post was not correct explanation.
Here is correct one. going by my 1st post
a = 2*(k2b) ===> a is even
from (1) a -b = 2k1 ==> even - b = even ==> b = even [must]

But if b is even that does not mean b/2 is odd i.e if b = 4 ==> b/2 = 2 (even)
Hence option B is ruled out.

now a/2 = k2*b Since b is even hence a/2 is even [must]. ==> option A ruled out.
Hence option D a/2 +1 must be an odd.

Option C: a/2 + b/2 ==> even + (odd or even it depends on value b) ==> even or odd
Hence option C is also ruled out.

Option E : b/2 + 1 ==> again (odd or even) + odd ==> odd or even.

In one of my previous post I ruled out option D on the basis of ( a = -2 )
in fact a will not eq to -2 otherwise a/b will not be even. [ if a = -2 then b = 1 or -1 to become a/b even. then in this case a-b will become odd]
Hence correct Answer should be D.

Going by the expl above Option D also gets eliminated right? am i missing something?

(a/2) can be even or odd...

What is the source of this question?

We know that a and b are both even numbers;

A/B = n = an even number also. This tells us that B goes into A an even number of times. Also B must be 2 or 4 or a multiple of either of those. The important point being here that if B goes into A an even number of times and B is 4 or a multiple of four, then if B is a multiple of 4 the number of times 2 would go into A would be 2n thus. If B is 2 or a multiple of two we know from the question that n is a even number.Thus,

a/2 = n = even number, thus,

a/2 +1 = odd number, or

a/2+2/2 = a/2+1, which is equivalent to saying

(a+2)/2 = must equal an odd number.

We can't be sure about [(b+2)/2)] because considering b could be 2 or 4 or a multiple of either,

2/2+2/2=2 even.

4/2+2/2=3 odd.

E is not correct because if B is 6, then

(6/2)+(2/2)=3+1=4

4 is even. Thus E does not have to be true.

Hope this helps.

Thanks,

Jared

i am still not convinced....

and i am confused why some posters are assuming a=-2 when question clearly says a and b are POSITIVE integers...
am i missing some thing

from trail and error method...
it seems that both a and b needs to be even to fullfill both cases....
so as both are positive numbers
if we eliminate E we can eliminate D as well

even+2/2 answer is odd....then both D and E are correct

a-b= even int, tell us that a and b must be either odd +- odd or even +- even.

a/b= even integer, tells us that even/even or even/odd.

Because both must be true we can deduce that a and b are both even.

When we think about b, we can think that b must equal 2, 4, 6 or an multiple of one those.


For any number that is divisible by 6 any even number of times;

36/6=6, and 36/2=18

or;

48/6=8, 48/2=24

or;

72/6=12, 72/2=36

Thus, any number that will be divisible by 6 an even number of times will also be divisible by 2 an even number of time.

For any number that is divisible by 4 and even number of time;

16/4=4, and 16/2=8

40/4=10, and 40/2=20

8/4=2, and 8/2=4.


Thus, any number that is divisible by 4 and even number of times, will also be divisible by 2 an even number of times.

The question says that a/b is even so no need to prove the case if b is 2.

Thus, any even number that is divided by another even number an even number of times, that number divided by 2 will be an even number.

So;

(a+2)/2 = (a/2)+(2/2) = (a/2) + 1

As we noted above if a/2 is always an even number, then (a/2)+1 must always be an odd number.

Hope this clarifies.

Thanks,

Jared