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

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If a and b are positive integers such that a b and a/b are [#permalink] New post 22 Apr 2006, 19:49
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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?

1) a/2

2) b/2

3) (a +b)/2

4) (a+2)/2

5) (b+2)/2
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 [#permalink] New post 22 Apr 2006, 22:30
Ans -> 4


a-b even: Thus both a, b are both even or both odd
a/b even: Thus a is always even, b could be even or odd!

Combining, if a is even, b has to be even also!

Thus, a = rb where r is even!

1. Always even (rb/2 = even )
2. Could be even or odd.
3. Always integer (even or odd)
4. ODD! (the expression reduces to a/2 +1. a/2 is always even, as shown above, hence even + 1 is odd)
5. reduces to b/2 +1. Result could be even or odd
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 [#permalink] New post 22 Apr 2006, 22:30
What is the OA? And what is the source of the question?


It took me a while to figure out! How on earth do u do this in <2 min?
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 [#permalink] New post 22 Apr 2006, 23:04
sm176811 wrote:
Ans -> 4


a-b even: Thus both a, b are both even or both odd
a/b even: Thus a is always even, b could be even or odd!

Combining, if a is even, b has to be even also!

Thus, a = rb where r is even!

1. Always even (rb/2 = even )
2. Could be even or odd.
3. Always integer (even or odd)
4. ODD! (the expression reduces to a/2 +1. a/2 is always even, as shown above, hence even + 1 is odd)
5. reduces to b/2 +1. Result could be even or odd


John,
(a+2)/2 could be even too.

if a = 6, result is even. If a = 8, result is odd.
I tried this problem & I'm not seeing any choice that must be odd :(
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 [#permalink] New post 22 Apr 2006, 23:08
Okay... I guess I shud have been more clear.

b is even... as proved above!

a/b = n (given, where n is even)


this a = nb where n has a factor or 2 and b also has a factor of 2!

This when a/2, the result shud have a factor of 2. Cos 2 will cancel out either the factor in n or in b. But the other term will still have a factor of 2 left. Thus, a/2 is ALWAYS even!

Last edited by sm176811 on 23 Apr 2006, 03:59, edited 1 time in total.
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 [#permalink] New post 22 Apr 2006, 23:27
hik...I got my mistake after I wrote to you...anyway, thanks John :)
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 [#permalink] New post 22 Apr 2006, 23:37
Took me a while 2 solve this tho... wonder how u are supposed 2 get it in < 2 mins! Am quite concerned...!
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 [#permalink] New post 23 Apr 2006, 00:43
I still don't get it, can you please elaborate more?

from the Q. condition we know that:

i) a - b is even => ( a AND b are even) OR ( a AND b are odd )
ii) a/b is even => a is even

from (i) AND (ii) => a AND b are even

hence both 4 AND 5 should satisfy.

please correct me if i am wrong
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 [#permalink] New post 23 Apr 2006, 01:20
Please see my explanation above... only 4 will satisfy!
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 [#permalink] New post 23 Apr 2006, 01:51
I guess the explaination given is the best. You need to also try with different combinations. The only thing to note here is that the a>b and that a-b=4 or multiple of 4 else a/b will be odd.
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 [#permalink] New post 23 Apr 2006, 03:56
hmm... I'm lost...

Can you explain me how b can be odd ? Since othervise
b/2 + 1 is always odd.

I just don't get why did you say that when you combined the conditions
you got that b is even or odd?
As I can see it, since b is even when a is even AND since a is indeed even so b is also even.

Can you please explain your point?
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 [#permalink] New post 23 Apr 2006, 04:01
deowl wrote:
hmm... I'm lost...

Can you explain me how b can be odd ? Since othervise
b/2 + 1 is always odd.

I just don't get why did you say that when you combined the conditions
you got that b is even or odd?
As I can see it, since b is even when a is even AND since a is indeed even so b is also even.

Can you please explain your point?


b can be even or odd!! See above.

Its a which is a factor or 4. hence a/2 is always even!
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 [#permalink] New post 23 Apr 2006, 16:35
sm176811 wrote:
Ans -> 4


a-b even: Thus both a, b are both even or both odd
a/b even: Thus a is always even, b could be even or odd!

Combining, if a is even, b has to be even also!

Thus, a = rb where r is even!

1. Always even (rb/2 = even )
2. Could be even or odd.
3. Always integer (even or odd)
4. ODD! (the expression reduces to a/2 +1. a/2 is always even, as shown above, hence even + 1 is odd)
5. reduces to b/2 +1. Result could be even or odd


John,
I'm not convinced with your explanation.

Given that

a + b = E -------1

and a/b is E ----2

For 1 to be Even,
either a and b both will be even or both will be odd.
For 2 to be even a must be even but b can be even or odd.

Here only even even case satisfies both condition, that is both a and b must be even.
Hence all answer choice are irrelevant.

Altough OA is D, but I'm not convince.
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 [#permalink] New post 23 Apr 2006, 18:29
gmat_crack wrote:
sm176811 wrote:
Ans -> 4


a-b even: Thus both a, b are both even or both odd
a/b even: Thus a is always even, b could be even or odd!

Combining, if a is even, b has to be even also!

Thus, a = rb where r is even!

1. Always even (rb/2 = even )
2. Could be even or odd.
3. Always integer (even or odd)
4. ODD! (the expression reduces to a/2 +1. a/2 is always even, as shown above, hence even + 1 is odd)
5. reduces to b/2 +1. Result could be even or odd


John,
I'm not convinced with your explanation.

Given that

a + b = E -------1

and a/b is E ----2

For 1 to be Even,
either a and b both will be even or both will be odd.
For 2 to be even a must be even but b can be even or odd.

Here only even even case satisfies both condition, that is both a and b must be even.
Hence all answer choice are irrelevant.

Altough OA is D, but I'm not convince.


Solve for a, b: a - b = even; a/b = even.

Observe that the above equations can only be satisfied for even a and even b.

We would never get an even number on dividing two odd numbers. Also, we would never get an even number if we subtract two odd numbers and divide the two odd numbers and intend to finish with an even number!!

So a = even (say 2m, m is an integer, b = even (say 2n, n is an integer).

Now try the options.

i. a/2 = even*b/2 (as a = even) = even*2n/2 = even (always).
ii. b/2 = even/2 => could be odd or even.
iii. (a+b)/2 = (even + even)/2 = even
iv. (a+2)/2 = (a/2) + 1 = even (from i) + 1 = odd
v. (b+2)/2 = (b/2) +1 = even or odd (as from ii we get b/2 as odd or even)

So the answer is iv. The trick is to note (as John has pointed out) that not only are a and b even but a = even*b = even*even. This implies a/2 is always even.
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 [#permalink] New post 23 Apr 2006, 20:01
John, Zooroopa

Thanks for nice explanation. I appreciate it.

Indeed OA is D.
  [#permalink] 23 Apr 2006, 20:01
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