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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 17 Dec 2009, 03:58
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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?

A. a/2
B. b/2
C. (a+b)/2
D. (a + 2)/2
E. (b+2)/2
[Reveal] Spoiler: OA
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Re: Even and Odd : GMATPrep [#permalink] New post 17 Dec 2009, 04:32
zaarathelab wrote:
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


is the question correct?? none of the options seem to satisfy :|

here a>b since a/b is even integer

let a= 24 and b = 4.

A. a/2 = 12 => even
B. b/2 = 2 => even
C. (a+b)/2 = 14 => even
D. (a+2)/2 = 13 => odd
E. (b+2)/2 = 3 =>odd

now if a = 20 and b = 2 then a-b = 18 and a/b = 10
A. a/2 = 10 => even
B. b/2 = 1 => odd
C. (a+b)/2 = 11 => odd
D. (a+2)/2 = 13 => odd
E. (b+2)/2 = 2 => even
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Re: Even and Odd : GMATPrep [#permalink] New post 17 Dec 2009, 04:38
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zaarathelab wrote:
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


\(a-b\) even --> either both even or both odd

\(\frac{a}{b}\) even --> either both even or \(a\) is even and \(b\) is odd.

As both statements are true --> \(a\) and \(b\) must be even.

As \(\frac{a}{b}\) is an even integer --> \(a\) must be multiple of 4.

Options A is always even.
Options B can be even or odd.
Options C can be even or odd.
Options D: \(\frac{a+2}{2}=\frac{a}{2}+1\), as \(a\) is multiple of \(4\), \(\frac{a}{2}\) is even integer --> even+1=odd. Hence option D is always odd.
Options E can be even, odd.

Answer: D.
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Re: Even and Odd : GMATPrep [#permalink] New post 17 Dec 2009, 04:41
kp1811 wrote:
zaarathelab wrote:
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


is the question correct?? none of the options seem to satisfy :|

here a>b since a/b is even integer

let a= 24 and b = 4.

A. a/2 = 12 => even
B. b/2 = 2 => even
C. (a+b)/2 = 14 => even
D. (a+2)/2 = 13 => odd
E. (b+2)/2 = 3 =>odd

now if a = 20 and b = 2 then a-b = 18 and a/b = 10
A. a/2 = 10 => even
B. b/2 = 1 => odd
C. (a+b)/2 = 11 => odd
D. (a+2)/2 = 13 => odd
E. (b+2)/2 = 2 => even


what the heck.... :x ...D it will be...sorry for oversight from my end :|
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Re: Even and Odd : GMATPrep [#permalink] New post 17 Dec 2009, 06:30
Quote:
As \frac{a}{b} is an even integer --> a must be multiple of 4.


Bunuel, you are an absolute whiz

Yep the catch is that a has to be a multiple of 4 for a/b to be even

OA is D
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Re: Even and Odd : GMATPrep [#permalink] New post 17 Dec 2009, 08:10
I didnt get why a has to be multiple of 4

Now if u take a as 4 and b as 2 then a-b is even a/b is even and a+b/2 is odd which is option C.

Please explain where i am going wrong?
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Re: Even and Odd : GMATPrep [#permalink] New post 17 Dec 2009, 08:20
sagarsabnis wrote:
I didnt get why a has to be multiple of 4

Now if u take a as 4 and b as 2 then a-b is even a/b is even and a+b/2 is odd which is option C.

Please explain where i am going wrong?


its a "MUST be" question. (a+b)/2 will not always be ODD.

also a-b and a/b is even and a and b will be even satisfying this condition. Since a/b is even so a will be multiple of 4.

Any set of values satisfying this will have a as 4n [n = natural number]
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Re: Even and Odd : GMATPrep [#permalink] New post 17 Dec 2009, 08:35
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sagarsabnis wrote:
I didnt get why a has to be multiple of 4

Now if u take a as 4 and b as 2 then a-b is even a/b is even and a+b/2 is odd which is option C.

Please explain where i am going wrong?


First question: Why \(a\) has to be multiple of 4?

As we concluded \(a\) and \(b\) have to be even integers to meet both conditions in the stem: \(a=2m\) and \(b=2n\).

Then we have \(\frac{a}{b}\) is an even integer: \(\frac{a}{b}=2k\) --> \(a=2bk=4kn\) --> \(a\) is a multiple of \(4\).

Second question: Option C also COULD give an even result, so why not C?

The question asks: "which of the following MUST be an odd integer?"

If we take \(a=16\) and \(b=4\), then \(\frac{a+b}{2}=10=even\). So option C may or may not be odd and we are asked to determine which option is ALWAYS odd, hence C is out.

In fact:
Options A is always even.
Options B can be even or odd.
Options C can be even or odd.
Options D is always odd.
Options E can be even or odd.

Only option which is always odd is D.

Hope it helps.
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Re: Even and Odd : GMATPrep [#permalink] New post 17 Dec 2009, 12:00
@Bunuel what a explanation man!!!!

Thanks a ton!!!!
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Re: Even and Odd : GMATPrep [#permalink] New post 09 Aug 2010, 13:57
great explanation Bunuel.
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Re: If a and b are positive integers [#permalink] New post 26 Oct 2010, 07:45
Given:
a > 0, b > 0 and a, b are Integers

a - b and a/b are both even intergers
==> \(a - b = 2m_1\)
\(\frac{a}{b} = 2m_2\) ==> \(a = (b)(2m_2)\)
==> a & b are both even, otherwise the above conditions won't be satisfied.

Now lets go with options one by one ...
A) a/2
==> can be even or odd ( a = 6 or a = 8)

B) b/2
==> Can be either even or odd ( b = 6 or b = 8)

C) (a+b) /2
==> Since, \(a = (b)(2m_2)\), \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2) + b}{2}\)
==> \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2 + 1)}{2}\)
==> Can be even or odd

D) (a+2)/2
==> \(\frac{a + b}{2}\) =\(\frac{2bm_2 + 2}{2}\)
==> \(\frac{a + b}{2}\) =\(bm_2 + 1\)
==> Since, b is even, \(bm_2\) is even and \(bm_2 + 1\) is odd always

E) (b+2)/2
==> \(\frac{b+2}{2} = 1 + \frac{b}{2}\)
==> Can be even or odd
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Re: Even and Odd : GMATPrep [#permalink] New post 27 Oct 2010, 21:40
good explanation from Bunuel
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even [#permalink] New post 21 Feb 2012, 22:11
a and b both have to be even to fulfill the condition.
a could be = 4, 6, 8 ...., b = 2,4, ....
so a - b = 2 (Even)
a/b= 2k (Even)
a/2=2k
a = 4k
so a must be multiple of 4
now try the options
ans. D
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Re: Even and Odd : GMATPrep [#permalink] New post 21 May 2012, 05:13
kp1811 wrote:
zaarathelab wrote:
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


is the question correct?? none of the options seem to satisfy :|

here a>b since a/b is even integer

let a= 24 and b = 4.

A. a/2 = 12 => even
B. b/2 = 2 => even
C. (a+b)/2 = 14 => even
D. (a+2)/2 = 13 => odd
E. (b+2)/2 = 3 =>odd

now if a = 20 and b = 2 then a-b = 18 and a/b = 10
A. a/2 = 10 => even
B. b/2 = 1 => odd
C. (a+b)/2 = 11 => odd
D. (a+2)/2 = 13 => odd
E. (b+2)/2 = 2 => even



small correction : (a+2)/2 = 11 as a= 20 ,so 22/2= 11 not 13
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Re: If a and b are positive integers such that a – b and a/b are [#permalink] New post 19 Jun 2013, 03:52
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Re: If a and b are positive integers such that a – b and a/b are [#permalink] New post 19 Jun 2013, 04:21
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zaarathelab wrote:
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


Given that (a-b) = even

Thus, \(b(\frac{a}{b}-1)\) = even --> as \(\frac{a}{b}\) is even, \((\frac{a}{b}-1)\) will be odd. Thus, b*odd = even--> b must be even

Thus, as a = b*2p --> a = 2l*2p = 4lp , where l,p are non-zero positive integers.

Thus,\(\frac{a}{2}\) + 1 = 2lp+1, which will always be an odd number.

D.
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Re: If a and b are positive integers such that a – b and a/b are [#permalink] New post 21 Jul 2014, 04:44
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Re: If a and b are positive integers such that a – b and a/b are   [#permalink] 21 Jul 2014, 04:44
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