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

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Manager
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If a and b are positive integers such that a – b and a/b are [#permalink]

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17 Dec 2009, 04:58
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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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Senior Manager
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Re: Even and Odd : GMATPrep [#permalink]

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17 Dec 2009, 05:32
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]

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17 Dec 2009, 05:38
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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.

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Re: Even and Odd : GMATPrep [#permalink]

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17 Dec 2009, 05:41
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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.... ...D it will be...sorry for oversight from my end

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Re: Even and Odd : GMATPrep [#permalink]

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17 Dec 2009, 07:30
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]

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17 Dec 2009, 09:10
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]

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17 Dec 2009, 09:20
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]

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17 Dec 2009, 09:35
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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]

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17 Dec 2009, 13:00
@Bunuel what a explanation man!!!!

Thanks a ton!!!!

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Re: Even and Odd : GMATPrep [#permalink]

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09 Aug 2010, 14:57
great explanation Bunuel.

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Re: If a and b are positive integers [#permalink]

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26 Oct 2010, 08:45
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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]

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27 Oct 2010, 22:40
good explanation from Bunuel

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21 Feb 2012, 23:11
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]

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21 May 2012, 06:13
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]

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19 Jun 2013, 04:52
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Re: If a and b are positive integers such that a – b and a/b are [#permalink]

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19 Jun 2013, 05:21
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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]

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21 Jul 2014, 05:44
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Re: If a and b are positive integers such that a – b and a/b are [#permalink]

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26 Jul 2015, 22:56
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Re: If a and b are positive integers such that a – b and a/b are [#permalink]

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Re: If a and b are positive integers such that a – b and a/b are   [#permalink] 20 Oct 2016, 08:14
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