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Re: Even and Odd : GMATPrep [#permalink]
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
Re: Even and Odd : GMATPrep [#permalink]
17 Dec 2009, 04:38
15
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Expert's post
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This post was BOOKMARKED
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.
Re: Even and Odd : GMATPrep [#permalink]
17 Dec 2009, 04:41
1
This post was BOOKMARKED
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
Re: Even and Odd : GMATPrep [#permalink]
17 Dec 2009, 08:35
3
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Expert's post
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.
Re: If a and b are positive integers [#permalink]
26 Oct 2010, 07:45
1
This post received KUDOS
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 _________________
Cheers! Ravi
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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 _________________
Re: Even and Odd : GMATPrep [#permalink]
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
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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Re: If a and b are positive integers such that a – b and a/b are [#permalink]
26 Jul 2015, 21:56
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