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If a and b are positive integers such that a b and a/b are [#permalink]
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06 Aug 2008, 12:50
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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



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Re: positive integers  PS [#permalink]
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06 Aug 2008, 12:58
chan4312 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 C ab= 2k a/b= 2l  > a= 2bl (a+b)/2 > 2bl+b/2 > b/2(2l+1) (2l+1) is always odd.. odd * b/2 > odd
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Last edited by x2suresh on 06 Aug 2008, 13:39, edited 2 times in total.



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Re: positive integers  PS [#permalink]
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06 Aug 2008, 13:55
chan4312 wrote: x2suresh wrote: chan4312 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 C ab= 2k a/b= 2l  > a= 2bl (a+b)/2 > 2bl+b/2 > b/2(2l+1) (2l+1) is always odd.. odd * b/2 > odd Wrong answer. Try again oops you are right.. odd * even  > even. It should be E then (by trial and error)..
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Re: positive integers  PS [#permalink]
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06 Aug 2008, 14:31
chan4312 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 IMO D Take a = 4,5,6 and b = 2,1,2; perform the operations accordingly (ab) = 2, 4, 4, all even (a/b) = 2,5,3, even and odd Now only even A and B gives even for both the operations; The numbers that satisfy are a = 4,8 b = 2,4 Hence D is the answer



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Re: positive integers  PS [#permalink]
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06 Aug 2008, 14:40
ab = even . i.e either both are EVEN or ODD  1
a/b=even; this suggests 'a' has to be even and 'b' might be odd or even  2
from 1 and 2 both a and b are even
Now at the options :
1 a/2 ; No , as a might be 6(2*3) or 12 (2*6) 2 b/2 ; No 3 (a+b)/2 = a/2 + b/2 ; cannot be , as odd + odd OR even + even is again even 4 (a+2)/2 = a/2 + 1 ; Yes this should be the answer as a/b = even so a/2 has to be even , and even + odd is definitely odd. 5 NO



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Re: positive integers  PS [#permalink]
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06 Aug 2008, 16:53
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chan4312 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 is even indicates a is even ,ab is even integer only when a and b are both even integers. also most importantly a must be multiple of 4 so that a/b is even inspite of b being even hence say a=4k where k is integer. now eliminate (a) a/2 is even eliminate (b) b/2 is even or odd say b=4 ,say b=6 then b/2 varies eliminate (c)a+b /2 is again even or odd (D)(a+2)/2 is always odd > (a+2)/2 = (4k+2)/2 =2k+1 => hence always odd eliminate (E)b+2 /2 is even or odd IMO D
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Re: positive integers  PS [#permalink]
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07 Aug 2008, 06:01
OA is D
Thank you all.. pretty good explanation spriya.!



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Re: positive integers  PS [#permalink]
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07 Aug 2008, 16:58
Oh thanks
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Re: positive integers  PS [#permalink]
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30 Aug 2011, 22:01
stallone wrote: ab = even . i.e either both are EVEN or ODD  1
a/b=even; this suggests 'a' has to be even and 'b' might be odd or even  2
from 1 and 2 both a and b are even
Now at the options :
1 a/2 ; No , as a might be 6(2*3) or 12 (2*6) 2 b/2 ; No 3 (a+b)/2 = a/2 + b/2 ; cannot be , as odd + odd OR even + even is again even 4 (a+2)/2 = a/2 + 1 ; Yes this should be the answer as a/b = even so a/2 has to be even , and even + odd is definitely odd. 5 NO What if we take a=6 then option 4 would become> 6+2/2 which is even... pls help me understand this.



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Re: positive integers  PS [#permalink]
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30 Aug 2011, 23:38
DeeptiM wrote: stallone wrote: ab = even . i.e either both are EVEN or ODD  1
a/b=even; this suggests 'a' has to be even and 'b' might be odd or even  2
from 1 and 2 both a and b are even
Now at the options :
1 a/2 ; No , as a might be 6(2*3) or 12 (2*6) 2 b/2 ; No 3 (a+b)/2 = a/2 + b/2 ; cannot be , as odd + odd OR even + even is again even 4 (a+2)/2 = a/2 + 1 ; Yes this should be the answer as a/b = even so a/2 has to be even , and even + odd is definitely odd. 5 NO What if we take a=6 then option 4 would become> 6+2/2 which is even... pls help me understand this. Hello, You cannot consider a= 6 because you cannot find another integer "b" such that a/b is even and a  b is also even The possible values for a,b = {(4,2),(8,2),(8,4),(4,4)} and so on regards Raghav.V Consider kudos if you find my post helpful:



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Re: positive integers  PS [#permalink]
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30 Aug 2011, 23:50
DeeptiM wrote: stallone wrote: ab = even . i.e either both are EVEN or ODD  1
a/b=even; this suggests 'a' has to be even and 'b' might be odd or even  2
from 1 and 2 both a and b are even
Now at the options :
1 a/2 ; No , as a might be 6(2*3) or 12 (2*6) 2 b/2 ; No 3 (a+b)/2 = a/2 + b/2 ; cannot be , as odd + odd OR even + even is again even 4 (a+2)/2 = a/2 + 1 ; Yes this should be the answer as a/b = even so a/2 has to be even , and even + odd is definitely odd. 5 NO What if we take a=6 then option 4 would become> 6+2/2 which is even... pls help me understand this. Hello, Just to add 1. a and b need to be even 2. the number of 2's in the prime factorization of "a" must be greater than number of 2's in the prime factorization of "b" regards Raghav.V Consider kudos if you find my post helpful:



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Re: positive integers  PS [#permalink]
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01 Sep 2011, 19:40
Here, ab = +ve integer and a/b = +ve integer In order to satisfy both condition, a and b both must be even integer. Given the answer choice, it is clear that the average of two even integer is always Odd.
i.e. (a+b)/2 is always odd




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