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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

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

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

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 (a-b) = 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

a-b = 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

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 ,a-b 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

a-b = 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...

a-b = 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

a-b = 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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