If a and b are positive integers such that a b and a/b are : PS Archive
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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]

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30 Apr 2006, 07:19
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If a and b are positive integers such that a â€“ 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

VP
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Re: PS: a&b, odd integer? [#permalink]

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01 May 2006, 11:57
M8 wrote:
If a and b are positive integers such that a â€“ 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

we know that a and b both are even integers.
a, b and c are easily ruled out.
D and E, both, could be true but E is not.
D is always true cuz if (a-b) and (a/b) are evens, then a is an even multiple of even integer ([.e. 2(b)]. for example.

if b = 4, a has to be 2b or muktiple of 2b. if so,
a = 8
a-b = 8-4
a/b=8/4=2

if so, then only (a+2)/2 is odd. so D is correct....
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Re: PS: a&b, odd integer? [#permalink]

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01 May 2006, 11:57
M8 wrote:
If a and b are positive integers such that a â€“ 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

doesn't work when a is odd because a/b will be always odd

so a must be even, => b is even

pick numbers
(A) a/2
a could be 4
b could be 2
a-b and a/b are even hold
a/2 = 2,
so this is not the answer

(B) b/2
a could be 8
b could be 4
a-b and a/b are even hold
b/2 = 4
so this is not the answer

(C)
a could be 8
b could be 4
a-b and a/b are even hold
(a+b)/2 = 6
so this is not the answer

(E)
a could be 4
b could be 2
a-b and a/b are even hold
(b+2)/2 = 2
so this is not the answer

so we are left with:
(D) (a+2)/2

by elimination this is the answer a proper proof might be to time consuming, so just stick to elimination.

however for a-b and a/b to be even integers at the same time a-b must be a multiple of 4, if you add 2 to a then it becomes the middle point between two even, so is an odd
VP
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01 May 2006, 12:16
Okay,... A,B,C,E for reasons above

now a, b are even integers

a/b=even
so a = b* even.

Hence, a is even
as a-b=even, b has to be even

thus a = even*even = (2*m)*(2*n) = 4*mn

thus (a+2)/2 = (4*mn+2)/2 = 2(2*mn+1)/2 = 2*mn+1.
2*mn is always even.
So 2*mn + 1 is always odd!

OED!
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07 May 2006, 19:20
I remember the rule of odd / odd = odd & odd - even or odd + even = odd , so we know a and b are even.

now, we just have to find the value that holds true for an odd value . The main difference between D and E is A has to be greater than b since we are talking about positive integers for the results of a - b and a/b. Therefore even + 2 / even. always be odd.

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GMAT the final frontie!!!.

07 May 2006, 19:20
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