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

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Senior Manager
Joined: 20 Feb 2006
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If a and b are positive integers such that a b and a/b are [#permalink]  12 Dec 2006, 16:25
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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should be D

take a=64 and b=16
Manager
Joined: 18 Nov 2006
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D

from the stem: a=b * 2^m (m>=1)

take D) (a+2)/2 = a/2+1 = b*2^(m-1) + 1 ( m>=1)

= even num + 1 = odd num

all others, can be proven wrong:
a) a/2 must be even
b) b/2 can be even or odd (b =4 or 6)
c) (a+b)/2 = b* (2^m+1) /2 based on if b is div by 4 or 2 ..this can be even or odd
e) (b+2)/2= b/2 +1 ..can be odd or even ..b=3*2 =6 or b=2
Senior Manager
Joined: 05 Oct 2006
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hi bliss,

take a=10 and b= 2

a-b is even;a/b is even.
but (a+2)/2 is even.

so this must not be the choice.

in ur solution, u've taken a= b*2^m
i htink it should be a = b*2m

BLISSFUL wrote:
D

from the stem: a=b * 2^m (m>=1)

take D) (a+2)/2 = a/2+1 = b*2^(m-1) + 1 ( m>=1)

= even num + 1 = odd num

all others, can be proven wrong:
a) a/2 must be even
b) b/2 can be even or odd (b =4 or 6)
c) (a+b)/2 = b* (2^m+1) /2 based on if b is div by 4 or 2 ..this can be even or odd
e) (b+2)/2= b/2 +1 ..can be odd or even ..b=3*2 =6 or b=2
Manager
Joined: 18 Nov 2006
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from the stem

a/b is even

in your example, 10/2=5 is odd..

so, I will still hold onto my answer until proven wrong
Senior Manager
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yeas,

i missed out on this.
10/2 is odd
Manager
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basically a has to be multiple of 4 for a/b and a-b to be even

{(multiple of 4) + 2}/2 is always odd
Senior Manager
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jainan24 wrote:
basically a has to be multiple of 4 for a/b and a-b to be even

{(multiple of 4) + 2}/2 is always odd

Exactly. One can also plug pairs of #s such as 16 and 4, 36 and 6, etc, and discard answer choices till one of them is left.
Manager
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If we put a=20 and b=10 then
a-b=10(even) a/b=2(even)

then (a+b)/2 =15 this could be the answer.

as well as (a+2) /2= (20+2)/2 =11

whats wrong in this?
VP
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I got E.

From the question:
a - b = even ----------- (1)
AND
a/b = even ----------- (2)
=> a = b*even -------- (2a)

From (1) and (2a) , a and b are even.

Puting (2a) into (1)
b(even) - b = even
=>b(even - 1) = even
=>b(odd) = even

b = (even/odd)

This means that b MUST be an even multiple of 2.

so E: (b + 2)/2 = b/2 + 2/2 = even + 1 = odd
SVP
Joined: 01 May 2006
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(D) for me

Since we have :
o a - b = even, implying that
> a and b are both even
or
> a and b are both odd

and

o a / b = even, implying that
> a and b are both even with a = 2*k * b (k an integer)
or
> a even and b odd with a = 2*k * b (k an integer)

, thus we can conclude that:

o a/2 = 2*k*b / 2 = k*b >> No, Even as b is an even

o b/2 = even/2 >> No, Even or Odd

o (a + b) / 2
= (2*k*b + b) / 2
= (2*k+1)*(2*j) / 2
= (2*k+1)*j >> No, Even or Odd (depending on j)

o (a+2)/2
= (2*k*b + 2) / 2
= k*b + 1 >> Bingo, b is even, so k*b is even and even + odd = odd

o (b+2)/2
= (2*j + 2) / 2
= j + 1 >> No, Even or Odd (depending on j)

Last edited by Fig on 12 Jan 2007, 08:24, edited 1 time in total.
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johnycute wrote:
If we put a=20 and b=10 then
a-b=10(even) a/b=2(even)

then (a+b)/2 =15 this could be the answer.

as well as (a+2) /2= (20+2)/2 =11

whats wrong in this?

Nothing ..... (a+b) / 2 could be odd or even... So, in your case, yes it's an odd, but we cannot conclude in a general case that it's all time an odd
Senior Manager
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D

a â€“ b = even => possible combinations (a even and b even), (a odd and b odd)
a / b = even => a = even * b
So a has to be even and b could be even or odd

Together, both a and b has to be even

Point to be noted, b / 2 could be even or odd
But a / 2 = (even * b) / 2 = (even * even )/ 2 = even, so a / 2 has to be even

A) a / 2 = even
B) b / 2 = even or odd
C) (a + b) / 2 = a / 2 + b / 2 = (even) + (even or odd) = result is even or odd
D) (a + 2) / 2 = a / 2 + 1 = (even) + 1 = result is odd
E) b + 2 / 2 = b / 2 + 1 = (even or odd) + 1 = result is even or odd
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Re: Number properties [#permalink]  12 Jan 2007, 08:44
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

What we know
a/b=2m (a is even); a-b=2k (a and b are both odd or both even)
=> both a and b are even
let b=2s, then a=2mb=2m*2s=4ms
i.e. a is multiples of 4

a/2=2ms even
a/2=s, could be even or odd, depending on s
(a+b)/2=(4ms+2s)/2=(2m+1)s, could be even or odd, depending on s
(a+2)/2=(4ms+2)/2=2ms+1, always odd
(b+2)/2=s+1, could be even or odd, depending on s

It can be very helpful if you are able to express "even" and "odd" algebrally.
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VP
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Fig, D is right given your approach. However, what is wrong with my approach? I can't see anything wrong with it.
SVP
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kripalkavi wrote:
I got E.

From the question:
a - b = even ----------- (1)
AND
a/b = even ----------- (2)
=> a = b*even -------- (2a)

From (1) and (2a) , a and b are even.

Puting (2a) into (1)
b(even) - b = even
=>b(even - 1) = even
=>b(odd) = even

b = (even/odd)

This means that b MUST be an even multiple of 2.

so E: (b + 2)/2 = b/2 + 2/2 = even + 1 = odd

This part is wrong .... an even devided by 2 can give either an odd or an even

We can take the following examples
o If b = 2, then b/2 = 1 : odd
o If b = 4, then b/2 = 2 : even
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