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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

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

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

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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