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Bunuel
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IMO answer is option B

This is how I solved:

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Bunuel
If k is an integer, is k even?

(1) k/2 is not an even integer.

(2) 5 - k is an even integer.

#1
k/2 result in fraction /odd value insufficient
#2
5-k is an even integer only when k is odd
IMO B
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sumi747 I don't think we can test 4 as you did for statement 1 since we are told \(\frac{k}{2}\) is not an even integer. Please correct me if I am wrong.

sumi747
IMO B
Statement 1, if k = 4, k/2=2. If k= 6, k/2= 3. Both cases k is even. Not sufficient.
Statement 2, 5-k= even, odd-odd=even, hence we know k is odd. Sufficient.
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St 1: if k = 3 ==> \(\frac{k}{2}\) = 1.5 (not an even integer). If k = 6 ==> \(\frac{k}{2}\)= 3 (not an even integer). Therefore, st 1 is not sufficient since we are able to show that k could be 3 (odd) or 6(even)

St 2: k must be odd since odd - odd = even, therefore st 2 is sufficient.

Answer is \(B\)
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