1000 DS section 7 #23

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by Morgoth » Thu Jul 09, 2009 5:01 am
odd + odd = even
even + odd = odd

odd*even = even
odd*odd = odd


the equation is (x+p)(x+q)

if you look at the above odd-even relations, then you will know that we need to know the values of p, q & x, whether they are odd or even.

Statement I - p is even
Statement II- q is even

but nothing is mentioned about x,

if we taken x as even, the resulting number will be even, but if we take x as odd, the resulting number will be odd, hence neither statements are sufficient.

Thus, E.

Hope this helps.