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sarahw_gmat
- Junior | Next Rank: 30 Posts
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it will be great next time to hide answer is spoiler
is c even if c and d are integers
(1) c(d+1)=cd+c=even, possible cases are
even(c)*even(d)+even(c)=even here c is even
odd(c)*odd(d)+odd(c)=odd+odd=even, here c is odd, insuff
(1) (c+2)(d+4)=even
cd+4c+2d+8=even the value of 4c+2d+8 does not matter itis simply even
cd+even=even
even(c)*even(d)+even=even. here c is even
odd(c)*even(d)+even=even+even=even here c is odd
again insuff
both
from 1 st cd+c=even, cd=even-c, and insert st2, cd+even=even
even-c+even=even
even-c=even
so c must be even
both are suff
