If c and d are integers, is c even?
1. c(d+1) is even
2. (c+2)(d+4) is even
Focus on what must be true and can be deduced. Keep in mind that (even)(any integer) = even, and (odd)(odd) = odd.
From statement (1), if c(d+1) is even, then it must be true that either c = even or (d+1)=even, or possibly both.
d+1=even means that d would be odd.
So from statement 1, there are two possibilities, which aren't mutually exclusive (meaning they could both be happening at the same time):
either c= even (and it doesn't matter what d is) or d=odd (and it doesn't matter what c is).
It's possible that c is even. It's possible that c is odd. That's insufficient to answer the original question.
From statement (2), it must be true that (c+2) is even or (d+4) is even.
If (c+2) = even, then c= even. If (d+4) = even, then d= even.
From statement (2), either c=even (with d being even or odd), or d=even (with c being even or odd), or both. It's possible that c is even, it's possible that c=odd. That's insufficient to answer the original question.
Together, it may be a big confusing to think through all of this. Let's organize it this way.
Either d=odd or d=even. That's for sure.
If d=odd, then according to statement (2), c would HAVE to be even. That's because according to statement (2), at least one of the two variables HAS to be even. [Note that d=odd and c=even also satisfies statement (1).]
If d=even, the according to statement (1), c would HAVE to be even. That's because according to statement (1), if d=even, then c would have be even in order for statement (1) to hold true. [Note that d=even and c=even also satisfies statement (2).]
Therefore in both scenarios, c has to be even.
Hope that helps.
