yvonne12 wrote:if c and d are integers, is c even

1. c(d+1) is even
2. (c+2) (c+4) is even

Is the answer B.

(1) c(d+1) is even. This could be the case when c and d is even or when c and d is odd. Hence not sufficient

(2) (c+2)(c+4) is even. This means c should be even. Hence sufficient.

Another B

no the answer is c

Agree with C. (Though my first guess was also B )

1. Clearly INSUFF as explained by Prasanna.
2. This stmt guarantees that C is even in all the cases except when C = 0; this statement is true but it doesn't tells about C (because 0 is neither even nor odd)- so INSUFF.

But if we combine both, first statement tells that C is not zero. (because something multiplied by C is even).

Therefore C.

mendiratta wrote:Agree with C. (Though my first guess was also B )

1. Clearly INSUFF as explained by Prasanna.
2. This stmt guarantees that C is even in all the cases except when C = 0; this statement is true but it doesn't tells about C (because 0 is neither even nor odd)- so INSUFF.

But if we combine both, first statement tells that C is not zero. (because something multiplied by C is even).

Therefore C.

Hmm, OG quants (Green book) tells us that the sequence -4, -2, 0, 2, 4 ...
is even.

I just checked wikipedia and it says 0 is even as well.

https://en.wikipedia.org/wiki/Even_and_odd_numbers

Not sure how it can be C, given the above...

mendiratta,

To make (c+2)(c+4) zero, C must be either -2 or -4 which is also even... so I think it's B. confusing...

anyone confirm OA again??