If c and d are integers, is c even?
(1) c(d + 1) is even
(2) (c + 2)(d + 4) is even

My approach:
From (1): there are 3 cases:
A. c is even and (d + 1) is even
B. c is even and (d + 1) is odd
C. c is odd and (d + 1) is even
----> insufficient

From (2): 3 cases:
A. both (c + 2) and (d + 4) are even
B. (c + 2) is even and (d + 4) is odd
C. (c + 2) is odd and (d + 4) is even
-----> insufficient

From (1) and (2):
c(d + 1) is even ---> cd + c = 2n (k is an integer) (**)
(c + 2)(d + 4) is even ---> cd + 4c + 2d + 8 = 2m (*)
(*) - (**): 3c + 2(d + 4) = 2(m - n)
---> c = [2(m - n) - 2(d + 4)]/3
-----> insufficient

I chose E but the OA is C.

Could you explain how to combine (1) and (2) can make c even by algebraic approach? (not by picking numbers)
Thanks in advance!

Combining both Statements, we get:
1. cd+c=Even
2. cd+4c+2d+8=Even

By analyzing statement 2 we can see that, 4c and 2d have to be even in nature( since they are being multiplied by an even number). Furthermore, 8 is also even and we also know that
even +even+even=EVEN
Therefore, Statement 2 can be rephrased as
cd+Even=Even.
This means that cd has to be even(Even+Even=Even).

Substituting this information in Statement 1, we get,
Even+c=Even. This shows that c has to be even under any circumstance.
Hence the answer is C

Hope this helps

I like your approach, very clear and simple.
Thank you :)

I think I can use this method to solve the last equation:
3c + 2(d + 4) = 2(n - m)
That means: 3c + even = even ---> 3c must be even ---> c is even

If c and d are integers, is c even?
(1) c(d + 1) is even
(2) (c + 2)(d + 4) is even

A product is even only in the below two cases:

Even * Even = Even
Even * odd = Even

Statement -1 :

Worst case is Even * Odd

Either C is even and (d+1) is odd or
C is odd and (d+1) even

Not Sufficient to answer

Either C is even and (d+1) is odd or
C is odd and (d+1) even

Statement -2 :

Worst case is Even * Odd

Either C+2 is even and (d+4) is odd or ==> C is even
C+2 is odd and (d+4) even ==> C is odd

Not sufficient to answer

Take both statements together-->

Consider C is odd

Statement -1 becomes (d+1) is even ==> d is odd
Statment -2 , C odd and d is odd which does NOT conclude to (c+2) * (d+4) = Odd... Because odd * odd = odd

Consider C is even

Statement -1 becomes (d+1) is odd ==> d is even
Statment -2 , C even and d is even which MUST conclude to (c+2) * (d+4) = even... Because even * even = Even..

Hope this is understandable.

Regards

the problem here is that you aren't SIMPLIFYING the cases that you're getting out of the statements.
if you have a statement about even/odd with something like x + 3, x - 1, etc., you can ALWAYS translate that into a statement about even/odd with x itself.


translated:
(a) c = even, d = odd
(b) c = even, d = even
(c) c = odd, d = odd
insufficient.


translated:
(a) c = even, d = even
(b) c = even, d = odd
(c) c = odd, d = even
insufficient.

combine them:
the only cases that exist in both statements are
* c = even, d = even
* c = even, d = odd
so, c must be even.
sufficient
(c)

--

re: this


um ... wow

if you find yourself doing something like this, EVER, then you should quit immediately -- if you are doing this much busy work, then the train has already gone off the rails a long time ago, and you should therefore give up and start doing a number-picking method.

I dont know if this is the quickest method , but here it is anyways:
1. c(d+1) = E
put d=odd number , hence c=E or O N.S
2. (c+2)(d+4)= E
put d=even , hence c= E or O N.S
1 &2
simplify 2:
cd+(4c+2d+8)=E.....Part in brackets will always be Even
thus: cd = E-E=E
we have
c(d+1)=E
cd=E
possible only if c is even.

C

vicksikand, very interesting approach. I've walked through it and I can't fault any of it. I think Ron's approach is quite quick too.

Nice work.

_________________
Jamie Nelson
ManhattanGMAT Instructor