If c and d are integers, is c even? GMATPrep - DS

[trang.kieu.phung]

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!

[gokul_nair1984]

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

[trang.kieu.phung]

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

[gokul_nair1984]

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

Yes you can :)

[tejkumar.m]

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

[RonPurewal]

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.

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

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

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

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

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.

[vicksikand]

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

[jnelson0612]

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.

[AbhilashM94]

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.

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

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

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

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

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.

c = odd, d = odd work with both statements as well!

[tim]

Please let us know if you have a question here.

[AbhilashM94]

Please let us know if you have a question here.

c = odd, d = odd works with both statements.

So combining both c = odd & c= even prove conclusively that c is even?

[tim]

Your statement is false; can you explain the reasoning that led you there so we can help you further?

Your question makes no sense; what does it even mean to combine "both c = odd & c= even"?

[RonPurewal]

If both c and d are odd, statement 2 cannot possibly be true.

Odd + 2 and odd + 4 are still odd. Odd times odd is odd.

[AbhilashM94]

If both c and d are odd, statement 2 cannot possibly be true.

Odd + 2 and odd + 4 are still odd. Odd times odd is odd.

Yes, you are right. I goofed up my first grade math :(

[tim]

The GMAT writes a lot of problems that try to exploit our tendency to speed through "first grade math", so this is all the more reason to be very careful on those types of calculations! :)