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Re: If c and d are integers, is c even? [#permalink]
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Thanks Bunuel

OA is C

But i got lost in this number property question (was consuming too much time) and hence picked E on the test.

Is there any faster way to solve this?
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Re: If c and d are integers, is c even? [#permalink]
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zaarathelab wrote:
Thanks Bunuel

OA is C

But i got lost in this number property question (was consuming too much time) and hence picked E on the test.

Is there any faster way to solve this?


We have c(d+1)=even and c(d+4)=even (I say c as it's the same as to write c+2 in this case).

Now if c is not even than d+1 and d+4 must be even, but they cannot be even together, hence c is even.

Sorry no other faster way.
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Re: If c and d are integers, is c even? [#permalink]
uzzy12 wrote:
If c and d are integers, is c even?

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



Statement 1 is insufficient since c could be even or odd and (d+1) could also be even or odd and still c * (d+1) could be even.

Refer the below table.

c ***** d+1 ***** c (d+1)
E ***** E ***** E
E ***** O ***** E
O ***** E ***** E

Statement 2 is insufficient.
(c+2) (d+4) = cd + 2d + 4c + 8. (Here 2d, 4c and 8 are even).
Hence c * d should even, however we cannot say whether c is even.

Refer the below table.

c ***** d ***** c * d
E ***** E ***** E
E ***** O ***** E
O ***** E ***** E


Combining two statements.

c(d+1) and cd is even. Also if d is even then (d+1) is odd and vice versa.

Now set up a table.

c ***** d ***** (d+1) **** cd *********** c(d+1)
O ***** E ***** O **** Even *********** Odd --- This combination does not work
E ***** O ***** E **** Even *********** Even --- This works and hence c is even.

Hence combining both statements would answer the question. Answer C.
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Re: If c and d are integers, is c even? [#permalink]
I don't know the best way; but here's how I would solve it;

1.
c(d+1) is even
c even and d+1 even i.e. d odd
OR
c odd and d+1 even i.e. d odd
OR
c even and d+1 odd i.e. d even

We can see that c can be even or odd.
Not sufficient.

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

c+2 even and d+4 even i.e. c even and d even
OR
c+2 odd and d+4 even i.e. c odd and d even
OR
c+2 even and d+4 odd i.e. c even and d odd

C can be even or odd.
Not sufficient.

If you see the odd case for c in both statements;
c odd and d+1 even i.e. d odd
c+2 odd and d+4 even i.e. c odd and d even
You see that for c=odd; 1 statement says d=odd; 2nd statement says d=even; Conflict; D can't be odd and even at the same time.

If you consider c=even;
c even and d+1 even i.e. d odd
c+2 even and d+4 odd i.e. c even and d odd
Both statements match.

c even and d+1 odd i.e. d even
c+2 even and d+4 even i.e. c even and d even
Both statements match.

Ans: "C"
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Re: If c and d are integers, is c even? [#permalink]
I tried this approach:

Statement 1:

c(d+1) is even

Therefore, we have three possibilities.
a) c - E & (d+1) - E => c is even and d is odd
b) c - E & (d+1) - O => c is even and d is even
c) c - O & (d+1) - E => c is odd and d is odd

So c and be even or odd!

Thus statement is not sufficient.

Statment 2:
(c+2) (d+4) is even
Therefore, there are three possibilities.
a) (c+2) is even and (d+4) is even => c is even and d is even
b) (c+2) is even and (d+4) is odd => c is even and d is odd
c) (c+2) is odd and (d+4) is even => c is odd and d is even
So c can be even or odd!
Thus statement is not sufficient.

Both Statements together:
For the cases c even and d odd or even, both the statements will always be true!
Thus c is even!
Ans: 'C'
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Re: If c and d are integers, is c even? [#permalink]
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Subtracting 2 from 1;

c(d+1)-(c+2)(d+4) = even-even=even
cd+c-cd-4c-2d-8=even
c-4c-2d-8=even
4c;2d;8 are all even
c should be even.
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Re: If c and d are integers, is c even? [#permalink]
zaarathelab wrote:
Thanks Bunuel

OA is C

But i got lost in this number property question (was consuming too much time) and hence picked E on the test.

Is there any faster way to solve this?

No easy way to do this but you can jump to c quickly ruling out a,d and b.

Once you see which statements remain valid while taking into consideration 1 and 2, we can see that c is even.
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Re: If c and d are integers, is c even? [#permalink]
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zaarathelab wrote:
If c and d are integers, is c even?

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


Here I go, catch me

Statement 1

Not sufficient obviously


Statement 2

Same here

Both Statements together

Well, lets see of 'c' is not even then d+1 must be even that means that 'd' must be odd

Now if d is odd. d+4 will be odd as well that means that c+2 must be even, and thus 'c' must be even

So C is sufficient here

Gimme some Kudos ok?

Cheers!
J :)
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Re: If c and d are integers, is c even? [#permalink]
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If c and d are integers, is c even?

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

Here we have a number properties question. We will be dealing with evens/odds and multiplication/addition. A couple of things to keep in mind before we look at the statements:
1. If c is even it will have at least one factor of 2
2. The rules for addition for evens/odds: E+E = E; O+O = E; E+0 = O; O+E = O
3. The rules for multiplication for evens/odds: We will arrive at an even product EXCEPT when we have O*O which will produce an O result

1. Statement 1 is true if :
a. C is Even and D is Even E(E+1) = E*O = E
b. C is Even and D is odd E(O+1) = E*E = E
c. C is Odd and D is Odd O(O+1) = O*E = E
a and b give us an answer of yes; c gives us an answer of no - INSUFFICIENT

2. Statement 2 is true if:
a. C is Even and D is Even (E+2)(E+4) = E*E = E
b. C is Even and D is Odd (E+2)(O+4) = E*O= E
c. C is Odd and D is Even (O+2)(E+4) = O*E=E
a and b give us an answer of yes; c gives us an answer of no - INSUFFICIENT


When we take the statements together we notice that the two possible scenarios they have in common are a and b (C is Even and D is Even OR C is Even and D is Odd). In both scenarios C is Even, so we have one definitive answer to the question. SUFFICIENT the answer is C

This may seem like a lengthy explanation; I provided every step for clarity. Make sure to memorize rules for number properties, and you can solve a question similar to this one in two minutes or less.
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Re: If c and d are integers, is c even? [#permalink]
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Hi All,

It appears that all of the posters used Number Property rules to answer this question. You can also TEST VALUES; in that way, you can physically "see" how those same Number Property rules "work" in real life.

We're told that C and D are integers. We're asked if C is even. This is a YES/NO question.

Fact 1: (C)(D+1) is even

This means that one or the other (or both) of the "terms" must be even....

IF....
C = 2
D = 1
(2)(2) = 4
The answer to the question is YES

IF....
C = 1
D = 1
(1)(2) = 2
The answer to the question is NO
Fact 1 is INSUFFICIENT

Fact 2: (C+2)(D+4) = even

Just as in Fact 1, this means that one or the other (or both) of the "terms" must be even....

IF....
C = 2
D = 1
(4)(5) = 20
The answer to the question is YES

IF...
C = 1
D = 2
(3)(6) = 18
The answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we know....
(C)(D+1) is even
(C+2)(D+4) is even

Since D is an integer, ONLY ONE of the two terms - (D+1) and (D+4) - will be even; the other will be odd. As such, the other term in each of the products (the one with the C in it) MUST be even....

eg...
C = 2
D = 1

Combined, SUFFICIENT

Final Answer:

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Re: If c and d are integers, is c even? [#permalink]
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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
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Re: If c and d are integers, is c even? [#permalink]
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zaarathelab wrote:
If c and d are integers, is c even?

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


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)

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Re: If c and d are integers, is c even? [#permalink]
If c and d are integers, is c even?

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

1. For c even and d odd or even --> yes. For c odd and d odd --> no. Insufficient.
2. When d is even, this will be true, irrespective of whether c is even or odd. Insufficient.
On combining, odd c and even d fails 1 whereas odd c and odd d fails 2. Hence, c cannot be odd.
C.
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Re: If c and d are integers, is c even? [#permalink]
Bunuel wrote:
If C and D are integers, is C even?

(1) C*(D+1) is even. In order the product to be even C must be even or D+1 must be even, or both. Hence it's not necessary C to be even. Not sufficient.

(2) (C+2)(D+4) is even. Notice that this is the same as C*(D+4) is even. Again, in order the product to be even C must be even or D+4 must be even, or both. Hence it's not necessary C to be even. Not sufficient.

(1)+(2) Both D+1 and D+4 cannot be even, thus C must be even. Sufficient.

Answer: C.


Hi Bunuel

I agree that Combination will prove C is even. But how can it be claimed that (d+1) and (d+4) both would not be even?

Because in both statements, either both can be even or either one can be even. So combining both surely would mean that both 'C' and 'D' are even.

Please correct me if am wrong.
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Re: If c and d are integers, is c even? [#permalink]
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Anurag06 wrote:
Bunuel wrote:
If C and D are integers, is C even?

(1) C*(D+1) is even. In order the product to be even C must be even or D+1 must be even, or both. Hence it's not necessary C to be even. Not sufficient.

(2) (C+2)(D+4) is even. Notice that this is the same as C*(D+4) is even. Again, in order the product to be even C must be even or D+4 must be even, or both. Hence it's not necessary C to be even. Not sufficient.

(1)+(2) Both D+1 and D+4 cannot be even, thus C must be even. Sufficient.

Answer: C.


Hi Bunuel

I agree that Combination will prove C is even. But how can it be claimed that (d+1) and (d+4) both would not be even?

Because in both statements, either both can be even or either one can be even. So combining both surely would mean that both 'C' and 'D' are even.

Please correct me if am wrong.


(d+1) and (d+4) are 3 apart. How can both of them be even?

If d + 1 is even, so if d is odd, then d + 4 = odd + even = odd and if d + 4 is even, so when d is even, then d + 1 = even + odd = odd.
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If c and d are integers, is c even? [#permalink]
Bunuel wrote:
Anurag06 wrote:
Bunuel wrote:
If C and D are integers, is C even?

(1) C*(D+1) is even. In order the product to be even C must be even or D+1 must be even, or both. Hence it's not necessary C to be even. Not sufficient.

(2) (C+2)(D+4) is even. Notice that this is the same as C*(D+4) is even. Again, in order the product to be even C must be even or D+4 must be even, or both. Hence it's not necessary C to be even. Not sufficient.

(1)+(2) Both D+1 and D+4 cannot be even, thus C must be even. Sufficient.

Answer: C.


Hi Bunuel

I agree that Combination will prove C is even. But how can it be claimed that (d+1) and (d+4) both would not be even?

Because in both statements, either both can be even or either one can be even. So combining both surely would mean that both 'C' and 'D' are even.

Please correct me if am wrong.


(d+1) and (d+4) are 3 apart. How can both of them be even?

If d + 1 is even, so if d is odd, then d + 4 = odd + even = odd and if d + 4 is even, so when d is even, then d + 1 = even + odd = odd.


Oh Thank you.

I had done this considering the above logic actually without thinking that way.


The logic was similar to this - Here we have a number properties question. We will be dealing with evens/odds and multiplication/addition. A couple of things to keep in mind before we look at the statements:
1. If c is even it will have at least one factor of 2
2. The rules for addition for evens/odds: E+E = E; O+O = E; E+0 = O; O+E = O
3. The rules for multiplication for evens/odds: We will arrive at an even product EXCEPT when we have O*O which will produce an O result

1. Statement 1 is true if :
a. C is Even and D is Even E(E+1) = E*O = E
b. C is Even and D is odd E(O+1) = E*E = E
c. C is Odd and D is Odd O(O+1) = O*E = E
a and b give us an answer of yes; c gives us an answer of no - INSUFFICIENT

2. Statement 2 is true if:
a. C is Even and D is Even (E+2)(E+4) = E*E = E
b. C is Even and D is Odd (E+2)(O+4) = E*O= E
c. C is Odd and D is Even (O+2)(E+4) = O*E=E
a and b give us an answer of yes; c gives us an answer of no - INSUFFICIENT


When we take the statements together we notice that the two possible scenarios they have in common are a and b (C is Even and D is Even OR C is Even and D is Odd). In both scenarios C is Even, so we have one definitive answer to the question. SUFFICIENT the answer is C

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Re: If c and d are integers, is c even? [#permalink]
Given: c and d are integers
Asked: Is c even?

(1) c(d + 1) is even
Either c is even and d is odd; or
c is odd and d is even
NOT SUFFICIENT

(2) (c + 2)(d + 4) is even
Either c and d both are even; or
c and d both are odd

(1) + (2)
(1) c(d + 1) is even
(2) (c + 2)(d + 4) is even
(d+1) & (d+4) both can not be even
c must be even
SUFFICIENT

IMO C
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