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# If c and d are integers, is c even?

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

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30 Jan 2010, 13:15
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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
[Reveal] Spoiler: OA
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Re: GMATprep DS (Number properties) [#permalink]

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30 Jan 2010, 15:23
zaarathelab wrote:
If c and d are integers, is c even?

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

What's the fastest way to solve this?

Statement 1:

C(d+1) is even : Insufficient
3 cases: Case 1: C is even, d+1 is even => C is even, D is Odd.
Case 2: C is even, d+1 is Odd => C is even, D is Even.
Case 3: C is Odd, d+1 is even => C is Odd, D is Odd.

Statement 2:
(C+2)(D+4) is even: Insufficient
3 Cases: Case 1:(C+2)is even, (D+4) is even => C and D : Both even
Case 2: (C+2) is even, (D+4) is Odd => C is even, D is Odd
Case 3: (C+2) is Odd, (D+4) is even => C is Odd, D is Odd.

Together: Insufficient: As multiple cases exist for both the statements.
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Re: GMATprep DS (Number properties) [#permalink]

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30 Jan 2010, 15:39
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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.

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Re: GMATprep DS (Number properties) [#permalink]

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31 Jan 2010, 10:49
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: GMATprep DS (Number properties) [#permalink]

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31 Jan 2010, 11:43
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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: GMATprep DS (Number properties) [#permalink]

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01 Feb 2010, 14:46
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obviously, 1 or 2 separatly is not sufficient.

Both:

(c+2)(d+4)=cd +4c+2d+8=even => cd is even

now statement one says :

c(d+1)=cd+c is even. From statement 2, cd is even => for stament 2 to be true c has to be even.

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Re: GMATprep DS (Number properties) [#permalink]

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02 Feb 2010, 05:29
alexBLR has applied good strategy.

Intially I went to do it all in head. But it works better if you do it on the paper--substitution with number.

C even?

1. say NO and check: 5 *(7+1) = even,
say YES and check: 8*(7+1) 0 even -> not sufficient, clearly not A

2. say NO and check: (5+2)*(2+4) = even
say YES and check: (6+2)*(2+4) = even -> not sufficient, clearly not B

3. say No and check: 5*(7+1) = even; (5+2)*(7+4) = odd -> restricted
5*(8+1) = odd; (5+2)*(8+4) = even ->restricted
say YES and check: just one look and you know (1) and (2) both are satisfied

Maybe it is not that faster. With numbers there are too many things to assume if I do it in head.
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Re: GMATprep DS (Number properties) [#permalink]

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02 Feb 2010, 13:15
zaarathelab wrote:
If c and d are integers, is c even?

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

What's the fastest way to solve this?

S1:
c(d+1) is even means either c is even or d+1 is even..... not suff

S2:
(c+2)(d+4) is even means either (c+2) is even or (d+4) is even... not suff

Both together:
c(c+2)(d+1+3) = c(d+1) + 3c + 2(d+1+3) which implies Even + __ + Even = Even

Therefore 3c is even... Hence c is even...

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Re: GMATprep DS (Number properties) [#permalink]

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14 Mar 2010, 02:58
zaarathelab wrote:
If c and d are integers, is c even?

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

What's the fastest way to solve this?

1. c can be odd then d can be odd
c can be even then d can be odd or even
alone insufficient.

2. c can be odd then d is even
c can be even then d is even
alone insufficient.

combining both common is c has to even d can be odd or even.
So sufficient.
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22 Aug 2010, 19:27
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.

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

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22 Feb 2011, 01:58
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]

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22 Feb 2011, 02:00
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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22 Feb 2011, 02:09
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: GMATprep DS (Number properties) [#permalink]

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24 Feb 2011, 23:38
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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05 Jan 2014, 18:25
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

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

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06 Jan 2014, 09:43
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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08 Mar 2015, 12:22
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

[Reveal] Spoiler:
C

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If c and d are integers, is C even? 1) c(d+1) is even 2) (c+2)(d+4) [#permalink]

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28 Mar 2016, 13:32
If c and d are integers, is C even? 1) c(d+1) is even 2) (c+2)(d+4) is even
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Re: If c and d are integers, is C even? 1) c(d+1) is even 2) (c+2)(d+4) [#permalink]

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28 Mar 2016, 13:46
1) c(d+1) is even: c is even and d is odd OR c is even and d is even OR c is odd and d is odd - Not sufficient
2) (c+2)(d+4) is even: c even and d even OR c odd and d even OR c even and d odd - Not sufficient
Combine: Only common part: c is even and d is odd OR c even and d even => c is even - Sufficient
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Re: If c and d are integers, is c even? [#permalink]

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28 Mar 2016, 13:57
deepaks12988 wrote:
If c and d are integers, is C even? 1) c(d+1) is even 2) (c+2)(d+4) is even

Merging topics. Please refer to the discussion above.
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Re: If c and d are integers, is c even?   [#permalink] 28 Mar 2016, 13:57

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