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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
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Re: GMATprep DS (Number properties) [#permalink]
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30 Jan 2010, 15:39
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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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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: start with a second statement:
(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.
Answer C



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Re: GMAT PREP DS [#permalink]
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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. 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]
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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) = eveneven=even cd+ccd4c2d8=even c4c2d8=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 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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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 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If c and d are integers, is c even? [#permalink]
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06 Feb 2017, 12:20
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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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