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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.

(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.

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.

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...

Answer C...
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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.

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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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.

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....

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 Answer C