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yezz
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C

from 1 c can be odd / even
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gmatnub
yezz
c and d are intigers is c even??

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


C
statement 1: cd + C = even, insuff
statement 2: 4c + 2d + cd + 8 = even, insuff

together: subtract statement 1 from 2 to get 3c + 2d + 8 = even

2d + 8 must be even, so 3c must also be even, so c is even

Yeppers! You nailed it.
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Twoone
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Yep,

C is the answer.

St1: either C or D+1 is even.

St2: either C+2 or D+4 is even.

combining both, D+1 and D+4 both can't be even at one time but C and C+2 has to be even as multiplication is even.

Hence. C is even. and C is the answer
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Great approaches both Twoone and gmatnub !!
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yezz
c and d are intigers is c even??

1) c(d+1) = even
2) (c+2)(d+4) = even
Clearly C

1) says c=even,d=odd
c=odd,d=odd
c=even,d=even
c can be even or odd

2)c=even,d=odd
c=odd,d=even
c=even,d=even

INSUFFI c =even or odd

again combining both we get c=even and d=odd
HENCE SUFFI
IMO C



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