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16 Jul 2025, 20:47
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(1) c(d+1) is even
We can't say about c because d+1 can be even and any number n multiplied by even is always even
Insufficient
(2) (c + 2)(d + 4) is even
Similar to say (1), we can't say about c because d+4 can be even and any number n multiplied by even is always even
Insufficient
Combining (1) + (2),
We know that c(d+1) is even
Now, we can write (2) as
(c + 2)((d+1)+3) = c(d+1) + 3c + 6 + 2(d+1) = even
We know that 6 + 2(d+1) will always be even and from (1) we know that c(d+1) is also even
It implies that 3c has to be even and it is only possible when c is even
(C) is the answer
We can't say about c because d+1 can be even and any number n multiplied by even is always even
Insufficient
(2) (c + 2)(d + 4) is even
Similar to say (1), we can't say about c because d+4 can be even and any number n multiplied by even is always even
Insufficient
Combining (1) + (2),
We know that c(d+1) is even
Now, we can write (2) as
(c + 2)((d+1)+3) = c(d+1) + 3c + 6 + 2(d+1) = even
We know that 6 + 2(d+1) will always be even and from (1) we know that c(d+1) is also even
It implies that 3c has to be even and it is only possible when c is even
(C) is the answer
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