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2 cannot give you an answer, consider 1,2,3,4,5,6 or 3,4,5,6,7,8 both are not divisible by 9 whereas 2,3,4,5,6,7 is divisible and hence we cannot say conclusively.

D is the OA.
But I am unsure how stmt 2 plays out if you have -ve consecutive integers. There isn't really a proper OE but here's my reasoning:

Well, the sum S=6n+15=3(2n+5), where n is the first number

stmt 1) irrespective of the what n is, S will be always be odd. Suff.

stmt 2) says that the 6 digit number formed by each integer is divisible by 9 --> meaning the sum of the numbers is divisible by 9 --> 3(2n+5) = 9x
therefore, (2n+5)=3x . How do we conclude from this whether the sum is even or odd?

D is the OA. But I am unsure how stmt 2 plays out if you have -ve consecutive integers. There isn't really a proper OE but here's my reasoning:

Well, the sum S=6n+15=3(2n+5), where n is the first number

stmt 1) irrespective of the what n is, S will be always be odd. Suff.

stmt 2) says that the 6 digit number formed by each integer is divisible by 9 --> meaning the sum of the numbers is divisible by 9 --> 3(2n+5) = 9x therefore, (2n+5)=3x . How do we conclude from this whether the sum is even or odd?

Tank....

If you let 6n+15 equal to a multiple of 9, you will find out that the first term is an even number in the series. And from (1), we know that the sum is always odd. So (2) is sufficient.

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