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state 1....b is even......sp 10^a + b......if b = 2, then 10^a + 2 will always be divisible by 3 (as the sum of all digits will be 3) so ans is YES....
if b = 6, then 10^a + 6....if a is 1.....16 is not divisible by 3...insuff
state 2....b can only be from 1<b<10.....again insuff
The question is asking us whether 1+b is divisble by 3
Statement 1: b is an even integer.
if b is 2 b+1 will be divisbile by 3
if b is 6 b+1 will not be divisbile by 3,
So Statement 1 is not sufficient.
Statement 2: b < 10
if b is 8 b+1 will be divisbile by 3
if b is 6 b+1 will not be divisbile by 3
So Statement 2 is not sufficient.