manik11 wrote:If each term in an infinite sequence is found by the equation Sx=8x-6, and S1=2 , is every term in the sequence divisible by y?

(1) y is an even integer.

(2) At least 2 of the first 4 terms in the sequence are divisible by y

OA : B
Source : Veritas Prep

First, let's get a handle on the sequence by seeing what the first few term should be. If Sx=8x-6, then S1 = 2, and S2 = 10, and S3 = 18, and S4 = 26. So those are our first four terms: 2, 10, 18, 26.

Now statement 1 tells us that y is even. If y =2, then every terms is divisible by y and the answer to the question is YES. If y = 4, the terms above are not divisible by y, so the answer to the question is NO. Not sufficient.

Statement 2 tell us that at least 2 of the first four terms are divisible by y. So we can see that y could be 2, in which case, YES every term is divisible by y. The question, then, is whether there are any other possible values for y. Well, y could be 1, but that would still give us a YES, as every integer is divisible by 1. So the thing to see here is that the terms above don't share any other factors:

Factors of 2: 1, 2
Factors of 10: 1, 2, 5, 10
Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 26: 1, 2, 13, 26

Because y must be 1 or 2, and both 1 and 2 yield an answer of YES, this statement alone is sufficient. And the answer is B.

Another approach:

Since each term can be rewritten as 2 * (4x - 3), we know that 2 is a factor of EVERY term in the sequence.

S1 tells us y is even. If y is 2, then every term will divide by y, but if y = 4, the first term won't divide by y, since 2 is not divisible by 4. NOT SUFFICIENT

S2 tells us that (8*1 - 6), (8*2 - 6), (8*3 - 6), and (8*4 - 6) are all divisible by y. Since these terms are 2, 10, 18, and 26, we can see that their common factors are 1 and 2, so y must be 1 or 2. Since 2 divides every term in our sequence (as we saw above) and 1 is a factor of 2, y will divide EVERY term in our sequence, regardless of whether y is 1 or 2. SUFFICIENT