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Bunuel
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D.

1) I didn't know how to do this any other way but test numbers:

X=2, y=8
X =3, y = 14
X = 5, y = 32

Clearly not divisible by 3

2) y = 3x + 2

y-2/(3) = x or y/3 = x + 2

See the trick? It's a remainder set up.

This means that y has a remainder of 2 when dividing by 3. Sufficient

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Imo D
Taking 1 0 and -1 we can see that none of the expression is divisible by 3 .

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Bunuel
If x and y are integers, is y divisible by 3?

(1) y = x^2 + x + 2
(2) y = 3x + 2

Statement 1: y = x^2 + x + 2.

Now , x^2 + x = even ( all time ) regardless of the value of x .

even + 2 = even .

So, statement 1 is not divisible by 3.

Sufficient.

Statement 2: y = 3x + 2 .

Here 3x is divisible by 3 but 2 becomes remainder .

Thus , y is not divisible by 3.

Sufficient.

D is the correct answer.
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Bunuel - Can you please explain how equation 1 is sufficient?
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Step 1: Analyse Question Stem

x and y are integers.
We have to find out if y is divisible by 3.

Step 2: Analyse Statements Independently (And eliminate options) – AD / BCE

Statement 1: y = \(x^2\) + x + 2

The expression on the RHS can be rewritten as x(x + 1) + 2.
Therefore, y = x(x + 1) + 2.

If x = 1, y = 1 (2) + 2 = 4. Is y divisible by 3? NO

If x = -1, y = -1 (0) + 2 = 2. Is y divisible by 3? NO

If x = 2, y = 2(3) + 2 = 8. Is y divisible by 3? NO

If x = 3, y = 3 (4) + 2 = 14. Is y divisible by 3? NO

So, we see that, regardless of whether x is a multiple of 3 or not, y is NEVER divisible by 3.

The data in statement 1 is sufficient to answer the question with a definite NO.
Statement 1 alone is sufficient. Answer options B, C and E can be eliminated.

Statement 2: y = 3x + 2

This tells us that y will always leave a remainder of 2, when divided by 3. Therefore, y is NEVER divisible by 3.

The data in statement 2 is sufficient to answer the question with a definite NO.
Statement 2 alone is sufficient. Answer option A can be eliminated.

The correct answer option is D.
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thanbass
I think the fastest and most reliable approach is the following:

(1) y = x^2 + x + 2
If x is even:
even*even + even + 2 = even + even + 2 = even
If x is odd:
odd*odd + odd + 2 = odd + odd + 2 = even +2 = even

In both cases y is even, so not divisible by 3
Sufficient

(2) y = 3x + 2

For whatever value of x you put, y divided by 3 will always leave a remainder of 2 because it starts with y=2 (for x=0) and then goes on adding 3 forever.
Sufficient

Therefore, answer is D.

In statement 1-
Even numbers can be divisible by 3,

3x2 = 6
3x4 = 12
3x6 = 18

and so on..

so y can be even and yet divisible by 3
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