A: is sufficient. Think about what non-integer y could you cube and get an integer. If you think of a non-integer as represented by a decimal - then multiplication rules will show that you will always end up with a non-integer. So if y^3 is an integer, then y itself must be an integer

B: is not sufficient. i.e. if y = 2/3, then 3*2/3 = 2. So 3y was an integer (2), but y was not an integer.

IMO C

1 Insufficient...there are cube roots that are integers and cube roots that are not
2 Insufficient... 1/3 * 3 is an integer and so is any integer times 3

Now I'm just guessing. Together.... I can't think of any non integers that can be cubed to make an integer and also multiplied by three to make an integer. So, I'm guessing by the two it would be an integer...so C, taken together they are sufficient

jeffedwards wrote:IMO C

1 Insufficient...there are cube roots that are integers and cube roots that are not
2 Insufficient... 1/3 * 3 is an integer and so is any integer times 3

Now I'm just guessing. Together.... I can't think of any non integers that can be cubed to make an integer and also multiplied by three to make an integer. So, I'm guessing by the two it would be an integer...so C, taken together they are sufficient

You are right. In the beginning, I thought A is the ans.....I wasn't able to understand how C could be the ans.
The explanation for this Q was given as:

1. y^3 is an integer.
if y^3 = 2, then y is not an integer. But if y^3 = 8, then y =2, which is an integer....Insufficient

2. The same explanation.....Insufficient

When 1 & 2 taken together....only for integer values of y, will both y^3 and 3y simultaneously be integers.

Does that mean, we need to solve it in the reverse order (i.e. by considering the root) rather than substituting for y, coz cube of any integer, undoubtedly, will be an integer!!

bharatishiv wrote: Does that mean, we need to solve it in the reverse order (i.e. by considering the root) rather than substituting for y, coz cube of any integer, undoubtedly, will be an integer!!

Good point, first you proved that it could be possible, but then you want to see if you can disprove it. Is there a non-integer that would work in that statement? That's what I normally try to do is see if I can disprove it. Negatives, zeros, and fractions are standard numbers to try, in addition to positive integers.

OKAY! GOT IT

Thank you

For those who'd like to see possibilities:

A. y^3 is an integer:

y^3 = 8 => y = 2
Y^3 = 10 => y = cuberoot(10) i.e. NOT an integer.
INSUFFICIENT

B. 3y is an integer:

3y= 3 => y = 1
3y= 4 => y= 4/3 i.e. NOT an integer.
INSUFFICIENT

Both A and B:
You know from B that y is basically an integer (Lets call it N) divided by 3. (y = N/3, where N is an integer)
Let's add in A. If y^3 is also an integer, this means that (N/3)^3 is an integer. Since N is an integer, it is clear that this will only be the case if N/3 is ALSO an integer. i.e. y is an integer. SUFFICIENT.

Answer is C