Here is the solution to show why B is correct

Statement 2 says y is divisible by 6. so Y will be multiple of 6.
Hence Y can be written as ;Y=k*6=k*2*3(where K is constant)
If take cube of Y ,So y^3=k^3*2^3*3^3=k^3*2^3*3*9

You can see y^3 has 9 as factor, so it is divisible by 9

GhassanMBA wrote:If y is an integer, is y^3 divisible by 9?

1) y is divisible by 4
2) y is divisible by 6


The OA is B

The explanation in the book is too confusing, and I would appreciate if someone can clarify why B is the answer. I've generally been having difficulties with this kind of questions.
Thanks

If y is an integer then y^3 is a cube integer which will be divisible by 9 once y is divisible by 3, otherwise NOT.

(1) A number divisible by 4 may or may not be divisible by 3 or is NOT necessarily divisible by 3. Insufficient

(2) A number divisible by 6 is necessarily divisible by 3. Sufficient

So, [spoiler]B[/spoiler]

sanju09 wrote:

GhassanMBA wrote:If y is an integer, is y^3 divisible by 9?

1) y is divisible by 4
2) y is divisible by 6


The OA is B

The explanation in the book is too confusing, and I would appreciate if someone can clarify why B is the answer. I've generally been having difficulties with this kind of questions.
Thanks

If y is an integer then y^3 is a cube integer which will be divisible by 9 once y is divisible by 3, otherwise NOT.

(1) A number divisible by 4 may or may not be divisible by 3 or is NOT necessarily divisible by 3. Insufficient

(2) A number divisible by 6 is necessarily divisible by 3. Sufficient

So, [spoiler]B[/spoiler]

_______________________
Hey sanju09,
I see that you have pointed out 2 divisibility rules.
My question is that :Isn't it better to solve the question using the long approach as given by ramsharma.
I personally don't gain any confidence until I solve the questions involving both the "power" and the "divisibility rules".

Going by the long approach atleast I can be sure of my answer or should I try to save some time on these question types by just remembering the rules and applying shortcut techniques.

Seeking advice??

harsh.champ wrote:

Hey sanju09,
I see that you have pointed out 2 divisibility rules.
My question is that :Isn't it better to solve the question using the long approach as given by ramsharma.
I personally don't gain any confidence until I solve the questions involving both the "power" and the "divisibility rules".

Going by the long approach atleast I can be sure of my answer or should I try to save some time on these question types by just remembering the rules and applying shortcut techniques.

Seeking advice??

Ramsharma's approach is to show why statement (2) by itself is sufficient and he did justice with that. While dealing with small positive integer numbers and powers simultaneously, his approach is quite handy. This question could be answered well within 2 minutes with his approach, no doubt. But that never means that a GMAT aspirant can deal with its quantitative section effectively without learning and practicing the most required strategies and short cuts that have been tried and tested so many times on the test so far. Divisibility rules of very a few positive integers, like 2, 3, 4, 5, 6, 8, 9, and 11 are very important to ponder and practice before writing the main test.

It's true that going by the long approach at least they can be sure of their answer but that won't work for them throughout the section. You'll start getting faith in strategies and short cuts too, once you give it few chances, and that would, in fact, turn out to be really productive for you on real test.

ATB

Hi sanju,

where can I get information on divisibility of numbers that you mentioned?

Thanks.

dxgamez wrote:Hi sanju,

where can I get information on divisibility of numbers that you mentioned?

Thanks.

see if this page could be of any help...


https://en.wikipedia.org/wiki/Divisibility_rule


ATB

Hello,

here is the explanation in the book:

1) Some multiples of 4 are divisible by 3, such as 12(y=12 is divisible by 3 hence y^3=12^3 is divisible by 27, and so y^3 = 12^3 is divisible by 9), and some multiples of 4 are not divisible by 3(y=8 is not divisible by 3, hence y^3 = 8^3 is not divisible by 3, and so y^3=8^3 certainly cannot be divisible by 9); NOT SUFFICIENT

Now my question is; how can one understand whether a power of a number is divisible by some other number or not without doing a long calculation each time. For example, I know that 12 is divisible by 3 so I take 12 but how do I know if 12^3 is divisible by 9 without doing the calculation as 12x12x12/9? What is the short way to know this??

Thanks

ozlemmetje wrote:Hello,

here is the explanation in the book:

1) Some multiples of 4 are divisible by 3, such as 12(y=12 is divisible by 3 hence y^3=12^3 is divisible by 27, and so y^3 = 12^3 is divisible by 9), and some multiples of 4 are not divisible by 3(y=8 is not divisible by 3, hence y^3 = 8^3 is not divisible by 3, and so y^3=8^3 certainly cannot be divisible by 9); NOT SUFFICIENT

Now my question is; how can one understand whether a power of a number is divisible by some other number or not without doing a long calculation each time. For example, I know that 12 is divisible by 3 so I take 12 but how do I know if 12^3 is divisible by 9 without doing the calculation as 12x12x12/9? What is the short way to know this??

Thanks

Hi ozlemmetje,

Once you know that 3 is a factor of 12, you are in every position to accept that (3)(3) is a factor of (12)(12). Hence we can say that 9 is a factor of (12)^2 and the greater positive integer exponents of 12 beyond 2, such as, YES, 9 is a factor of (12)^3 as well.