Question: Divisible by 3

Comment on Divisible by 3

I don't understand this.

If you make x=3 and y=3, then the equation in the question works out, but so do all the answer choices. Because the product of all the binomial expressions will always have a 3 in its prime factorisation.

Why isn't this legitimate? Or is it just a case of "might rather than must"? In which case, is your choice of x=2, y=1 instead just an inspired guess or based on something one should be able to spot? I chose x=3, y=3 because that was the easy way to ensure (8x + 11y) had a factor of 3.
greenlight-admin's picture

If a question asks "What MUST be true?" then that is the same as asking "What MUST ALWAYS be true?"

So, when you used x=3 and y=3, you learned that all of the answer choices are SOMETIMES true.

So, for example, when x=3 and y=3, you learned that x+y (answer choice A) is SOMETIMES true. This, however, doesn't mean that the correct answer is A.

Here's an analogous question: If x and y are integers, which of the following MUST be an even number.
A) x + y
B) 2xy
C) 3x - y
If we let x = 1 and y = 1, then A, B and C are all even. However, does this mean that answer choice A MUST be even?
No. If x = 1 and y = 2, then A is NOT even.

We can see, however, that B MUST be even, regardless of what integer values we use for x and y.

Does that help?

So what if I chose X=3 and Y=3 and found out all the answer choices work, then I work on X=3 and Y=6 for which all the answer choices are correct.

If that is the case should I keep selecting numbers to plug until I find a result which only works for one answer choice.

So how should we select the number to plug for these type of questions.
greenlight-admin's picture

Try choosing x-values and y-values that are not already divisible by 3 (e.g., 3, 6, 9 etc). Otherwise, all of the answer choices will continue to work.

Why is this? Well, this is one of the properties of divisors (see

So, if x is divisible by 3, then 2x, 3x, 7x and 8x will all be divisible by 3. Likewise, if y is divisible by 3, then 2y, 5y, 4y and 9y will all be divisible by 3.

There's also another rule that says, if j and k are both divisible by 3, then j+k and j-k will also be divisible by 3.

So, by choosing values that are already divisible by 3, you are ensuring that all of the answer choices will be divisible by 3.

greenlight-admin's picture

Our goal is to find an x-value and y-value so that 8x + 11y is divisible by 3.

So, let's try x = 1 (not divisible by 3).
So, 8x = 8(1) = 8
We need to add 11y to 8 AND get a sum that's divisible by 3.
If y = 1, then 11y = 11(1) = 11, which means 8x + 11y = 8 + 11 = 19 - NOT divisible by 3.

If y = 2, then 11y = 11(2) = 22, which means 8x + 11y = 8 + 22 = 3o, which IS divisible by 3.

Now test the answer choices by plugging in x=1 and y=2

When we do so, only answer choice B is divisible by 3.


Thank you for the clarification It is Clear now.

However, when 8x+11y is divisible by 3, then 8x must also be divisible by 3 and 11y should also. then 8x+11y can be divisible. but 8 is not divisible by 3 therefore x must have value that is divisible by 3 right?, even this is the case in 11y also.
so i chose x and y values as multiples of 3, i was trapped.

seems like this hard questions have to be learned with the experience, your first approach is just awesome.
greenlight-admin's picture

I think you're misinterpreting one of the divisor rules from

The rule says: If k is a divisor of BOTH M and N, then k is a divisor of M+N

This rule goes one way: If X then Y. We cannot conclude that the opposite direction is true (If Y then X)

That is, we CANNOT say "If k is a divisor of M+N then k is a divisor of BOTH M and N

I believe this is what you are doing. You are taking the information that says 8x + 11y is divisible by 3 and concluding that 8x and 11y are BOTH divisible by 3.

We can show that this conclusion is incorrect. For example, if x = 1 and y = 2, then 8x + 11y is divisible by 3, BUT 8x and 11y are NOT both divisible by 3.

oh! yeah, converse doesn't work.

You are genius.

I tried the plugging of value method, and got the answer pretty fast.
Does this approach always work provided we select non-multiple numbers?
greenlight-admin's picture

I'm hesitant to say "ALWAYS" works.

Yes, the approach often works quite nicely. That said, there's a bit of luck involved in testing values. In some cases, the first value(s) you plug in may allow you to eliminate 4 answer choices, in which case you're done.

In other cases, the first value(s) you plug in may allow you to eliminate only 2 or 3 answer choices, in which case you must test ANOTHER value (or values).

Also note that some GRE quant questions are NUMERIC ENTRY, in which case you won't have any answer choices to work with.


Hi Brent,

I tried also another approach, based also on your previous courses from this module.

We know that 8x + 11y is divisible by 3.
We will rewrite all answer choices using this formula:

A) x + 2y = (8x - 7x) +( 11y -9y)= (8x + 11y) -7x -9y
From here we know that 8x + 11y and -9y are divisible by 3.
-7x is not necessarily divisible by 3 so the answer cannot be A

B) 2x + 5y = (8x - 6x) + (11y-6y) = (8x + 11y) + (-6x) + (-6y)
Since all factors now are divisible by 3 we know that this is the answer. Thanks a million, great lessons!
greenlight-admin's picture

Perfect! Great work!!

Given 8x + 11y / 3 gives rem as 0

SO 8x+11y =3

As 11-8=3 x can be -1 and y can be 1

then plug in values in 2x+5y => 2(-1)+5(1)=3..
BOOM!! that is it!!
greenlight-admin's picture

Boom goes the dynamite!

I love how in so many integer problems, you can literally just put in some numbers that satisfy whatever conditions the problem is stating.
greenlight-admin's picture

I agree! I'd say at least 50% of all integer properties questions can be solved by testing values that satisfy the given information.

What if I break the upper function?
(8x + 11y) = (6x + 6y + 2x + 5y)/3
2x + 2y + (2x + 5y)/3
Here 2x + 5y must be divisible by 3, otherwise, it won't be an integer.

is it a good technic or was I just lucky?
greenlight-admin's picture

Your technique works perfectly. Great work!!

greenlight-admin's picture

Your technique works perfectly. Great work!!

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