# Question: Divisible by 3

## Comment on Divisible by 3

### can we use x=1 and y=2

can we use x=1 and y=2

### Definitely.

Definitely.
When x = 1 and y = 2, the given condition (8x + 11y is divisible by 3) is met, so we can definitely plug those values into the answer choices to see which one (or ones) yield a result that is divisible by 3.

### Hi I want to know how we know

Hi I want to know how we know that 3x + 3y is divisible by 3. I am not sure where that equation came from. Is there a video here you can refer me to. Also, what difficulty level would this question be?

### We can take 3x + 3y and

We can take 3x + 3y and factor out a 3 to get: 3(x + y)
Since x and y are integers, we know that x+y is an integer, which means 3(x + y) is a multiple of 3.

If 3(x + y) is a multiple of 3, then we can also say that 3(x + y) is divisible by 3.

Alternatively, there's a rule that says "If x is divisible by k, and y is divisible by k, then (x + y) is divisible by k. Since 3x is divisible by 3 and 3y is divisible by 3, then we know that 3x + 3y is divisible by 3.

This is a very tricky question. Probably in the 700+ range.

### I'm still a bit confused...I

I'm still a bit confused...I understand the expression 8x + 11y has a hidden 3 either in x or y or both since there's no 3 in 8 or 11. Is that the basis for the logic that (x+ y) must then also be divisible by 3, regardless of which variable hides the 3? Or is it just because literally ANY integer multiplied by 3 would be divisible by 3 thus your second expression 3x+3y?

### There are 2 important divisor

There are 2 important divisor rules that start at 3:35 in the following video: https://www.gmatprepnow.com/module/gmat-integer-properties/video/831

#1) If k is a divisor of M and N, then k is a divisor of M+N and a divisor of M-N
#2) If k is a divisor of M, but NOT a divisor of N, then k is NOT a divisor of M+N and k is NOT a divisor of M-N

So, if 8x + 11y is divisible by 3, then:
EITHER 8x and 11y are both divisible by 3
OR
8x is not divisible by and 11y is not divisible by 3.

Your question: Is that the basis for the logic that (x+y) must then also be divisible by 3....?
I'm not saying that (x+y) must be divisible by 3; I'm saying that 3x+3y must be divisible by 3.
If (3x+3y) and (8x+11y) are BOTH divisible by 3, then we can apply Rule #1 to conclude that [(3x+3y)+(8x+11y)] is divisible by 3.
Likewise, [(8x+11y)-(3x+3y)] is divisible by 3.

And so on.

Does that help?

Cheers,
Brent

### Thank you. If I understand

Thank you. If I understand correctly then if the question stated for example 5x+2y was divisible by 7, I could create the expression 7x+7y is divisible by 7 and then use the same rule #1 to show that 7 is divisible by 5x+ 2y and 7x +7y therefore the sum or difference of both expressions is also divisible by 7.

### I think you meant to say "5x

I think you meant to say "5x+2y and 7x+7y are divisible by 7, and therefore the sum or difference of both expressions is also divisible by 7."

If so, then you're absolutely right :-)

Cheers,
Brent

### Dear Brent, I used the

Dear Brent, I used the different approach and shockingly find intriguing answer : all of the possible answers are divisible by 3.

1. First we break the given equation into prime factorization : we get (2^3)X + (11)Y.
2. Since 2 and 11 both not divisible by 3, we must conclude that X and Y contain both number that must divisible by 3.
3. From that, I chose the number 3 as minimum number both X and Y must have. So, we can write : (2^3)(3 X n1 X n2) + (11)(3 X n1 x n2).
4. This automatically means that whatever the coefficient of X and Y, the equation must be divisible by 3.

Now I start to think, what is wrong with my approach?

### Your second step (2. Since 2

Your second step (2. Since 2 and 11 both not divisible by 3, we must conclude that X and Y contain both number that must divisible by 3.) is not correct. x and y need not be divisible by 3.

For example, if x = 1 and y = 2, then 8x + 11y = 8(1) + 11(2) = 8 + 22 = 30. In this example, 8x + 11y is divisible by 3, BUT x and y are not divisible by 3.

### Hi Brent. How did you know

Hi Brent. How did you know that should select 2 and 1 as your x and y, respectively? Do you just try random small integers until you can satisfy the ‘8x+11y is divisable by 3’ condition? Because at first I tried both x and y to equal 3, but that obviously resulted in all options being divisable by 3, and only after a while I figured that x=4 and y=5 to satisfy the condition. Is there a fast way to find the right x and y values or is random plugging and checking the only way? Thanks!

### There's no fast way to find

There's no fast way to find values of x and y such that 8x+11y is divisible by 3, but it shouldn't take long to do so.

NOTE: When plugging in values, you shouldn't expect to eliminate 4 of the 5 answer choices in one step.

If you happen to eliminate 4 of the 5 answer choices in one step, that's a bonus. But if you can eliminate 2 or 3 answer choices on the first step, and then eliminate all but 1 answer choice on the second step, then the entire solution will likely take less than 1 minute.

That said, when I tried to come up with two more pairs of values (x and y) such that 8x+11y is divisible by 3, I found that both pairs of values allowed me to eliminate 4 of the 5 answer choices.

The pairs I tried were:
x = -1, y = 1
x = 1, y = 2

Cheers,
Brent

### Hey Brent,

Hey Brent,

We can find the answer in less time by noticing that x is accompanied with an even integer(8) and y is with an odd integer(11), so as we analyze answer choices we see that only 2x+5y have x with even integer and y with odd!
Does this approach work on divisibility questions?

### Great idea, but that approach

Great idea, but that approach won't always work.

For example, 4x + 3y is not necessarily divisible by 3.

And, conversely, x + y IS definitely divisible by 3.

Cheers,
Brent

### https://gmatclub.com/forum

https://gmatclub.com/forum/the-difference-between-positive-two-digit-integer-a-and-the-smaller-tw-196634.html

### Hi Brent,

Hi Brent,

I happened to choose different numbers to plug in x=3, y=6, 8x3+11x6=90, 90 is divisible 3. But when I plug in these numbers into answer choice A), x+2y I get 3+2x6=15, which is divisible by 3. And as a matter of fact, if I will plug these numbers in the rest of the answer choices each of the answer choice will be divisible by 3 (we even can guess that all the choices will be divisible by 3 because we picked X and Y to be divisible by 3). So I am a bit confused here.. Can we really use the 2nd method of plugging in values???

### As you noted, the problem

As you noted, the problem with your solution is that the values you chose (3 and 6) are both divisible by 3. So, every answer choice will be divisible by 3.

The question doesn't state that x and y must be divisible by 3. It says that the sum, 8x + 11y, must divisible by 3. So, try to find values of x and y that satisfy that condition, yet are not each divisible by 3.

If you do that, you'll find that the strategy works.

Cheers,
Brent

### I picked x=3 and y=3 and the

I picked x=3 and y=3 and the first expression x+2y = 3+6=9 is also divisible by 3

### The key word here is MUST.

The key word here is MUST.
By testing x = 3 and y = 3, you've shown that x+2y COULD be divisible by 3, but that doesn't mean that x+2y MUST be divisible by 3.
When we test x = 2 and y = 1, we see that x+2y isn't divisible by 3.
This means x+2y isn't always divisible by 3, so we can eliminate answer choice A.

Does that help?