Are both of the integers x and y divisible by 3? (1) x + y and x – y

Question: Are both x and y divisible by 3?

Statement 1: x + y and x – y are divisible by 3.

x+y = 3a
x-y = 3b

Adding the equations

2x = 3(a+b)
ie.. x = 3(a+b)/2 SInce x is an Integer therefore x will be a multiple of 3
Similar will be the derivation of divisibility of y by 3 hence

SUFFICIENT

Statement 2: x^2 and y^2 are divisible by 9.

SInce x and y are Integers therefore x and y both will be multiples of 3

SUFFICIENT

Answer: Option D

(1) x + y and x – y are divisible by 3.
x+y =3 Z : where Z is quotient
x-y = 3 P, where P is any quotient
Adding two : 2x = 3 (z+p),
As equation is multiple of 3, it will be divisible by 3.
Another way can be picking of numbers, But you will not find numbers whose difference and addition is divisible by 3, Only satisfied case would be when numbers are multiple of 3 e.g
10 -7 =3 : but 10+7 =17 not divisible, so doesn't satisfy equation 2
18-15 = 3: and 18 +15 =33, both equation satisfied. and individually numbers divisible by 3

Sufficient

(2) x^2 and y^2 are divisible by 9.

As per number theory, any number which is divisible by 9 is divisible by 3, So square of it will be also be divisible by 3,
Clearly sufficient

IMO D

If X and y are a combination of 3 and 0, then this is not true? 0 is not divisible by 3. Can someone explain why this logic is wrong? Because of this, I picked B.

GMATNinja

0 is divisible by 3. In fact, 0 is divisible by all numbers

In general, if \(\frac{a}{b}=integer\) then \(a\) is divisible by \(b\)

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(1) If the sum of two numbers and the difference of two numbers are divisible by 3, then each number is divisible by 3. SUFFICIENT.

(2). \(x^2\) and \(y^2\) are divisible by \(3^2\). Therefore x and y are divisible by 3. SUFFICIENT.

Answer is D.
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When we consider Statement 2, why can't x^2 = 18? It just says that it's divisible by 9. So x would not come out to be a number that is divisible by 3. In that case, statement 2 would be insufficient. I ended up picking A as the right choice. Please help!

If x^2 = 18, then x is not an integer (x would equal 3√2, which is roughly 4.24), but the question tells us x is an integer, so x^2 cannot equal 18.
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Oh understood. So basically, while considering either of the statements, one must also assume that everything given as conditions in the question stem itself is the entire sample set?

10+17/3 = 27/3, here x and y is not divisible by 3.
17-8/3, here also x and y is not divisible by 3.

So, I thought the rule is x+y can either both be divisible by 3 or both of them not divisible by 3.

Here, the questions says that x+y and x-y is simultaneously divisible by 3. Is this the reason why A is sufficient??
To satisfy both the equation, x and y has to be divisible by 3??

Can some one pls explain here?

In (1), x and y cannot be 10 and 17 because 10 - 17 = -7 is not divisible by 3. Similarly, x and y cannot be 17 and 8 because 17 + 8 = 25 is not divisible by 3. When choosing numbers for x and y, they must satisfy both of the conditions: x + y and x – y must be divisible by 3.

Next, here is why (1) is sufficient. We are given that both x + y and x – y are divisible by 3:
x + y = (a multiple of 3)
x - y = (a multiple of 3)

Sum the above to get 2x = (a multiple of 3). Since the right hand side is a multiple of 3, then the left hand side must also be a multiple of 3, which means that x is a multiple of 3. If x is a multiple of 3, then from x + y = (a multiple of 3) it follows that y must also be a multiple of 3.

Hope it helps.
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