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# M18-03

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Joined: 02 Sep 2009
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16 Sep 2014, 01:02
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75% (hard)

Question Stats:

56% (01:57) correct 44% (01:37) wrong based on 241 sessions

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If $$x$$ and $$y$$ are integers, is $$xy$$ divisible by 3 ?

(1) $$(x + y)^2$$ is divisible by 9

(2) $$(x - y)^2$$ is divisible by 9

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16 Sep 2014, 01:02
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Official Solution:

Statement (1) by itself is insufficient. S1 gives that $$x + y$$ is divisible by 3. Consider $$x = 1$$, $$y = 2$$ (the answer is "no") and $$x = 3$$, $$y = 3$$ (the answer is "yes").

Statement (2) by itself is insufficient. S2 gives that $$x - y$$ is divisible by 3. Consider $$x = 4$$, $$y = 1$$ (the answer is "no") and $$x = 6$$, $$y = 3$$ (the answer is "yes").

Statements (1) and (2) combined are sufficient. If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3. Because $$x$$ is an integer, it must be divisible by 3. Therefore, $$xy$$ is divisible by 3 as well.

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09 Dec 2016, 17:31
If we say x+y is divisible by 3 and x-y is divisible by 3, does that mean x is divisible by 3? Is it alright to combine the statements?
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09 Dec 2016, 18:17
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cuhmoon wrote:
If we say x+y is divisible by 3 and x-y is divisible by 3, does that mean x is divisible by 3? Is it alright to combine the statements?

In the present way it is written ans is NO..
Say x is 9/2 and y is 3/2...
x+y=9/2+3/2=12/2=6, div by 3..
-x-y= 9/2-3/2=6/2=3, div by 3..
But x, 9/2 is div by 3,.. NO

But if it is given x and y are integers..YES
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19 Jan 2017, 19:55
If x=3 and y=0, then both statements hold true but xy is not divisible by 3. Am I missing something or is the question incorrect?
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19 Jan 2017, 19:56
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 gives that $$x + y$$ is divisible by 3. Consider $$x = 1$$, $$y = 2$$ (the answer is "no") and $$x = 3$$, $$y = 3$$ (the answer is "yes").

Statement (2) by itself is insufficient. S2 gives that $$x - y$$ is divisible by 3. Consider $$x = 4$$, $$y = 1$$ (the answer is "no") and $$x = 6$$, $$y = 3$$ (the answer is "yes").

Statements (1) and (2) combined are sufficient. If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3. Because $$x$$ is an integer, it must be divisible by 3. Therefore, $$xy$$ is divisible by 3 as well.

If x=3 and y=0, then both statements hold true but xy is not divisible by 3. Am I missing something or is the question incorrect?
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19 Jan 2017, 20:06
Cez005 wrote:
If x=3 and y=0, then both statements hold true but xy is not divisible by 3. Am I missing something or is the question incorrect?

Hi

0 is div by all numbers..
So xy =3*0=0 and 0 is div by 3..
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05 Jun 2017, 06:23
2
Alternatively we can simply proceed with Algebra:

Statement (1) Insufficient. Let, (x + y)^2 = 9P where P is a perfect square. If you take square root on both sides, (x + y) is divisible by 3. But that has nothing to do with divisibility of xy

Statement (2) Insufficient. Let, (x - y)^2 = 9Q where Q is also a perfect square. Again by similar logic, It says that (x - y) is divisible by 3. But that has nothing to do with divisibility of xy

Now the Combination of both Statements:

Subtracting (2) from (1) we get, 4xy = 9(P - Q). So, xy = 9.(P - Q)/4
As x and y are both integers (and even P and Q are also integers), xy actually is divisible by 9 (so, of course divisible by 3 as well).

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05 Jun 2017, 07:36
1
The answer has to be option C. My take:

Statement 1 - Not sufficient. We don't know if xy is a factor of 3.
Statement 2 - Not sufficient. We don't know if xy is a factor of 3.

Combine 1 & 2 - subtract "2" from "1". We get 4xy=9 * (some integer). 4 and 9 have no common factor, the inference in such a case is that xy and 9 have a common factor. Hence 3 is a factor of xy.

Hence option C.
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05 Jun 2017, 10:11
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 gives that $$x + y$$ is divisible by 3. Consider $$x = 1$$, $$y = 2$$ (the answer is "no") and $$x = 3$$, $$y = 3$$ (the answer is "yes").

Statement (2) by itself is insufficient. S2 gives that $$x - y$$ is divisible by 3. Consider $$x = 4$$, $$y = 1$$ (the answer is "no") and $$x = 6$$, $$y = 3$$ (the answer is "yes").

Statements (1) and (2) combined are sufficient. If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3. Because $$x$$ is an integer, it must be divisible by 3. Therefore, $$xy$$ is divisible by 3 as well.

Hi, i just want you to confirm that in any case we cannot consider 0 as a interger, right?

0 is only even number not even interger number.

Thanks.

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05 Jun 2017, 10:50
goalMBA1990 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 gives that $$x + y$$ is divisible by 3. Consider $$x = 1$$, $$y = 2$$ (the answer is "no") and $$x = 3$$, $$y = 3$$ (the answer is "yes").

Statement (2) by itself is insufficient. S2 gives that $$x - y$$ is divisible by 3. Consider $$x = 4$$, $$y = 1$$ (the answer is "no") and $$x = 6$$, $$y = 3$$ (the answer is "yes").

Statements (1) and (2) combined are sufficient. If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3. Because $$x$$ is an integer, it must be divisible by 3. Therefore, $$xy$$ is divisible by 3 as well.

Hi, i just want you to confirm that in any case we cannot consider 0 as a interger, right?

0 is only even number not even interger number.

Thanks.

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Hi

'0' is an Integer
.

On the basis of their divisibility by 2, ALL integers can be classified into two types:
Even (those which are divisible by 2) and Odd (those which are NOT divisible by 2)

Since 0 is divisible by 2 (actually 0 is divisible by all non-zero numbers), Hence 0 is considered an Even integer.
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05 Jun 2017, 11:01
amanvermagmat wrote:
goalMBA1990 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 gives that $$x + y$$ is divisible by 3. Consider $$x = 1$$, $$y = 2$$ (the answer is "no") and $$x = 3$$, $$y = 3$$ (the answer is "yes").

Statement (2) by itself is insufficient. S2 gives that $$x - y$$ is divisible by 3. Consider $$x = 4$$, $$y = 1$$ (the answer is "no") and $$x = 6$$, $$y = 3$$ (the answer is "yes").

Statements (1) and (2) combined are sufficient. If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3. Because $$x$$ is an integer, it must be divisible by 3. Therefore, $$xy$$ is divisible by 3 as well.

Hi, i just want you to confirm that in any case we cannot consider 0 as a interger, right?

0 is only even number not even interger number.

Thanks.

Sent from my XT1663 using GMAT Club Forum mobile app

Hi

'0' is an Integer
.

On the basis of their divisibility by 2, ALL integers can be classified into two types:
Even (those which are divisible by 2) and Odd (those which are NOT divisible by 2)

Since 0 is divisible by 2 (actually 0 is divisible by all non-zero numbers), Hence 0 is considered an Even integer.

Hi,
Thank you very much but now i am more confuse.
Since 0 is an integer then answer to this question should be E. Can you please confirm?

Also zero is divisible by all numbers then why we consider it as an even number?

Thanks.
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06 Jun 2017, 04:46
goalMBA1990 wrote:
amanvermagmat wrote:
goalMBA1990 wrote:
[quote="Bunuel"]Official Solution:

Statement (1) by itself is insufficient. S1 gives that $$x + y$$ is divisible by 3. Consider $$x = 1$$, $$y = 2$$ (the answer is "no") and $$x = 3$$, $$y = 3$$ (the answer is "yes").

Statement (2) by itself is insufficient. S2 gives that $$x - y$$ is divisible by 3. Consider $$x = 4$$, $$y = 1$$ (the answer is "no") and $$x = 6$$, $$y = 3$$ (the answer is "yes").

Statements (1) and (2) combined are sufficient. If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3. Because $$x$$ is an integer, it must be divisible by 3. Therefore, $$xy$$ is divisible by 3 as well.

Hi, i just want you to confirm that in any case we cannot consider 0 as a interger, right?

0 is only even number not even interger number.

Thanks.

Sent from my XT1663 using GMAT Club Forum mobile app

Hi

'0' is an Integer
.

On the basis of their divisibility by 2, ALL integers can be classified into two types:
Even (those which are divisible by 2) and Odd (those which are NOT divisible by 2)

Since 0 is divisible by 2 (actually 0 is divisible by all non-zero numbers), Hence 0 is considered an Even integer.

Hi,
Thank you very much but now i am more confuse.
Since 0 is an integer then answer to this question should be E. Can you please confirm?

Also zero is divisible by all numbers then why we consider it as an even number?

Thanks.[/quote]

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06 Jun 2017, 09:40
goalMBA1990 wrote:
Hi, i just want you to confirm that in any case we cannot consider 0 as a interger, right?

0 is only even number not even interger number.

Thanks.

On the basis of their divisibility by 2, ALL integers can be classified into two types:
Even (those which are divisible by 2) and Odd (those which are NOT divisible by 2)

Since 0 is divisible by 2 (actually 0 is divisible by all non-zero numbers), Hence 0 is considered an Even integer.

Hi,
Thank you very much but now i am more confuse.
Since 0 is an integer then answer to this question should be E. Can you please confirm?

Also zero is divisible by all numbers then why we consider it as an even number?

Thanks.

Sent from my XT1663 using GMAT Club Forum mobile app[/quote]

You should brush up fundamentals.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: http://gmatclub.com/forum/number-proper ... 74996.html
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07 Jun 2017, 10:11
hi Bunuel
from stat 1: (x+y)^2 is divisible by 9,,

on expanding we get x^2 + 2xy + y^2..so is it not that each term shud be divisible by 9???
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07 Jun 2017, 14:38
mohshu wrote:
hi Bunuel
from stat 1: (x+y)^2 is divisible by 9,,

on expanding we get x^2 + 2xy + y^2..so is it not that each term shud be divisible by 9???

Take an easy example: (7 + 2)^2 = 9^2, so it's divisible by 9 but if you expand you'll get 49 + 28 + 4, the sum is still divisible by 9, while individual terms are not.
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29 Jun 2018, 01:56
Hi Bunuel,

I solved with different approach, please tell me if it's correct.

(1) (x+y)^2 is divisible by 9
Which means x^2+2xy+y^2 is divisible by 9, which alone is not sufficient.

(2) (x−y)^2 is divisible by 9
Again on expanding, x^2-2xy+y^2 is divisible by 9, which alone is insufficient.

Statements (1) and (2) combined
If 2 numbers are divisible by 9 then their difference must be divisible by 9. Hence (x+y)^2-(x-y)^2 is divisible by 9, which gives 4xy is divisible by 9 then xy must be divisible by 9 and so by 3. Sufficient.
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18 Nov 2018, 00:42
hrishipatil72 wrote:
Hi Bunuel,

I solved with different approach, please tell me if it's correct.

(1) (x+y)^2 is divisible by 9
Which means x^2+2xy+y^2 is divisible by 9, which alone is not sufficient.

(2) (x−y)^2 is divisible by 9
Again on expanding, x^2-2xy+y^2 is divisible by 9, which alone is insufficient.

Statements (1) and (2) combined
If 2 numbers are divisible by 9 then their difference must be divisible by 9. Hence (x+y)^2-(x-y)^2 is divisible by 9, which gives 4xy is divisible by 9 then xy must be divisible by 9 and so by 3. Sufficient.

That's perfect.
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28 Feb 2019, 00:39
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 gives that $$x + y$$ is divisible by 3. Consider $$x = 1$$, $$y = 2$$ (the answer is "no") and $$x = 3$$, $$y = 3$$ (the answer is "yes").

Statement (2) by itself is insufficient. S2 gives that $$x - y$$ is divisible by 3. Consider $$x = 4$$, $$y = 1$$ (the answer is "no") and $$x = 6$$, $$y = 3$$ (the answer is "yes").

Statements (1) and (2) combined are sufficient. If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3. Because $$x$$ is an integer, it must be divisible by 3. Therefore, $$xy$$ is divisible by 3 as well.

Hi Bunuel, chetan2u

I am not aware about following rule, could you please elaborate it little.

"If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3"
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28 Feb 2019, 00:48
Gmatprep550 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 gives that $$x + y$$ is divisible by 3. Consider $$x = 1$$, $$y = 2$$ (the answer is "no") and $$x = 3$$, $$y = 3$$ (the answer is "yes").

Statement (2) by itself is insufficient. S2 gives that $$x - y$$ is divisible by 3. Consider $$x = 4$$, $$y = 1$$ (the answer is "no") and $$x = 6$$, $$y = 3$$ (the answer is "yes").

Statements (1) and (2) combined are sufficient. If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3. Because $$x$$ is an integer, it must be divisible by 3. Therefore, $$xy$$ is divisible by 3 as well.

Hi Bunuel, chetan2u

I am not aware about following rule, could you please elaborate it little.

"If both $$x + y$$ and $$x - y$$ are divisible by 3 then $$x + y + x - y = 2x$$ is also divisible by 3"

Basically since both x + y and x - y are multiples of 3, then their sum and their difference will also be a multiple of 3.

WHEN THE SUM OR THE DIFFERENCE OF NUMBERS IS A MULTIPLE OF AN INTEGER
1. If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

2. If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

3. If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

For more on this check:

5. Divisibility/Multiples/Factors

For other subjects:
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# M18-03

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