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Danke Bunuel
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Press +1 Kudos if my post helped you a little and help me to ulcock the tests Wish you all success

I'd appreciate learning about the grammatical errors in my posts

Please let me know if I'm wrong somewhere and help me to learn Math Expert V
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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"

That would be true for all numbers...
So we have x-y is divisible by 3, thus x-y=3a, where a is an integer. Also x+y is divisible, so x+y=3b..
(x-y)+(x+y)=3a+3b=3(a+b).... 3(a+b) will also be divisible by 3..

If it were any other number other than 3, say 5, 7 etc, it will still be true.
So, "If both $$x + y$$ and $$x - y$$ are divisible by 7 then $$x + y + x - y = 2x$$ is also divisible by 7"
_________________ Re: M18-03   [#permalink] 28 Feb 2019, 02:12

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