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M18-03

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Joined: 21 Jul 2018
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Re: M18-03  [#permalink]

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New post 28 Feb 2019, 01:57
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 :-)
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Re: M18-03  [#permalink]

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New post 28 Feb 2019, 02:12
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.


Answer: C


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"
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Re: M18-03   [#permalink] 28 Feb 2019, 02:12

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