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Re: M18-03 [#permalink]
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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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M18-03 [#permalink]
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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M18-03 [#permalink]
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Gmatprep550 wrote:
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"


Essentially, since both x + y and x - y are multiples of 3, their sum and their difference will also be multiples 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 multiples 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 the other is not, then their sum and difference will NOT be multiples 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 multiples 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



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Re: M18-03 [#permalink]
This is how I approached the sum .. I expeanded (x+y)^2 = x*x +y*y + 2*x*y .. Since each term must be divided by 9 to ensure (x+y)^2 is divisble by 9... xy is divible by 9 and hence 3.

Clearly I was wrong.. But want to know which concept of mine was faltered. Can someone help?
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Re: M18-03 [#permalink]
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Seekingredemption wrote:
This is how I approached the sum .. I expeanded (x+y)^2 = x*x +y*y + 2*x*y .. Since each term must be divided by 9 to ensure (x+y)^2 is divisble by 9... xy is divible by 9 and hence 3.

Clearly I was wrong.. But want to know which concept of mine was faltered. Can someone help?



The red part is not correct. (x + y)^2 can be divisible by 9 without x and y being 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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Re: M18-03 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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