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1/ If the operation # is defined for all integers a and b by

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SVP
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1/ If the operation # is defined for all integers a and b by [#permalink]

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New post 15 Oct 2006, 08:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1/ If the operation # is defined for all integers a and b by
a#b = a + b - ab which of the followin statements must be true for all integers a,b, and c?

A) a#b= b#a
B) a#0= a
C) (a#b)#C = a#(b#c)

it is abvious that A,B are true lets check three

a#b = a+b-ab , b#c = b+c - bc

(a+b-ab)#c = a+b-ab+c - ac-bc+abc

(b+c - bc)#a = a+b+c -bc - ab-ac+abc

thus my answer is A,B ONLY
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New post 15 Oct 2006, 09:12
First, thanks for seperating the posts...
sorry, next time will be careful!!

In fact for this question, the answer is A,B a,d C!!!
The question was on GMATprep so I only have the answer and not the explanations...

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New post 15 Oct 2006, 09:35
so wut is the answer
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Re: Zamzim separate posts [#permalink]

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New post 29 Mar 2008, 18:25
a#b = a + b - ab

We already know that (A) and (B) are true.

C) (a#b)#C = a#(b#c)

LS
= (a+b-ab) + c - (a+b-ab)*c
= a + b - ab + c - ac - bc + abc
= a + b + c - ab - ac - bc + abc (reordering)

RS
= a + (b+c-bc) - a(b+c-bc)
= a + b + c - bc -ab -ac + abc
= a + b + c - ab - ac - bc + abc

Therefore, LS=RS and all of A, B, and C are true.

I think it's the confusion caused when multiplying all the terms. We just gotta be extra careful, which is tough when we're rushed to do these questions in 2 minutes!
Re: Zamzim separate posts   [#permalink] 29 Mar 2008, 18:25
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1/ If the operation # is defined for all integers a and b by

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