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# If the operation @ is defined for all integers a and b

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Manager
Joined: 14 Mar 2014
Posts: 148
GMAT 1: 710 Q50 V34
Re: If operation # for all integers is defined as a#b=a+b-ab, which of the [#permalink]

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02 Aug 2015, 03:18
1
KUDOS
visram04 wrote:
If operation # for all integers is defined as a#b=a+b-ab, which of the following is must be ture for all integers a,b and c?
i) a#b=b#a
ii) a#0=a
iii) (a#b)#c=a#(b#c)

Answer is all 3. Can someone help me understand i) and iii)?

Given: a#b=a+b-ab

(i) a#b=b#a
a#b = a+b-ab
b#a = b+a-ba --> we can rewrite this as a+b-ab
which is same as a#b. Thus it is true

(ii) a#0=a
a#0 = a+0-(a*0) = a. Thus true

(iii) (a#b)#c=a#(b#c)
(a#b)#c
Consider a#b = x. Thus (a#b)#c ==> x#c => x+c-(x*c)
Now a#b = x => a+b-ab =x. Sub in the above equation
a+b-ab+c-[(a+b-ab)*c) = a+b+c-ab -[ac+bc-abc] = a+b+c-ab-ac-bc+abc--(1)

a#(b#c)
Consider b#c =y. Thus a#(b#c) --> a#y -> a+y-ay
Now b#c=y --> b+c-bc =y. Sub in the above equation
a+b+c-bc-[a*(b+c-bc)] --> a+b+c-bc-[ab+ac-abc] --> a+b+c-bc-ab-ac+abc --(2)
(1) =(2)
Thus (a#b)#c=a#(b#c) is true.

Hence all are true
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I'm happy, if I make math for you slightly clearer
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Intern
Joined: 14 Oct 2014
Posts: 3
Re: If the operation is defined for all integers a and b [#permalink]

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25 Dec 2015, 14:32
I apologize for the rudimentary questions here, but can you break down the strategy for solving this problem? I always get confused with custom character problems. What are we trying to prove here?
Manager
Joined: 03 Jan 2017
Posts: 184
Re: If the operation is defined for all integers a and b [#permalink]

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28 Mar 2017, 10:20
I just picked random positive integers and started testing right away, all 3 worked for me

Though time consuming, but one can arrive pretty quickly at 1,2. 3 took much more time
Manager
Joined: 31 Dec 2016
Posts: 89
If the operation is defined for all integers a and b [#permalink]

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16 Jul 2017, 10:53
While Brunnel provides terrific answers the issues are that he sometimes assumes we have his level of knowledge of math, which we don't. Here is a more simple version of what he did in answer choice (C) which explains the steps.

Essentially, for operator questions one variable becomes another expression. So (A operator B) Operator C means A operator B becomes A and C becomes B. Hopefully, the logic flows through below.
Attachments

gmat answer.png [ 297.08 KiB | Viewed 723 times ]

Intern
Joined: 08 Jul 2016
Posts: 21
Location: United Arab Emirates
GMAT 1: 570 Q43 V25
Re: If the operation @ is defined for all integers a and b [#permalink]

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07 Feb 2018, 08:51
Bunuel wrote:
fguardini1 wrote:
Sorry for bumping up an old thread, I have a doubt: my approach for solving the question was to assume that the operation in this case was the union between two sets, a and b, and consequently the three points were the properties of the union of sets. Is that a correct approach or might it be too risky in the actual exam? (or is it even a wrong assumption, and I got it right out of luck?)

That's not correct. Stem defines some function @ for all integers a and b by a@b=a+b-ab. For example if a=1 and b=2, then a@b=1@2=1+2-1*2.

Hope it's clear.

Hey Bunuel,

Since I always tend to mix up the variables by the end of the problem, please tell if it would be wrong to test the function with random numbers (positive integers)?
Intern
Joined: 22 Nov 2016
Posts: 2
Re: If the operation @ is defined for all integers a and b [#permalink]

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08 Apr 2018, 18:59
Bunuel wrote:
cmugeria wrote:
If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a
II. a@0 = a
III. (a@b)@c = a@(b@c)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III

We have that: $$a@b=a+b-ab$$

I. $$a@b = b@a$$ --> $$a@b=a+b-ab$$ and $$b@a=b+a-ab$$ --> $$a+b-ab=b+a-ab$$, results match;

II. $$a@0 = a$$ --> $$a@0=a+0-a*0=0$$ --> $$0=0$$, results match;

III. $$(a@b)@c = a@(b@c)$$ --> $$(a@b)@c=a@b+c-(a@b)*c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc$$ and $$a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab-ac+abc$$, results match.

Hi,
FOr this particular questions, can we substitute numbers? WIll that apprach always work?
Re: If the operation @ is defined for all integers a and b   [#permalink] 08 Apr 2018, 18:59

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