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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: 146
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, 02:18
1
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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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, 13:32
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: 150
Re: If the operation is defined for all integers a and b  [#permalink]

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28 Mar 2017, 09:20
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: 73
If the operation is defined for all integers a and b  [#permalink]

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16 Jul 2017, 09:53
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 1355 times ]

Intern
Joined: 08 Jul 2016
Posts: 38
Location: Singapore
GMAT 1: 570 Q43 V25
GMAT 2: 640 Q42 V36
WE: Underwriter (Insurance)
Re: If the operation @ is defined for all integers a and b  [#permalink]

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07 Feb 2018, 07:51
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: 8
Re: If the operation @ is defined for all integers a and b  [#permalink]

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08 Apr 2018, 17:59
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?
Intern
Joined: 20 Aug 2018
Posts: 27
If the operation @ is defined for all integers a and b  [#permalink]

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Updated on: 24 Nov 2018, 18:26
A video explanation can be found here:

The fastest way to solve the problem is to pick numbers for a, b, and c that are small and easy to work with - for example, 1, 2 and 3

You can see pretty quickly that statements I and II must be true

Testing statement 3 is more time-consuming, but statement 3 turns out to be true.

_________________

Originally posted by CrushGMAT on 24 Nov 2018, 14:28.
Last edited by CrushGMAT on 24 Nov 2018, 18:26, edited 1 time in total.
Intern
Joined: 04 Sep 2018
Posts: 31
GPA: 3.33
Re: If the operation @ is defined for all integers a and b  [#permalink]

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08 Jan 2019, 12:45
Can someone please comment if picking numbers, and checking if the Left hand side equals Right Hand side for the equations is a good way to approach the problem?

I think that approach makes it very easy for me personally to attempt the question.

Re: If the operation @ is defined for all integers a and b &nbs [#permalink] 08 Jan 2019, 12:45

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