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# How many ordered triples (a, b, c) of non-zero real numbers have the p

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Manager
Status: students
Joined: 08 Mar 2019
Posts: 93
GPA: 3.9
WE: Account Management (Accounting)
How many ordered triples (a, b, c) of non-zero real numbers have the p  [#permalink]

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26 May 2019, 06:03
3
00:00

Difficulty:

95% (hard)

Question Stats:

29% (02:30) correct 71% (01:36) wrong based on 35 sessions

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How many ordered triples (a, b, c) of non-zero real numbers have the property that each number is the product of the other two?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5
ISB School Moderator
Joined: 08 Dec 2013
Posts: 594
Location: India
Concentration: Nonprofit, Sustainability
Schools: ISB '21
GMAT 1: 630 Q47 V30
WE: Operations (Non-Profit and Government)
Re: How many ordered triples (a, b, c) of non-zero real numbers have the p  [#permalink]

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26 May 2019, 07:14
2
1
Rubina11 wrote:
How many ordered triples (a, b, c) of non-zero real numbers have the property that each number is the product of the other two?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Common tricky question,

a b c
1 1 1
1 -1 -1
-1 1 -1
-1 -1 1.

4 distinct triplets.
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Joined: 19 Oct 2018
Posts: 961
Location: India
Re: How many ordered triples (a, b, c) of non-zero real numbers have the p  [#permalink]

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26 May 2019, 07:58
1
a=bc, b=ca and c=ab
Multiply 3 equations, we get
{abc}^2 = abc
abc=1 or abc=0
Because a,b and c are non-zero, abc can't equal to 0
abc=1
1. This is possible if all 3 of them are equal to 1
Arrangements possible= 3!/3!=1
2. when two of them equal to -1 and one of them equal to 1
Arrangements possible= 3!/2!=3

Total ordered triplets possible= 1+3=4
Re: How many ordered triples (a, b, c) of non-zero real numbers have the p   [#permalink] 26 May 2019, 07:58
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