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18 Apr 2020, 04:55
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E
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Difficulty:
95%
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Question Stats:
25% (01:51) correct
75%
(01:24)
wrong
based on 89
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An ordered triplet of integers (a, b, c) is called friendly, if each number is equal to the product of the other 2. How many (distinct) friendly triplets are there?
Note: a,b and c may or may not be distinct.
A. 0
B. 1
C. 2
D. 4
E. 5
Note: a,b and c may or may not be distinct.
A. 0
B. 1
C. 2
D. 4
E. 5
18 Apr 2020, 05:30
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nick1816
Ordered triplet means that the order of elements is important..
Now a=bc, b=ac, c=ab means following cases
1) abc=0, that is {0, 0, 0}....1 way
2) a=b=c=1, that is {1, 1, 1}...1 way
3) another way using 1 is {1, -1, -1}.....Since it is ordered triplet, each of three can take value of 1...Hence 3 ways.
Total 1+1+3=5 ways
E
GraemeGmatPanda
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19 Apr 2020, 02:31
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We could solve this by counting the obvious candidates but in order to prove there are no more we need to use some equations (useful if there were higher numbers among the solution choices for example!)
We have:
(1) a = b x c
(2) b = c x a
(3) c = a x b
Combining (1) and (2) we get a = (a x c) x c or \(a = a c^2\) which is equivalent to \(c^2 = 1\) OR a = 0.
Similarly:
Using (2) and (3) we get \(a^2 = \)1 OR b = 0
Using (3) and (1) we get \(b^2 = 1\) OR c = 0
We can set aside the case when any of a, b or c = 0 straight away. In this case using all equations we can prove that the other 2 variables must be 0 too. We have (0,0,0) as one solution.
In all other cases we have that \(a^2 = b^2 = c^2 = 1\) which means that (a,b,c) are either -1 or 1.
Here (1,1,1) becomes a second obvious answer, by plugging in the numbers.
Finally, some simple calculations show that if we have a b or c =-1 then we have to have a combination of -1, -1 and 1 (3 times in 3 different orders)
In total: 5 solutions - answer E
(0,0,0)
(1,1,1)
(1,-1,-1) ; (-1,1,-1) and (-1,-1,1)
We have:
(1) a = b x c
(2) b = c x a
(3) c = a x b
Combining (1) and (2) we get a = (a x c) x c or \(a = a c^2\) which is equivalent to \(c^2 = 1\) OR a = 0.
Similarly:
Using (2) and (3) we get \(a^2 = \)1 OR b = 0
Using (3) and (1) we get \(b^2 = 1\) OR c = 0
We can set aside the case when any of a, b or c = 0 straight away. In this case using all equations we can prove that the other 2 variables must be 0 too. We have (0,0,0) as one solution.
In all other cases we have that \(a^2 = b^2 = c^2 = 1\) which means that (a,b,c) are either -1 or 1.
Here (1,1,1) becomes a second obvious answer, by plugging in the numbers.
Finally, some simple calculations show that if we have a b or c =-1 then we have to have a combination of -1, -1 and 1 (3 times in 3 different orders)
In total: 5 solutions - answer E
(0,0,0)
(1,1,1)
(1,-1,-1) ; (-1,1,-1) and (-1,-1,1)
