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triplets

Expert replies
by shashank.ism » Wed Feb 10, 2010 9:41 am
If |(a − b + c)(b − c + a)(c − a + b)| = 15 ,where |x| has its usual meaning , then the possible number of ordered integral triplets (a,b,c) are

A) 36
B) 33
C) 66
D) 72
E) none of these
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Source: — Problem Solving |

by harsh.champ » Wed Feb 10, 2010 10:48 pm
shashank.ism wrote:If |(a − b + c)(b − c + a)(c − a + b)| = 15 ,where |x| has its usual meaning , then the possible number of ordered integral triplets (a,b,c) are

A) 36
B) 33
C) 66
D) 72
E) none of these
Hey shashank,
Can you post the OA and the solution approach ??
I guess this is way too difficult.What-say tutors?

I guess the question uses some concept pertaining to cyclic numbers??
Can someone point out the concept or does it requires some other concept??
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by sparky_paris » Thu Feb 11, 2010 5:21 pm
I think its none of these option E,

I got 13 possible ordered triplets, since a-b+c, b-c+a, c-a+b should be either plus or minus (1, 1, 15) or plus or minus (1, 3, 5)

Taking into account repetitions in (1,1,15) I get 13 possible triplets.
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by Mom4MBA » Sun Feb 14, 2010 2:12 pm
need help on this, anybody got the solution ??????
Stay focused
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by kaustubh_b » Mon Feb 15, 2010 1:31 am
As per my understanding, we can have 2 sets
Set-1: (1, -1 and 15, -15)
Set-2: (1, -1, 3, -3 and 5, -5)

This gives us the possible combinations as:
(4 x 4 x 2) + ( 6 x 4 x 2) = 80

Answer: E (none of the above)

OA plz??
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by ajith » Mon Feb 15, 2010 6:19 am
shashank.ism wrote:If |(a − b + c)(b − c + a)(c − a + b)| = 15 ,where |x| has its usual meaning , then the possible number of ordered integral triplets (a,b,c) are

A) 36
B) 33
C) 66
D) 72
E) none of these
(a − b + c)(b − c + a)(c − a + b) =+/-15
1*1*15 (3 solutions)
1*-1*15 (6solutions)
-1*1*15 (6solutions)
1*1*-15(6solutions)
-1*-1*15(3solutions)
-1*1*-15(6solutions)
1*-1*-15(6solutions)
-1*-1*-15(3solutions)

39 solutions

1*3*5 (6)
1*-3*5 (6)
.
.
.
.
-1*-3*-5 (6)

Total 6*8 = 48 solutions

Total no of solutions = 48+39 = 87 solutions
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by m&m » Mon Feb 15, 2010 8:13 am
Factors of |15| are 1, 3, 5, 15, -1, -3, -5, -15

so we can either have combination of form 15*1*1 or combination of form 3*5*1


1) if 15 is not inclded then we have

x y z
6*4*2
=48

2) if 15 is included then we have

15*1*1
15*-1*1
15*1*-1
15*-1*-1
-15*1*1
-15*-1*1
-15*1*-1
-15*-1*-1
------------------
8 solutions * 3 = 24

48 + 24 = 72
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by harsh.champ » Thu Feb 18, 2010 5:06 am
Hey ppl,
I am confused between m&m soln. and ajith's soln.

1*1*15 (3 solutions)
1*-1*15 (6solutions)
-1*1*15 (6solutions)
1*1*-15(6solutions)
-1*-1*15(3solutions)
-1*1*-15(6solutions)
1*-1*-15(6solutions)
-1*-1*-15(3solutions)

39 solutions

1*3*5 (6)
1*-3*5 (6)
.
.
.
.
-1*-3*-5 (6)

Total 6*8 = 48 solutions
My guess some soln. are getting repeated in the 39 and the 48 soln.(probably 15 since that will yield the answer as 72)


I guess using these 2 cases is sufficient to solve the question:-
1) if 15 is not included
2) if 15 is included
Hey shashank,
Can you post the formal answer??
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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"Keep Walking" - Johnny Walker :P
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