Problem Solving

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

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??

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.

need help on this, anybody got the solution ??????

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??

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

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

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??