An ordered triple of real numbers
(a,b,c) is called friendly, if each
number is equal to the product of
the other 2. How many friendly
triples are there?
Details and assumptions
The numbers need not be equal to
each other.
The numbers need not to be
pairwise distinct (which
means that no two of them
are the same). I got 31 but it isn't
right ...
brilliant na??
stuck on that one also :((
To find the number of friendly triples, we can start by considering the possible values of each number in the triple. Let's assume the ordered triple is (a, b, c).
Since each number is equal to the product of the other two, we can set up the following equations:
a = b * c
b = a * c
c = a * b
Solving these equations will give us the possible values for (a, b, c).
Let's start with the first equation:
a = b * c
If we assume b = 0, then a would always be 0. However, this violates the condition that the numbers need not be pairwise distinct. So, we can conclude that b cannot be equal to 0. Similarly, c cannot be equal to 0.
If we assume b = 1, then a = c = 1. This gives us one friendly triple: (1, 1, 1).
Now, let's consider the case when b = c = 1. This gives us a = 1, which means another friendly triple: (1, 1, 1).
Next, let's consider the case when b = 1 and c ≠ 1. Substituting these values into the second equation (b = a * c), we get:
1 = a * c
This means a = 1/c. Since a and c are real numbers, the values of a and c depend on each other. For each value of c ≠ 1, we will get a unique value for a. So, in this case, there are infinitely many friendly triples.
Finally, let's consider the case when b ≠ 1 and c ≠ 1. Substituting these values into the third equation (c = a * b), we get:
c = a * b
This means a = c/b. Again, since a and b are real numbers, the values of a and b depend on each other. For each value of b and c ≠ 1, we will get a unique value for a. So, in this case, there are also infinitely many friendly triples.
In summary, we have found three types of friendly triples:
1. (1, 1, 1)
2. (1, c, 1/c) for any c ≠ 1
3. (a, b, c) where a = c/b and b ≠ 1, c ≠ 1 (infinitely many possibilities)
Therefore, the number of friendly triples is infinite, and we cannot give a finite value for it.