- #1
maxkor
- 84
- 0
Let $n \equiv 3 (\mod{6})$ objects $a_1, a_2, \dots, a_n$, show one can find $\frac{\binom{n}{2}}{3}$ triples $(a_i,a_j,a_k)$ such that every pair $(a_i,a_j) (i \ne j)$ appears in exactly one triple.
Anyone who (like me) has struggled with this problem might like to look at these links:maxkor said:Let $n \equiv 3 (\mod{6})$ objects $a_1, a_2, \dots, a_n$, show one can find $\frac{\binom{n}{2}}{3}$ triples $(a_i,a_j,a_k)$ such that every pair $(a_i,a_j) (i \ne j)$ appears in exactly one triple.
maxkor said:Let $n \equiv 3 (\mod{6})$ objects $a_1, a_2, \dots, a_n$, show one can find $\frac{\binom{n}{2}}{3}$ triples $(a_i,a_j,a_k)$ such that every pair $(a_i,a_j) (i \ne j)$ appears in exactly one triple.
MarkFL said:A week has now gone by...perhaps you would like to share with us your solution to this problem. :D
maxkor said:I do not have a solution
This forum is for the posting of problems and puzzles which our members find challenging, instructional or interesting and who wish to share them with others. As such, the OP should already have the correct solution ready to post in the event that no correct solution is given within at least 1 week's time.
This phrase refers to finding a set of three objects from a group of $n$ objects that satisfy a specific mathematical condition, known as the "mod 6" condition. Mod 6 is a mathematical operation that finds the remainder when a number is divided by 6.
To satisfy the "mod 6" condition, an object must have a remainder of either 0, 2, or 4 when divided by 6. This can be determined by performing the division or by using a calculator.
Finding exactly one triple with the "mod 6" condition has applications in various fields of mathematics, including number theory, combinatorics, and graph theory. It also has practical uses in computer science, such as in error-correcting codes and data encryption.
The number of possible triples that can be found depends on the value of $n$ and the specific objects being considered. In general, the number of possible triples increases as $n$ increases. However, there may be cases where no triples can be found, or where multiple triples can be found.
Yes, there are many other variations of this problem that use different mathematical operations, such as mod 3, mod 4, or mod 7. These variations may have different conditions for what constitutes a "triple" and can have different applications and solutions.