text: ### Master0fPuppets's blog
  
  By Master0fPuppets, history, 5 weeks ago,
  
  I have been trying to solve this problem and I did with just observation but when I read the solution I didn't understand the proof behind why there is no answer for an even number.
  
  • 0
  
   » 5 weeks ago, # |   0 In this problem, we need to find permutations $A$, $B$ and $C$ such that $A_i + B_i = C_i$ $mod$ $n$.Note that this implies that $\sum A_i + \sum B_i = \sum C_i$ $mod$ $n$. However, for an even value of $n$, the left side of this equation is different from the right side, meaning that no permutation triple ($A$, $B$, $C$) can satisfy the problem's conditions.
  •  » » 5 weeks ago, # ^ |   0 "for an even value of n, the left side of this equation is different from the right side"I am sorry but could you elaborate on this part more.
  •  » » » 5 weeks ago, # ^ |   +5 $\sum A_i = \sum B_i = \sum C_i = \frac{n \cdot (n - 1)}{2}$For odd values of $n$:$\frac{n \cdot (n - 1)}{2} = \left[\frac{n - 1}{2} \cdot n\right] = 0 \; mod \; n$Therefore, in this case, $\sum A_i + \sum B_i = \sum C_i \; mod \; n$For even values of $n$:$\frac{n \cdot (n - 1)}{2} = \left[\frac{n}{2} \cdot (n - 1)\right] = \left[\frac{n}{2} \cdot (-1)\right] = \frac{-n}{2} = n - \frac{n}{2} = \frac{n}{2} \; mod \; n$Therefore, in this case, $\sum A_i + \sum B_i = 2 \cdot \frac{n}{2} = 0 \; mod \; n$, while $\sum C_i = \frac{n}{2} \; mod \; n$
explanation: The text extract is a blog post from a user named Master0fPuppets, discussing a problem involving permutations and modular arithmetic. The post includes a question about why there is no solution for even values of \( n \), and an explanation is provided in the comments.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ### Master0fPuppets's blog
  
  By Master0fPuppets, history, 5 weeks ago,
  
  I have been trying to solve this problem and I did with just observation but when I read the solution I didn't understand the proof behind why there is no answer for an even number.
  
  • 0
  
   » 5 weeks ago, # |   0 In this problem, we need to find permutations $A$, $B$ and $C$ such that $A_i + B_i = C_i$ $mod$ $n$.Note that this implies that $\sum A_i + \sum B_i = \sum C_i$ $mod$ $n$. However, for an even value of $n$, the left side of this equation is different from the right side, meaning that no permutation triple ($A$, $B$, $C$) can satisfy the problem's conditions.
  •  » » 5 weeks ago, # ^ |   0 "for an even value of n, the left side of this equation is different from the right side"I am sorry but could you elaborate on this part more.
  •  » » » 5 weeks ago, # ^ |   +5 $\sum A_i = \sum B_i = \sum C_i = \frac{n \cdot (n - 1)}{2}$For odd values of $n$:$\frac{n \cdot (n - 1)}{2} = \left[\frac{n - 1}{2} \cdot n\right] = 0 \; mod \; n$Therefore, in this case, $\sum A_i + \sum B_i = \sum C_i \; mod \; n$For even values of $n$:$\frac{n \cdot (n - 1)}{2} = \left[\frac{n}{2} \cdot (n - 1)\right] = \left[\frac{n}{2} \cdot (-1)\right] = \frac{-n}{2} = n - \frac{n}{2} = \frac{n}{2} \; mod \; n$Therefore, in this case, $\sum A_i + \sum B_i = 2 \cdot \frac{n}{2} = 0 \; mod \; n$, while $\sum C_i = \frac{n}{2} \; mod \; n$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a blog post from a user named Master0fPuppets, discussing a problem involving permutations and modular arithmetic. The post includes a question about why there is no solution for even values of \( n \), and an explanation is provided in the comments.
  
  Label: question-and-answer