### 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 $\sum C_i = \frac{n}{2} \; mod \; n$