We can get $ (c,b,a)$ from $ (a,b,c)$: $ (a,b,c)\rightarrow(c,a-\frac{1}{10},b+\frac{1}{10})$, $ (b+\frac{1}{10},a-\frac{1}{10},c)\rightarrow(c,b,a)$ We can see that we are able to change places of any two numbers. So we change 2 with 31, 3 with 30, 4 with 29, ... , 16 with 17 Done

I actually got (a,b,c) from (a,b,c)

Bachukas wrote: We can get $ (c,b,a)$ from $ (a,b,c)$: $ (a,b,c)\rightarrow(c,a - \frac {1}{10},b + \frac {1}{10})$, $ (b + \frac {1}{10},a - \frac {1}{10},c)\rightarrow(c,b,a)$ We can see that we are able to change places of any two numbers. So we change 2 with 31, 3 with 30, 4 with 29, ... , 16 with 17 Done Wrong! $ (a,b,c)\rightarrow(c,a - \frac {1}{10},b + \frac {1}{10})$, $ (b + \frac {1}{10},a - \frac {1}{10},c)\rightarrow(c,b,a)$ They follow that $ (a,b,c)\rightarrow (a,b,c)$ (arrange positions in triples correctly). And nothing in your post is helpful.