Solutions for $(x)_n+(y)_n+(z)_n=(30)_n$ for odd $x,y,z$ Suppose we choose $x,y,z$  from $\left\{1,3,5,7,9,11,13,15\right\}$ are there any solutions for $$x+y+z=30$$ 
It is obvious that adding three odd numbers is never an even number. So this equation does not have $\color{red} {\text{a  solution}}$ .
now change the point of view : 
 If we change base of numbers for example taking $7$ vs $10$ natural base . one solution is $$\color{red} {(11)_7+(11)_7+(5)_7=(30)_7}$$ 
$$(7+1)+(7+1)+5=21=(30)_7 \color{blue}{\checkmark}$$
now my question : Are there any more solution ,or a method to find other solutions like this...
I am thankful for your hint ,guide or any solution  In advanced.
 A: I think that the new title does not respect the original spirit of the question. I think you are not thinking of odd numbers, but to numbers that seem odd if though as expressed in base ten, but that are expressed in another base. Looking at your example in base $7$, you use $(11)_7$ which is in fact an even number. I think it is not hard to find solutions like yours, you just have to take care that you cannot use all of the elements of the set you have given: for instance "$9$" is not a valid number in base $7$. A few solutions I found in zero time (I chose an odd base, I started with $(11)_n$ which is even, I added a bigger even number, and then found the last needed odd number; but there are other strategies):
$$(13)_9 + (11)_9 + (5)_9 = (30)_9$$
$$(15)_9 + (11)_9 + (3)_9 = (30)_9$$
$$(13)_7 + (11)_7 + (3)_7 = (30)_7$$
$$(13)_5 + (11)_5 + (1)_5 = (30)_5$$
$$(13)_{11} + (11)_{11} + (7)_{11} = (30)_{11}$$
If you instead want to really use odd numbers, it is even easier to find a lot of solutions. Think of the easiest (in any odd base $\gt 3$): $(10)_n + (10)_n + (10)_n =(30)_n$. Then you can move around even numbers at will, e.g. $(10+2)_n + (10-2)_n + (10)_n =(30)_n$.
