Represent every Natural number as a summation/subtraction  of distinct power of 3 I have seen this in a riddle where you have to chose 4 weights to calculate any weight from 1 to 40kgs. 
Some examples,
$$8 = {3}^{2} - {3}^{0}$$
$$12 = {3}^{2} + {3}^{1}$$
$$13 = {3}^{2} + {3}^{1}+ {3}^{0}$$
Later I found its also possible to use only 5 weights to calculate any weight between 1-121. 
$$100 = {3}^{4} + {3}^{3} - {3}^{2} + {3}^{0}$$ 
$$121 = {3}^{4} + {3}^{3} + {3}^{2} + {3}^{1} + {3}^{0}$$      
Note: It allows negative numbers too. how I represent 8 and 100. 
I want to know if any natural number can be represented as a summation of power of 3. I know this is true for 2. But is it really true for 3? What about the other numbers? Say $4, 5, 6, ... $

You can represent any number $n$ as $a_k 3^k + a_{k-1} 3^{k-1} + \dots + a_1 3 + a_0$, where $a_i \in \{-1,0,1\}$. This is called balanced ternary system, and as Wikipedia says, one way to get balanced ternary from normal ternary is to add ..1111 to the number (formally) with carry, and then subtract ..1111 without carry. For a generalization, see here.
