<s>[INST] Finding $a + b + c$ for the $101$st number of the form $3^a + 3^b + 3^c$ when these numbers are listed in increasing order $a,b,c\neq0$ & $a,b,c$ are distinct & $a,b,c\in \mathbb{N}$
numbers in form of
$3^a+3^b+3^c$
If we order them in increasing order, what is the sum of 
$a+b+c=?$
for the $101$st such number?
There are some things I need to clarify about the question a number is represented by $3^a+3^b+3^c$, not $3^a,3^b,3^c$. These do not represent an individual number. 
That being said we could accept $a>b>c$ and then acknowledge that $a,b,c\in \left\{ 0,1,2,3,\cdots,n \right\}$. There are $\dbinom{n+1}{3}$ choices possible for $n$. Therefore, $n$ must give a number greater than $101$. I have chosen $\dbinom{10}{3}=120$, so $n=9$. But I also used $\dbinom{9}{3}=84$ to know which number is $3^9+3^1+3^0$ (it is the $84$th).  I couldn't spot how I would proceed in getting the $101$st number? What do you propose?  
 [/INST] Without loss of generality we can assume that $a > b > c$.  
Let's consider the lexicographic ordering;
we will write $(a,b,c) > (a',b',c')$, 
if one of the following occures:  


*

*$a > a'$ ; or

*$a=a' $ and $b > b'$; or

*$a=a' $ and $b=b' $ and $c > c'$. 


Remark: 
Let's consider two ordered triples $(a,b,c) , \  (a',b',c')$;
i.e. $a > b > c$ and $a' > b' > c'$. 
Then: 
$$ 
3^a+3^b+3^c > 3^{a'}+3^{b'}+3^{c'} 
\Longleftrightarrow 
(a,b,c) > (a',b',c').
$$

As you have been said;
$3^9+3^1+3^0$ is greater than $3^a+3^b+3^c $ 
for every $\{ a,b,c \} \subset \{ 0, 1, 2, ..., 8 \}$.
Also notice that there are $84 = {9 \choose 3}$ choices for $\{ a,b,c \}$ ; 
so $3^9+3^1+3^0$ is in the $85^{\text{th}}$-place.  

$3^9+3^6+3^0$ is greater than $3^9+3^b+3^c $ 
for every $\{ b,c \} \subset \{ 0, 1, 2, ..., 5 \}$.
Also notice that there are $15 = {6 \choose 2}$ choices for $\{b,c \}$ ; 
so $3^9+3^6+3^0$ is in the $1+15+84=100^{\text{th}}$-place.  

Finally one can easilly check that $3^9+3^6+3^1$
is in the $101^{\text{th}}$-place.




How to find the number, at the $m^{\text{th}}$-place?
An effective algorithm for the above problem is as follows:  


*

*Find greatest integer $a \in \mathbb{N}_0$ such that 
${a \choose 3} \leq m-1$.  

*Find greatest integer $b \in \mathbb{N}_0$ such that 
${b \choose 2} \leq m-{a \choose 3}-1$.  

*Find greatest integer $c \in \mathbb{N}_0$ such that 
${c \choose 1} \leq m-{a \choose 3}-{b \choose 2}-1$;
i.e $c=m-{a \choose 3}-{b \choose 2}-1$.  
One can check that $3^a+3^b+3^c$
is in the $1+{c \choose 1}+{b \choose 2}+{a \choose 3}=m^{\text{th}}$-place.  
</s>