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?  
 A: This question is equivalent to finding the 101st base 3 number consisting of only 0's and 1's (since a b and c are distinct) with exactly three 1s, for the reason that $3>2$ and binary covers every single integer once. Thus, this is also equivalent to finding the 101st binary number with three 1s.
Now to find this number, we will need to bash it out. The ways of choosing three 1's out of 10 0's or 1's is the first to exceed 100 with a value of 120, so now we can just count backwards.
There are 8 numbers (120-113) of the form
$$11xxxxxxxx$$
7 of the form (112-106)
$$101xxxxxxx$$
So now we can count backwards.
$$105:1001100000$$
$$104:1001010000$$
$$103:1001001000$$
$$102:1001000100$$
$$101:1001000010$$
Now we need to find a+b+c, which is the sum of the positions, so we have
$$1+6+9=16$$
Which is our answer if a,b,c can be 0. Otherwise, it is 16+3=19.
A: 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.  
A: At the top of your question you say $a,b,c\neq 0$ but then the rest of what you say seems to include the possibility that one of them can be $0$. I'll assume that $0$ is not allowed in what follows. Obviously if you include $3^0$ as a possible power the answer will be different, but the method is the same. 
As you say, $\binom 93=84$ and $\binom{10}3=120$. So there are $84$ numbers of the form $3^a+3^b+3^c$ where $a,b,c$ are distinct elements of $\{1,2,...,9\}$, and another $36$ if we allow $10$. All of these are less than $3\times3^{10}=3^{11}$, so these are the first $120$ numbers of the form $3^a+3^b+3^c$ where $a,b,c$ are distinct positive integers.
Any choice of $a,b,c$ where the largest is $10$ will be bigger than any choice where they are all less than $10$, so the $84$ where all powers are at most $9$ come first, and then you want the $17$th number of the form $3^{10}+3^b+3^c$, where $b,c$ are distinct positive integers less than $10$.
Hint for finishing from here: how many choices for $b,c$ are there if $b,c\in\{1,2,...,6\}$?
