<s>[INST] A positive integer $n$ is a balanced number if the sum of the digits of $n$ equals the sum of the digits of $2n$. In this problem, we will consider only $3$-digit numbers. Let $d$, $e$, and $f$ be (not necessarily distinct) digits of a $3$-digit balanced number. Let $n=100d+10e+f$ be the $3$-digit number. We have that $2n = 2(100d+10e+f) = 200d+20e+2f$; so, $2f\leq 18$ or $f\in\{0,1,2,3,4\}$. Now, we have $5$ cases:
Case 1: If $f=0$, then the sum of the digits of $n$ is $d+e$, while the sum of the digits of $2n$ is $2d+2e$, so $d+e=2d+2e$ and we have $d=e=0$, but then we have $n=000$, which is not a $3$-digit number, so this case is impossible for $n$. Case 2: If $f=1$, then the sum of the digits of $n$ is $d+e+1$, while the sum of the digits of $2n$ is $2d+2e+2$, so $d+e+1=2d+2e+2$ and we have $d+e=1$. Also, we have $2(100d+10e+1)=200d+20e+2$, which implies $2d+2e+2\leq 18$ and so $d+e\leq 8$. Since $d,e\in\{0,1,2,3,4,5,6,7,8,9\}$, we have $(d,e)\in\{(0,1),(1,0)\}$. Case 3: If $f=2$, then the sum of the digits of $n$ is $d+e+2$, while the sum of the digits of $2n$ is $2d+2e+4$, so $d+e+2=2d+2e+4$ and we have $d+e=2$. Since $d,e\in\{0,1,2,3,4,5,6,7,8,9\}$, we have $(d,e)\in\{(0,2),(2,0),(1,1)\}$. Case 4: If $f=3$, then the sum of the digits of $n$ is $d+e+3$, while the sum of the digits of $2n$ is $2d+2e+6$, so $d+e+3=2d+2e+6$ and we have $d+e=3$. Since $d,e\in\{0,1,2,3,4,5,6,7,8,9\}$, we have $(d,e)\in\{(0,3),(3,0),(1,2),(2,1)\}$. Case 5: If $f=4$, then the sum of the digits of $n$ is $d+e+4$, while the sum of the digits of $2n$ is $2d+2e+8$, so $d+e+4=2d+2e+8$ and we have $d+e=4$. Since $d,e\in\{0,1,2,3,4,5,6,7,8,9\}$, we have $(d,e)\in\{(0,4),(4,0),(1,3),(2,2),(3,1)\}$.
Thus, the possible values for $n$ are $101,110,202,212,220,303,312,330,404,412,420,434$.
The question asks the following: What is the sum of all the 3-digit balanced numbers?[/INST]