AMC 12 Mock Exam 1 Problem 19

Write $n$ with three digits $a,b,c$, allowing leading zeros (so $n=0$ must be excluded at the end). When $n$ is doubled digit by digit, a digit $d$ together with an incoming carry of $0$ or $1$ produces a carry out exactly when $2d+(\text{carry in})\ge10$, i.e., exactly when $d\ge5$ (because $2d+1\ge10$ iff $d\ge5$ for integers). Each carry replaces a $10$ in one place by a $1$ in the next, lowering the digit sum by $9$. Therefore $$S(2n)=2S(n)-9k,$$ where $k$ is the number of digits of $n$ that are at least $5$. So $S(2n)=S(n)$ iff $S(n)=9k$.

$k=0$: $S(n)=0$ forces $n=0$, excluded.

$k=1$: one digit $b\in\{5,\dots,9\}$ in any of $3$ positions, the other two digits in $\{0,\dots,4\}$ with sum $9-b\in\{4,3,2,1,0\}$; there are $5,4,3,2,1$ such pairs respectively, i.e., $15$ in all. Count: $3\cdot15=45$.

$k=2$: one digit $s\in\{0,\dots,4\}$ in any of $3$ positions, and two digits in $\{5,\dots,9\}$ with sum $18-s\in\{18,17,16,15,14\}$; there are $1,2,3,4,5$ ordered pairs respectively, i.e., $15$ in all. Count: $3\cdot15=45$.

$k=3$: $S(n)=27$ forces $n=999$: $1$ number.

The total is $45+45+1=\boxed{91}$.