fprosk wrote: Let $n$ be a positive integer. We call $C_n$ the number of positive integers $x$ less than $10^n$ such that the sum of the digits of $2x$ is less than the sum of the digits of $x$. Show that $C_n\geq\frac{4}{9}(10^{n}-1)$. Let $n_k(x)$ where $k\in\{1,2,3,...,9\}$ be the number of digits $k$ in the decimal representation of the positive integer $x$ Let $S(x)=\sum_{k=1}^9kn_k(x)$ be the sum of digits of $x$ Let $C_n$ the number of integers in $(1,10^n)$ such that $S(2x)<S(x)$ Let $D_n$ the number of integers in $(1,10^n)$ such that $S(2x)>S(x)$ Let $E_n$ the number of integers in $(1,10^n)$ such that $S(2x)=S(x)$ $C_n+D_n+E_n=10^n-1$ $S(2x)=\sum_{k=1}^92kn_k(x)-9\sum_{k=5}^9n_k(x)$ $S(x)-S(2x)=\sum_{k=5}^8(9-k)n_k(x)-\sum_{k=1}^4kn_k(x)$ If $0<x<10^n$, the above formula shows that $S(x)-S(2x)=S(2(10^n-x))-S(10^n-x)$ and so $C_n=D_n$ $S(2x)=S(x)$ implies $S(x)=9\sum_{k=5}^9n_k(x)$ and so $9|x$ and so $E_n\le\frac{10^n-1}9$ And so $C_n=D_n=\frac{10^n-1-E_n}2\ge \frac{(10^n-1)-\frac{10^n-1}9}2$ $=\frac 49(10^n-1)$ Q.E.D.