$\left[\begin{array}{l}\text { Equations in integers } \\ {[\text { Case enumeration }}\end{array}\right]$

Let $S(x)$ be the sum of the digits of a natural number $x$. Solve the equation $x+S(x)=2001$.
If $x>1999$, then $x+S(x) \geq 2002$, if $x<1000$, then $x+S(x)<1000+3 \cdot 9=1027$. Therefore, $x-$ is a four-digit number with the first digit 1.

Let $a$ be the hundreds digit, $b$ be the tens digit, and $c$ be the units digit, then $(1000+100a+10b+c)+1=2001$, or $101a+11b+2c=1000$. If $a<9$ the last sum is less than $101 \cdot 8+11 \cdot 9+2 \cdot 9<1000$. Thus, $a=9$. Similarly, by trial and error, we get that $\mathrm{x}=1977$.