3. Find the four-digit number $\overline{\mathrm{abcd}}$, if $\overline{\mathrm{abcd}}+\overline{\mathrm{abc}}+\overline{\mathrm{ab}}+\mathrm{a}=1990$.
Solution. From the condition of the problem, we get $1000 a+100 b+10 c+d+100 a+10 b+c+10 a+b+a=1990$, i.e., $1111a+111b+11c+d=1990$. Given that $\mathrm{a}=1$, otherwise the sum would be greater than 1990. Then $111 \mathrm{~b}+11 \mathrm{c}+\mathrm{d}=879$. The digit $\mathrm{b}879$. Therefore, $b \leq 7$. Similarly, $b>6$, because otherwise $111 \mathrm{~b}+11 \mathrm{c}+\mathrm{d}<879$. Finally, $b=7$. According to this, for $b=7$, we get $11 c+d=102$, from which it follows that $c=9$ and $d=3$.