19. (6 points) The five-digit number $\overline{186 \mathrm{ab}}$, when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 3, and when divided by 11 leaves a remainder of 0, then $\overline{a b}=$
【Analysis】For the five-digit number $186 \mathrm{ab}$, when divided by 3, the remainder is 2, so $10 a+b=3 k+2$. When divided by 5, the remainder is 3, so $b$ is 3 or 8. When divided by 11, the remainder is 0, so $1+6+b-8-a=11 n$. Since $a$ and $b$ are both less than 10, then $n=0, b=a+1$. Therefore, $a=2, b=3$, and the five-digit number is 18623. According to this, $\overline{\mathrm{ab}}$ can be found.

【Solution】Solution: For the five-digit number $186 \mathrm{ab}$, when divided by 3, the remainder is 2, so $10 a+b=3 k+2$. When divided by 5, the remainder is 3, so $b$ is 3 or 8. When divided by 11, the remainder is 0, so $1+6+b-8-a=11 n$. Since $a$ and $b$ are both less than 10, then $n=0, b=a+1$,

so $a=2, b=3$, and the five-digit number is 18623,
so $\overline{\mathrm{ab}}=23$.
Therefore, the answer is: 23.