3. In a certain basketball tournament, Xiao Ming played 10 games. In the 6th, 7th, 8th, and 9th games, he scored 23 points, 14 points, 11 points, and 20 points, respectively. His average score in the first 9 games was higher than his average score in the first 5 games. If the average score of the 10 games he played exceeds 18 points, then the minimum score he could have in the 10th game is $\qquad$ .
3.29.

Let the average score of the first 5 games be $x$, then the average score of the first 9 games is
$$
\frac{5 x+23+14+11+20}{9}=\frac{5 x+68}{9} .
$$

According to the problem, $\frac{5 x+68}{9}>x$.
Solving this, we get $x<17$.
Therefore, the total score of the first 5 games is at most $5 \times 17-1=84$ points.
Let the score of the 10th game be $y$, then the total score of the 10 games he played is at least $18 \times 10+1=181$ points.
Thus, we have $y+84+68 \geqslant 181$, which means $y \geqslant 29$.
Therefore, the minimum score of the 10th game is 29.