5. If the three-digit number $\overline{a b c}$ satisfies $1 \leqslant a \leqslant b \leqslant c \leqslant 9$, then $\overline{a b c}$ is called an "uphill number". Then, the number of uphill numbers is $\qquad$
5. 165 .

For a fixed $b$, the number $c$ can take $10-b$ values, namely $b, b+1, \cdots, 9$; and for a fixed number $a$, the number $b$ can take $a, a+1, \cdots, 9$. Therefore, the number of three-digit increasing numbers is
$$
\begin{array}{l}
\sum_{a=1}^{9} \sum_{b=a}^{9}(10-b)=\sum_{a=1}^{9} \sum_{k=1}^{10-a} k \\
=\sum_{a=1}^{9} \frac{(10-a)(11-a)}{2} \\
=\frac{1}{2} \sum_{a=1}^{9}\left(a^{2}-21 a+110\right) \\
=\frac{1}{2}\left(\frac{9 \times 10 \times 19}{6}-21 \times \frac{9 \times 10}{2}+990\right)=165 .
\end{array}
$$