5. Arrange the numbers in the set $\left\{2^{x}+2^{y}+2^{z} \mid x 、 y 、 z \in \mathbf{N}, x<y<z\right\}$ in ascending order. The 100th number is $\qquad$ (answer with a number).


5.577.

Notice that the number of combinations $(x, y, z)$ such that $0 \leqslant x<y<z \leqslant n$ is $\mathrm{C}_{n+1}^{3}$.

Since $\mathrm{C}_{9}^{3}=84<100<120=\mathrm{C}_{10}^{3}$, the 100th number must satisfy $z=9$.

Also notice that the number of combinations $(x, y)$ such that $0 \leqslant x<y \leqslant m$ is $\mathrm{C}_{m+1}^{2}$.

Since $\mathrm{C}_{9}^{3}+\mathrm{C}_{6}^{2}=99$, the 100th number must satisfy $y=6, x=0$, which means the 100th number is
$$
2^{0}+2^{6}+2^{9}=577 .
$$