Problem 138

Since the function $f(x)=\sin ωx+a\cos ωx=\sqrt{1+a^2}\left(\frac{1}{\sqrt{1+a^2}}\sin ωx+\frac{a}{\sqrt{1+a^2}}\cos ωx\right)=\sqrt{1+a^2}\sin(ωx+φ)$ ($\tan φ=a$) is symmetric about the point $M(\frac{π}{3},0)$,
we have $\frac{π}{3}ω+φ=kπ$ ($k∈\mathbb{Z}$), $(①)$

Also, the function has a minimum value at $x=\frac{π}{6}$, so $\frac{π}{6}ω+φ=\frac{3π}{2}+2kπ$ ($k∈\mathbb{Z}$), $(②)$

Solving the system of equations $(①)$ and $(②)$, we obtain $ω=-9-6k$ and $φ=kπ+3π$ ($k∈\mathbb{Z}$).

Thus, $a=\tan(kπ+3π)=0$,

And $a+ω=-9-6k$ ($k∈\mathbb{Z}$).

Given that $a+ω∈[0,10]$, we have $k=-2$ and $a+ω=3$.

Therefore, the answer is $\boxed{3}$.

By analyzing the function and utilizing its symmetry, we can derive $ω=-9-6k$ and $φ=kπ+3π$ ($k∈\mathbb{Z}$). From this, we can calculate $a$ and subsequently $a+ω=-9-6k$ ($k∈\mathbb{Z}$), which enables us to find the answer. This problem primarily evaluates the understanding of the symmetry point, symmetry axis, and period of a sine and cosine function, making it a moderately difficult question.