jmc

First, let us convert $629_{10}$ to each of the two bases. To convert to base 7, we realize $7^4>629_{10}>7^3$. So, we can tell that $629_{10}$ in base seven will have four digits. $7^3=343$, which can go into 629 only one time at most, leaving $629-1\cdot 343=286$ for the next three digits. $7^2=49$ goes into 286 five times at most, leaving us with $286-5\cdot 49=41$. Then, $7^1=7$ goes into 41 five times at most, leaving $41-5\cdot 7=6$ for the ones digit. All together, the base seven equivalent of $629_{10}$ is $1556_7$.

To convert to base 8, we realize similarly that $8^4>629_{10}>8^3$. So, we can tell that $629_{10}$ in base eight will have four digits. $8^3=512$, which can go into 629 only one time at most, leaving $629-1\cdot 512=117$ for the next three digits. $8^2=64$ goes into 117 one time at most, leaving us with $117-1\cdot 64=53$. Then, $8^1=8$ goes into 53 six times at most, leaving $53-6\cdot 8=5$ for the ones digit. All together, the base eight equivalent of $629_{10}$ is $1165_8$.

Finally, comparing $1556_7$ and $1165_8$, we find that digits 1, 5, and 6 are present in both numbers. Thus, there are $\boxed{3}$ digits in common.