Express the product of $1001_2$ and $121_3$ in base 10.
We can first convert both numbers to base 10. For $1001_2$ we have $1\cdot 2^3+0\cdot 2^2+0\cdot 2^1+1\cdot 2^0$. When we evaluate this expression, we find that $1001_2=8+0+0+1=9$. Now for $121_3$ we have $1\cdot 3^2+2\cdot 3^1+1\cdot 3^0$. When we evaluate this expression, we find that $121_3=9+6+1=16$. So now to find the product of $1001_2$ and $121_3$, we just multiply 9 and 16 to find that $9\cdot 16=144$. So $1001_2$ times $121_3$ in base 10 is 144.

