### Instruction: Find the product of $1011_2 \cdot 101_2$.  Express your answer in base 2.### Response: I know that $1011_2$ means $1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0$, which is equivalent to $11_{10}$ in base 10.
Similarly, $101_2$ means $1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0$, which is equivalent to $5_{10}$ in base 10.
So, the problem is asking me to find the product of $11_{10}$ and $5_{10}$, and then convert the answer back to base 2.
I can use the distributive property to write $11_{10} \cdot 5_{10} = (10_{10} + 1_{10}) \cdot 5_{10} = 10_{10} \cdot 5_{10} + 1_{10} \cdot 5_{10}$.
I know that $10_{10} \cdot 5_{10} = 50_{10}$ and $1_{10} \cdot 5_{10} = 5_{10}$, so the sum is $50_{10} + 5_{10} = 55_{10}$.
Now, I need to convert $55_{10}$ to base 2.
I can use repeated division by 2 and write down the remainders to find the binary digits.
55 divided by 2 is 27 with a remainder of 1.
27 divided by 2 is 13 with a remainder of 1.
13 divided by 2 is 6 with a remainder of 1.
6 divided by 2 is 3 with a remainder of 0.
3 divided by 2 is 1 with a remainder of 1.
1 divided by 2 is 0 with a remainder of 1.
The remainders, read from bottom to top, form the binary digits of 55, so $55_{10} = 110111_2$.
Therefore, the product of $1011_2 \cdot 101_2$ is $110111_2$.