$\displaystyle (110101)_{two}/(101)_{two}$
I keep getting 101 which is wrong. How do I do this problem then?
Exactly as you do decimal division: take the first three digits in the first number (110), then 101 fits once in it, so multiply 101 times 1, put below 110 and substract:
110

101

001
Next bring down the next digit, 0: as 0010=10 is less than the divisor 101, we put a zero besides the 1 in the quotient and bring down 1, and etc.
At the end you get 1010, which is correct: 110101 (53) divided by 101(5) equals 1010 (10) and gives a residue of 11 (3).
Tonio
Hello, dwsmith!
Tonio is absolutely correct.$\displaystyle 110101_2 \div 101_2$
The division looks like this:
. . $\displaystyle \begin{array}{cccccccc}
&&&& 1 & 0 & 1 & 0 \\
& &  &  &  &  &  &  \\
1\quad0\quad1 & \bigg) & 1 & 1 & 0 & 1 & 0 & 1 \\
& & 1 & 0 & 1 \\ & &  &  &  \\
&&&& 1 & 1 & 0 \\
&&&& 1 & 0 & 1 \\
&&&&  &  &  \\
&&&&&& 1 & 1
\end{array}$
$\displaystyle \text{We have: }\:110101_2 \div 101_2 \;=\;1010_2,\text{ rem }11_2$
.$\displaystyle \text{That is: }\;53 \div 5 \;=\;10, \text{ rem }3$