1. ## [SOLVED] Binary divsion

$\displaystyle (110101)_{two}/(101)_{two}$

I keep getting 101 which is wrong. How do I do this problem then?

2. Originally Posted by dwsmith
$\displaystyle (110101)_{two}/(101)_{two}$

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

3. Hello, dwsmith!

$\displaystyle 110101_2 \div 101_2$
Tonio is absolutely correct.

The division looks like this:

. . $\displaystyle \begin{array}{cccccccc} &&&& 1 & 0 & 1 & 0 \\ & & -- & -- & -- & -- & -- & -- \\ 1\quad0\quad1 & \bigg) & 1 & 1 & 0 & 1 & 0 & 1 \\ & & 1 & 0 & 1 \\ & & -- & -- & -- \\ &&&& 1 & 1 & 0 \\ &&&& 1 & 0 & 1 \\ &&&& -- & -- & -- \\ &&&&&& 1 & 1 \end{array}$

$\displaystyle \text{We have: }\:110101_2 \div 101_2 \;=\;1010_2,\text{ rem }11_2$

.$\displaystyle \text{That is: }\;53 \div 5 \;=\;10, \text{ rem }3$