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Math Help - [SOLVED] Binary divsion

  1. #1
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    [SOLVED] Binary divsion

    (110101)_{two}/(101)_{two}

    I keep getting 101 which is wrong. How do I do this problem then?
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    (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
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  3. #3
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    Hello, dwsmith!

    110101_2 \div 101_2
    Tonio is absolutely correct.

    The division looks like this:


    . . \begin{array}{cccccccc}<br />
&&&& 1 & 0 & 1 & 0 \\<br />
& & -- & -- & -- & -- & -- & -- \\<br />
1\quad0\quad1 & \bigg) & 1 & 1 & 0 & 1 & 0 & 1 \\<br />
& & 1 & 0 & 1 \\ & & -- & -- & -- \\<br /> <br />
&&&& 1 & 1 & 0 \\<br />
&&&& 1 & 0 & 1 \\<br />
&&&& -- & -- & -- \\<br />
&&&&&& 1 & 1<br />
\end{array}




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

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

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