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Math Help - [SOLVED] Convert from base8 to base7

  1. #1
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    [SOLVED] Convert from base8 to base7

    How would i convert base8 number of 725 to a base7 number?
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  2. #2
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    Quote Originally Posted by Edbaseball17 View Post
    How would i convert base8 number of 725 to a base7 number?
    The easiest way I know is to convert the base 8 number to base 10, then convert the base 10 number to base 7.
    725_8 = 7 \cdot 8^2 + 2 \cdot 8^1 + 5 \cdot 8^0 = 469
    where the 469 is in base 10.

    Now to convert 469 to base 7. The largest power of 7 that divides 469 (without being larger than it) is 3:
    469 = 1 \cdot 7^3 + 126

    The next largest power of 7 is 2:
    469 = 1 \cdot 7^3 + 2 \cdot 7^2 + 28

    The next largest power of 7 is 1:
    469 = 1 \cdot 7^3 + 2 \cdot 7^2 + 4 \cdot 7^1 + 0

    And the last power of 7 is 0:
    469 = 1 \cdot 7^3 + 2 \cdot 7^2 + 4 \cdot 7^1 + 0 \cdot 7^0

    So 725_8 = 469_{10} = 1240_7

    I would not be surprised to find that the Euclidean Algorithm provides a way to do this directly from base 8 to base 7, but I don't know how to set it up.

    -Dan
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  3. #3
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    Hello, Edbaseball17!

    How would i convert 725_8 to a base 7 number?

    Convert to base-ten: . 725_8 \:=\:469_{10}

    Then convert to base-seven: . 469_{10} \:=\:1240_7



    I trust you know the necessary procedures.
    Otherwise, you shouldn't have been assigned this problem.
    .
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  4. #4
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    Hey, thanks alot. You made my day, now i understand it perfectly. I was trying to go directly from 8 to 7 instead of working with the base10 instead.
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  5. #5
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    Hello, Edbaseball17!

    I was trying to go directly from base 8 to base 7.

    This can be done IF you can divide in base 8 . . .


    \begin{array}{cccccc} & & 1 & 0 & 3 \\<br />
& & -- & -- & -- \\ 7 & ) & 7 & 2 & 5 \\ & & 7 \\<br />
& & --& -- \\<br />
& & & 2 & 5\\<br />
& & & 2 & 5 \\<br />
& & & --& -- \\<br />
& & & & 0<br />
\end{array} . . Remainder: 0



    . . \begin{array}{cccccc}& & & 1 & 1 \\<br />
& & -- & -- & --\\<br />
7 & ) & 1 & 0 & 3 \\<br />
& & & 7 \\<br />
& & & -- \\<br />
& & & 1 & 3 \\<br />
& &  & & 7 \\<br />
& & & --& -- \\<br />
& & & & 4 \end{array} . . Remainder: 4



    . . . . \begin{array}{cccccc} & & & 1 \\<br />
& & -- & -- \\<br />
7 & ) & 1 & 1 \\<br />
& & & 7 \\<br />
& & -- & -- \\<br />
& & & 2 \end{array} . . Remainder: 2


    . . . . . . \begin{array}{cccccc}& & 0 \\<br />
& & -- \\<br />
7 & ) & 1  \\<br />
& & 0 \\<br />
& & -- \\<br />
& & 1 \end{array} . . Remainder: 1



    Therefore: . \boxed{\;725_8 \;=\;1240_7\;}

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