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Math Help - Numbers in different bases

  1. #1
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    Numbers in different bases

    Hello Everyone. I need help starting this problem. I do not need anyone to do it all for me. Im just looking for some help getting pointed in the right direction. Thanks.

    If we write a number in a base other than base 10 (how we usually write numbers),
    it is customary to write the base as a subscript as a clarification. For example,
    345(
    16) signifies the number 345 in base 16. If we convert this to base 10, we find
    that this is equal to 837. In other words 345(
    16) = 837(10).

    Problem:
    Find all three digit sequences XY Z and two digit sequences AB such
    that
    AB(16) = XY Z(10) and AB(10) = XY Z(8).


    Last edited by mr fantastic; March 12th 2010 at 12:50 PM. Reason: Changed post title
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  2. #2
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    Hello, Twinsfan85!

    Find all three-digit numbers XYZ and two-digit numbers AB
    . . such that: .  AB_{16} \:=\:XYZ _{10}\;\text{ and } \;AB_{10} \:=\: XYZ_8
    Note that: . \begin{array}{ccccc}<br />
A,B & \in & \{0,1,2,3,\hdots,9\} & A \neq 0 \\ X,Y,Z & \in & \{0,1,2,\hdots,7\} & X \neq 0\end{array}


    We have: . \begin{array}{ccccccccccc} AB_{16} &=& XYZ_{10} && \Rightarrow && 16A + B &=& 100X + 10Y + Z & {\color{blue}[1]} \\<br />
AB_{10} &=& XYZ_8 && \Rightarrow && 10A + B &=& 64X + 8Y + Z & {\color{blue}[2]} \end{array}


    Subtract {\color{blue}[1] \text{ - } [2]}: . 6A \:=\:36X + 2Y \quad\Rightarrow\quad A \:=\:6X + \frac{Y}{3}

    Since A is a digit, Y must be a multiple of 3: . Y \,=\,3m\;\;{\color{blue}[3]}
    . . Hence: . A \:=\:6X + m \;\;{\color{blue}[4]}

    Substitute into {\color{blue}[2]}: . 10(6X + m) + B \:=\:64X + 8(3m) + Z \quad\Rightarrow\quad B \:=\:4X + 14m + Z\;\;{\color{blue}[5]}

    Since B \leq 9,\;m = 0

    Then in {\color{blue}[3]}\!:\;\;\boxed{Y \,=\, 0}

    And in {\color{blue}[4]}\!:\;\;A \:=\:6X
    . . Since A \leq 9\,\text{ and }\,X \neq 0,\,\text{ then: }\:\boxed{X \,=\, 1}\;\;\boxed{A \,=\, 6}

    Hence, {\color{blue}[5]} becomes: . B \:=\:Z+4

    And we have: . \begin{array}{c||cccccc}\hline <br />
Z & 0&1&2&3&4&5 \\ \hline B & 4&5&6&7&8&9 \\ \hline \end{array}


    The solutions are:

    . . \begin{array}{|ccc|ccc|}\hline<br />
64_{16} &=& 100_{10} & 64_{10} &=& 100_8 \\<br />
65_{16} &=& 101_{10} & 65_{10} &=& 101_8 \\<br />
66_{16} &=& 102_{10} & 66_{10} &=& 102_8 \\<br />
67_{16} &=& 103_{10} & 67_{10} &=& 103_8 \\<br />
68_{16} &=& 104_{10} & 68_{10} &=& 104_8 \\<br />
69_{16} &=& 105_{10} & 69_{10} &=& 105_8 \\ \hline<br />
\end{array}

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  3. #3
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    Thank You very much Soroban. I was a little confused at first when I read over your solution. A quick read over the information on base 10 again solved my problems though Thanks Again

    Edit: I take back my previous statement. Why did you subtract the second equation from the first? Also more importantly, why did you substitute [4] back into equation 2 and not equation 1?
    Last edited by Twinsfan85; March 12th 2010 at 07:20 PM. Reason: Error
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  4. #4
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    Hello again, Twinsfan85!

    Why did you subtract the second equation from the first?
    Thought it was obvious . . . good opportunity to eliminate B's and Z's.


    Alsowhy did you substitute [4] back into equation 2 and not equation 1?
    Um ... Equation 2 had smaller coefficients . . .

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