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Math Help - Division Question

  1. #1
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    Division Question

    How many different solutions are there to this division? different letters stand for different digits and no number begins with zero.

    JMO / UK = OK ( what are the numbers)
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  2. #2
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    Quote Originally Posted by nazz View Post
    How many different solutions are there to this division? different letters stand for different digits and no number begins with zero.

    JMO / UK = OK ( what are the numbers)
    One solution.
    JMO = OK*UK
    K cannot be one, because K^2 = 1 = O,. repetition is not allowed.
    Let K = 2. Then O = 4
    So 42* U2 = JMO.
    Product is three digits. U cannot be 3,neither 2. So try one
    So 42*12 = 504.
    Try other combination, if possible.
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  3. #3
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    Hello nazz
    Quote Originally Posted by nazz View Post
    How many different solutions are there to this division? different letters stand for different digits and no number begins with zero.

    JMO / UK = OK ( what are the numbers)
    I think the complete set of solutions is:
    \begin{array}{c c c c c}<br />
J&M&O&U&K\\<br />
5&0&4&1&2\\<br />
8&9&6&1&4\\<br />
7&4&1&3&9\\<br />
9&3&1&4&9\\<br />
\end{array}
    Grandad
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello nazzI think the complete set of solutions is:
    \begin{array}{c c c c c}<br />
J&M&O&U&K\\<br />
5&0&4&1&2\\<br />
8&9&6&1&4\\<br />
7&4&1&3&9\\<br />
9&3&1&4&9\\<br />
\end{array}
    Grandad
    In the last set there is a repetition which is not allowed.
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  5. #5
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    Quote Originally Posted by sa-ri-ga-ma View Post
    In the last set there is a repetition which is not allowed.
    This is true! Strike that line from the solutions!

    Grandad
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  6. #6
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    thanks for all the replies, is there a way solving it by equations, instead of solving it by trials
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