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Math Help - puzzle

  1. #1
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    puzzle

    Logic+logic=prolog what is the value of alphabets involved in it.


    I was trying the rebues puzzle but couldnt get anything
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  2. #2
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    Re: puzzle

    Hello, prasum!

    If you've had no experience with "alphametics",
    . . no wonder you're having difficulty.


    \begin{array}{cccccc} ^1 &^2 & ^3  & ^4 & ^5 & ^6\\  &L&\Theta&G&I&C \\ +&L&\Theta&G&I&C \\ \hline P&R&\Theta&L&\Theta&G \end{array} \quad \text{where each letter represents a different digit.}

    \text{I use }\Theta\text{ because the letter }"O"\text{ can be confused with the digit }"0".

    You should be able to see that P = 1.

    (If you can't, these problems are not for you.)

    In column-3, we see that \Theta + \Theta ends in \Theta.

    Then there are two possible cases:
    . . (1) \Theta = 0
    . . (2) \Theta = 9 with a "carry" from column-4.


    (1) Suppose \Theta = 0.

    We have: . \begin{array}{cccccc} ^1 &^2 & ^3  & ^4 & ^5 & ^6\\  &L& 0&G&I&C \\ +&L& 0&G&I&C \\ \hline 1&R&0 &L& 0&G \end{array}


    In column-6, C+C \,=\,G with no carry.

    In column-5: . I+I ends in 0 with a carry.
    . . Hence: I = 5.

    In column-4: . G+G+1 \,=\,L with no carry.

    In column-2: . L+L \,=\,10+R


    In column-6, let C = 2.
    . . Then G = 4.

    In column-4: L = 9.

    In column-2: R = 8.

    We have found one solution: . \begin{array}{cccccc} &9&0&4&5&2 \\ +&9&0&4&5&2 \\ \hline 1&8&0&9&0&4 \end{array}

    There may be other solutions.
    Thanks from topsquark
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