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Math Help - Finding algebra variables

  1. #1
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    Talking Thank you!

    Quote Originally Posted by JeWiSh View Post
    a=7
    b=6
    c=5
    d=2
    e=3
    f=4

    I did it with some brief trial and error, and I'm not sure if maybe there's a more efficient, concrete method...
    That is brilliant. Thank you so much. I figured it was possible with trial and error but I guess I am a bit tired/out of practice.

    Thanks again
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  2. #2
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    Post Finding algebra variables

    Hi,

    Trying to help daughter with homework and it has been so long! Can't get my head round this one:

    ab*c+d=200+ef

    We are told that a,b,c,d,e,f represent 2,3,4,5,6,7 but not in that order. Can someone tell me how to work this out please, it is driving me nuts. Thank you very much.
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  3. #3
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    Quote Originally Posted by myfi View Post
    Hi,

    Trying to help daughter with homework and it has been so long! Can't get my head round this one:

    ab*c+d=200+ef

    We are told that a,b,c,d,e,f represent 2,3,4,5,6,7 but not in that order. Can someone tell me how to work this out please, it is driving me nuts. Thank you very much.
    a=7
    b=6
    c=5
    d=2
    e=3
    f=4

    I did it with some brief trial and error, and I'm not sure if maybe there's a more efficient, concrete method...
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  4. #4
    Super Member

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    May 2006
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    Lexington, MA (USA)
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    Hello, myfi!

    Trying to help daughter with homework.
    . . AB*C+D\:=\:200+EF

    We are told that A,B,C,D,E,F represent 2,3,4,5,6,7 but not in that order.

    I don't understand that solution
    . . . or maybe I don't understand the problem.

    I would assume that AB represents a two-digit number, as does EF.

    Otherwise, why would they write it that way?

    So we have a math puzzle to solve:

    . . \begin{array}{ccc}<br />
& A & B \\ & \times & C \\ \hline<br />
? & ? & ? \\<br />
+ & & D \\ \hline<br />
2 & E & F \end{array}


    There are 6! = 720 possible placements of those digits.

    We can eliminate many possibilities,
    . . but it will still require Brute Force trial-and-error.


    I found a solution (and I'm not looking for any more).

    . . \begin{array}{ccc}<br />
& {\color{red}3} & {\color{red}6} \\ & \times & {\color{red}7} \\ \hline<br />
2 & 5 & 2 \\<br />
+ & & {\color{red}2} \\ \hline<br />
2 & {\color{red}5} & {\color{red}4} \end{array}


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