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Thread: How many different solutions of equation?

  1. #1
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    How many different solutions of equation?

    How many different solutions has this equation:

    $\displaystyle \overline{abc}+\overline{def}+\overline{ghi}=963$

    Here $\displaystyle \overline{abc}, \overline{def}, \overline{ghi}$ are three-digit numbers

    $\displaystyle a,b,c,d,e,f,g,h,i $- different digits; $\displaystyle a,d,g\neq 0$

    I could begin with:
    $\displaystyle (a+d+g)\cdot100+(b+e+h)\cdot10+(c+f+i)=963$

    Now, $\displaystyle c+f+i$ must be 3, 13 or 23...

    Not know how to connect with other.

    Could you help to me. Thank you!!!
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  2. #2
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    Re: How many different solutions of equation?

    Hey feferon11.

    You have nine independent variables to start off with.

    If c + f + i = *3 (* is a number from 0 to 2 since max is 9+9+9 = 27) then you have one constraint. Now tke the * and move it the 10's group. Let * = p.

    (b + e + h + p) = *6. Where again * = (0,1,2 for the same reason as above). If * = q we have

    (a + d + g + q) = 9

    So in total we have the following constraints:

    All numbers between 0 and 9 inclusive
    a, d, g > 0
    c + f + i = 10*p + 3 (p = 0,1,2)
    b + e + h + p = 10*q + 6 (q = 0,1,2)
    (a + d + g + q) = 9

    These constraints though do not make it easy so I will suggest another way.

    Consider the two variable case.

    Let abc + def = 963. Since a, d != 0 we start abc = 100.

    Now def = 963 - def. This means that you have 963 - 100 = 863 different combinations for both numbers.

    In the three variable case you do the same thing but twice.

    abc + def + ghi = 963. Fix abc = 100 which means (def + ghi) = 963 - 100 = 863. Now figure out how many possibilities for 863 (it will be 863 - 100 = 763). Then do it for abc = 101, 102, ...., 863.

    You should find a pattern in this and I suggest you write it all out to see it for yourself. (Hint: Try and find the relationship given that abc = some number for how many combinations between def and ghi and add up all possibilities for each abc).
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  3. #3
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    Re: How many different solutions of equation?

    Hello, feferon11!

    This is an Alphametic.
    It requires some intelligent trial-and-error.
    Algebra usually won't help.


    How many different solutions has this problem?

    . . $\displaystyle \begin{array}{cccc} & ^1 & ^2 & ^3 \\ & A&B&C \\ &D&E&F \\ + & G&H&I \\ \hline & 9&6&3 \end{array}$

    Each letter represents a different digit and $\displaystyle A,D,G \ne 0.$

    In column-3, $\displaystyle C+F+I$ ends in 3.
    So, the sum could be 3, 13 or 23.

    If the sum is 3, then $\displaystyle C,F,I$ must be $\displaystyle 0,1,2$ in some order.
    That leaves $\displaystyle 3,4,5,6,7,8,9$ for the other digits.

    But in column-1, $\displaystyle A+D+G$ must equal 7, 8 or 9.
    And that is impossible with the remaining digits.

    Using similar reasoning, I eliminated a few more possibilties.


    I found a solution.

    The general solution looks like this:

    . . $\displaystyle \begin{array}{|c|c|c|} ^1 & ^2 & ^3 \\ \hline 1&3&6 \\ 2&4&8\\ 5&7&9 \\ \hline 9&6&3 \\ \hline \end{array}$

    The digits in each column can be arranged in $\displaystyle 3!$ ways.

    Therefore, there are:.$\displaystyle 6\cdot6\cdot6 \,=\, 216$ solutions.
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