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Math Help - 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:

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

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

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

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

    Now, 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?

    . . \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 A,D,G \ne 0.

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

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

    But in column-1, 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:

    . . \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 3! ways.

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