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Math Help - 1998xABC=CBAx8991: Find A, B, C.

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    1998xABC=CBAx8991: Find A, B, C.

    Find A, B, C

    1998xABC=CBAx8991


    I appreaciate if someone could help. I tried some numbers and some stuff but worthless..
    Last edited by mr fantastic; October 10th 2011 at 11:03 AM. Reason: Re-titled.
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  2. #2
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    Re: Need help

    Quote Originally Posted by Trolo View Post
    Find A, B, C

    1998xABC=CBAx8991


    I appreaciate if someone could help. I tried some numbers and some stuff but worthless..
    Order doesn't matter in multiplication so ABC would cancel to give 1998=8991 which is clearly not the case.

    Therefore all we can say is that ABC = 0 and individual values cannot be determined
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  3. #3
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    Re: Need help

    Will it change something if i say that A,B,C are natural numbers.
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    Re: Need help

    Hello, Trolo!

    This is obviously an "alphametic",
    . . where ABC and CBA represent 3-digit numbers.


    \text{Find }A, B, C .

    . . 1998 \times ABC \:=\:CBA \times 8991

    I began an orderly (but lengthy) search
    . . and virtually tripped over the solution.


    The problem has this form:

    . . \begin{array}{ccccccc} &&& 1 & 9 & 9 & 8 \\ && \times && A & B & C \\ && - & - & - & - & - \\ && * & * & * & * & * \\ & * & * & * & * & * \\ * & * & * & * & * \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & * \end{array} \qquad\begin{array}{ccccccc} &&& 8 & 9 & 9 & 1 \\ && \times && C & B & A \\ && - & - & - & - & - \\ && * & * & * & * & * \\ & * & * & * & * & * \\ * & * & * & * & * \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & * \end{array}


    Suppose C = 1, then A = 8.
    Then we have:

    . . \begin{array}{ccccccc} &&& 1 & 9 & 9 & 8 \\ && \times && 8 & B & 1 \\ && - & - & - & - & - \\ &&  & 1 & 9 & 9 & 8 \\ & * & * & * & * & * \\ 1 & 5 & 9 & 8 & 4 \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & 8 \end{array} \qquad\begin{array}{ccccccc} &&& 8 & 9 & 9 & 1 \\ && \times && 1 & B & 8 \\ && - & - & - & - & - \\ && 7 & 1 & 9 & 2 & 8 \\ & * & * & * & * & * \\  & 8 & 9 & 9 & 1 \\ - & - & - & - & - & - & - \\ * & * & * & * & * & * & 8 \end{array}


    The suspicious "shape" of the two factors: . 1\,9\,9\,8\,\text{ and }\,8\,B\,1
    . . led me to try B = 9 first . . . and it worked!

    . . . . . 1998 \times 891 \;=\;8991 \times 198 \;=\;1,\!780,\!218


    Therefore: . \begin{Bmatrix}A \:=\:8 \\ B \:=\:9 \\ C\:=\:1\end{Bmatrix}

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  5. #5
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    Re: Need help

    Multiplication is commutative, which means that the order you multiply is unimportant - you'll get the same result regardless.

    Therefore, ABC = BCA = CAB = CBA

    If you divide any number by itself, you get 1:

    \frac {x^1}{x^1} = x^1^-^1 = x^0 = 1

    So,
    1998*ABC=CBA*8991
    \frac {1998 * ABC}{CBA} = 8991
    ABC and CBA cancel out, to give:
    1998 * \frac{1}{1} = 8991
    1998 = 8991
    Which, of course, makes no sense, so ABC must equal 0. They must all be zero.

    EDIT: Sorry, I missed your post and assumed they were variables... my mistake! I'll leave this here anyway.
    Last edited by FrameOfMind; October 10th 2011 at 09:06 AM. Reason: Missed OP's second post.
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  6. #6
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    Re: Need help

    Soroban, "search" needs not be "lengthy"; since 1998/8991 = 2/7, then:
    191A - 898C = 70B : B = (191A - 898C) / 70
    a few manipulations lead to A=8, C=1 ; so B=9
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