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Math Help - Multiplication

  1. #1
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    Multiplication

    Multiplication by 9 reverses a four-digit number (produces the four-digit number with the same digits in the reverse order.) What is the number?

    I know the answer is 1089*9 = 9801 but there is a method to this madness. Does anyone know how to solve this?
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  2. #2
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    Quote Originally Posted by matgrl View Post
    Multiplication by 9 reverses a four-digit number (produces the four-digit number with the same digits in the reverse order.) What is the number?

    I know the answer is 1089*9 = 9801 but there is a method to this madness. Does anyone know how to solve this?
    Solve what? You just did it. Do you mean to prove there's no other possible answer?
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    yes please and thank you i do not know how to prove this
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  4. #4
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    Why not try something like setting up the following equations:

    9(a + 10b + 100c + 1000d) = d + 10c + 100b + 1000a (digit reversing)
    a+b+c+d = 9n, for some integer n (divisibility by 9). Either n = 1, 2, 3, or 4, because of the following constraints:

    0 <= a,b,c,d <= 9.

    Maybe that could get you started?
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  5. #5
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    Hello, matgrl!

    \text{Multiplication by 9 reverses a four-digit number (produces}
    \text{a four-digit number with the same digits in reverse order.)}
    \text{What is the number?}

    I will assume that A,B,C,D are four different digits.


    . . \begin{array}{cccc}<br />
^1 & ^2 & ^3 & ^4 \\<br />
A & B & C & D \\<br />
\times &&& 9 \\ \hline<br />
D & C & B & A<br />
 \end{array}


    In column-1, we see that: . A = 1,\:D = 9
    And there is no "carry" from column-2.


    So we have:

    . . \begin{array}{cccc}<br />
^1 & ^2 & ^3 & ^4 \\<br />
1 & B & C & 9 \\<br />
\times &&& 9 \\ \hline<br />
9 & C & B & 1<br />
 \end{array}


    Since there is no "carry" frim column-2, \,B must be 0 or 1.
    Since A = 1, then: . B \,=\,0


    Then we have:

    . . \begin{array}{cccc}<br />
^1 & ^2 & ^3 & ^4 \\<br />
1 & 0 & C & 9 \\<br />
\times &&& 9 \\ \hline<br />
9 & C & 0 & 1<br />
 \end{array}


    In column-3, we have: . 9C + 8 ends in 0.
    . . Hence, \,9C ends in 2 \quad\Rightarrow\quad C = 8


    Therefore, the solution is:

    . . \begin{array}{cccc}<br />
1 & 0 & 8 & 9 \\<br />
\times &&& 9 \\ \hline<br />
9 & 8 & 0 & 1<br />
 \end{array}

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  6. #6
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    wonderful thank you!
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