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Math Help - Reverse digits problem

  1. #1
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    Reverse digits problem

    What five digit number when multiplied by 4 produces a five digit number whose digits are those of the original number but in reverse order?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Ian Moore View Post
    What five digit number when multiplied by 4 produces a five digit number whose digits are those of the original number but in reverse order?
    The leading digit can only be 1 or 2, and as when multiplied by 4 you get the
    digits reversed it cannout be 1 (as then we would have an even number ending in 1),
    so the number is of the form 2 . . . 8

    Now the next most significant digit must be 0, 1 or 2 (as there is no carry when
    multiplied by 4), and trial and error show that 21978 * 4 = 87912

    RonL
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  3. #3
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    Hello, Ian!

    Here is a start on the problem . . .


    What five-digit number, when multiplied by 4, produces a five digit-number
    whose digits are those of the original number but in reverse order?
    The problem looks like this:

    . . . \begin{array}{ccccc}^1 & ^2 & ^3 & ^4 & ^5 \\<br />
A & B & C & D & E \\<br />
\times & & & & 4 \\ \hline<br />
E & D & C & B & A \end{array}


    In column-1: . 4\times A \:\to \:E . . . Hence: . A \:=\:1\text{ or }2

    In column-5: . 4 \times E is even . . . Hence: . A \:=\:2

    . . . \begin{array}{ccccc}^1 & ^2 & ^3 & ^4 & ^5 \\<br />
2 & B & C & D & E \\<br />
\times & & & & 4 \\ \hline<br />
E & D & C & B & 2 \end{array}


    In column-1: . E \:=\:8\text{ or }9
    In column-1: . 4 \times E \:\to\:2 . . . Hence: . E \:= \:8

    . . . \begin{array}{ccccc}^1 & ^2 & ^3 & ^4 & ^5 \\<br />
2 & B & C & D & 8 \\<br />
\times & & & & 4 \\ \hline<br />
8 & D & C & B & A \end{array}


    In column-2: . 4 \times B \to D ... and there is no "carry" to column-1.
    . . Hence: . B \:=\:0\text{ or }1


    Can you finish it?

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  4. #4
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    Thankyou Soroban and Captain Black.
    Much appreciated.
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