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Math Help - Problem

  1. #1
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    Problem

    Find all 6-digit numbers x with the following property:

    If the first digit in x is deleted and added after the 5 other digits, you get a number that is 3 times as large as x.

    For instance, the number x=158249 gets turned into 582491.

    The number x=158249 does not meet the condition, because 3*158249 ≠ 582491

    I hope that someone can help me with this problem - I have no idea how to solve it..
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  2. #2
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    Awetuouncsygg
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    Re: Problem

    x=\overline{abcdef}, a, b, c, d, e, f are digits, a > 0

    3\cdot \overline{abcdef}=\overline{bcdefa} \Leftrightarrow 3\cdot a \cdot 10^5+3\cdot \overline{bcd} \cdot 10^2+3\cdot \overline{ef}=\overline{bcd}\cdot 10^3+\overline{ef}\cdot 10+a \Leftrightarrow (3\cdot 10^5-1)a=700 \overline{bcd}+7\overline{ef}\Leftrightarrow 299.999\cdot a=7 \overline{bcdef} \Leftrightarrow  42.857 \cdot a=\overline{bcdef}

    a is a digit and 42.857 \cdot a = \overline{bcdef} < 99.999, so a can be 1 or 2. If a is 1, then x is 142857, if a is 2, x is 242857.
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  3. #3
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    Lexington, MA (USA)
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    Re: Problem

    Hello, emilhp!

    This can be solved as an Alphametic.


    Find all 6-digit numbers x with the following property:

    If the first digit in x is deleted and added after the 5 other digits,
    you get a number that is 3 times as large as x.

    We have: . \begin{array}{cccccc} A&B&C&D&E&F \\ \times &&&&& 3 \\ \hline B&C&D&E&F&A \end{array}


    We see that: A \:=\:1,\,2
    If A \ge 3, the product will be a seven-digit number.


    Suppose A = 1.

    We have: . \begin{array}{cccccc} 1&B&C&D&E&F \\ \times &&&&& 3 \\ \hline B&C&D&E&F&1 \end{array}\qquad\text{We see that: }\,F = 7.


    We have: . \begin{array}{cccccc} 1&B&C&D&E&7 \\ \times &&&&& 3 \\ \hline B&C&D&E&7&1 \end{array} \qquad\text{We see that: }E = 5.


    We have: . \begin{array}{cccccc} 1&B&C&D&5&7 \\ \times &&&&& 3 \\ \hline B&C&D&5&7&1 \end{array} \qquad\text{We see that: }D = 8.


    We have: . \begin{array}{cccccc} 1&B&C&8&5&7 \\ \times &&&&& 3 \\ \hline B&C&8&5&7&1 \end{array} \qquad\text{We see that: }C = 2.


    We have: . \begin{array}{cccccc} 1&B&2&8&5&7 \\ \times &&&&& 3 \\ \hline B&2&8&5&7&1 \end{array} \qquad\text{We see that: }B = 4.


    We have: . \begin{array}{cccccc} 1&4&2&8&5&7 \\ \times &&&&& 3 \\ \hline 4&2&8&5&7&1 \end{array}



    Assume A = 2 .and using similar reasoning,

    . . we have: . \begin{array}{cccccc} 2&8&5&7&1&4 \\ \times &&&&& 3 \\ \hline 8&5&7&1&4&2 \end{array}

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