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Thread: Putnam Problem of the Day

  1. #1
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    Putnam Problem of the Day

    Hi , let's think about this problem :

    Find the smallest natural number with $\displaystyle 6$ as the last digit , such that if the final $\displaystyle 6$ is moved to the front of the number it is multiplied by $\displaystyle 4$ .


    Is the answer
    Spoiler:
    153846
    ?
    Last edited by simplependulum; Aug 14th 2010 at 08:00 PM.
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  2. #2
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    Let the number be $\displaystyle a_1a_2...a_n6$.

    Let $\displaystyle x=0.\overline{a_1a_2...a_n6}$.

    Then $\displaystyle 4x = 0.\overline{6a_1a_2...a_n} \Rightarrow 40x = 6.\overline{a_1a_2...a_n6} = 6+x$.

    Therefore $\displaystyle x = \frac{6}{39} = \frac{2}{13} = 0.\overline{153846}$.

    By backwards argument it's easy to see that all the cycles in $\displaystyle \frac{2}{13}$ are solutions.

    Thus we have found all the solutions, the smallest of which is $\displaystyle 153846$.

    This technique can be applied to this kind of problems in many cases.
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  3. #3
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    Hello, simplependulum!

    Just write out the multiplcation,
    . . and the digits will appear sequentially.


    Find the smallest natural number with $\displaystyle 6$ as the last digit,
    such that if the final $\displaystyle 6$ is moved to the front of the number it is multiplied by $\displaystyle 4$.

    We have: .$\displaystyle \begin{array}{cccccc}
    ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
    A&B&C&D&E&6 \\ \times &&&&& 4\\ \hline 6&A&B&C&D&E \end{array}$



    In column-6: .$\displaystyle 6 \!\times\! 4 \,=\,24 \quad\Rightarrow\quad E = 4$

    . . $\displaystyle \begin{array}{cccccc}
    ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
    A&B&C&D&4&6 \\
    \times &&&&& 4\\ \hline
    6&A&B&C&D&4 \end{array}$



    In column-5: .$\displaystyle 4 \!\times\! 4 + 2 \:=\:18 \quad\Rightarrow\quad D = 8$

    . . $\displaystyle \begin{array}{cccccc}
    ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
    A&B&C&8&4&6 \\
    \times &&&&& 4\\ \hline
    6&A&B&C&8&4 \end{array}$



    In column-4: .$\displaystyle 8 \!\times\! 4 + 1 \:=\:33 \quad\Rightarrow\quad C = 3 $

    . . $\displaystyle \begin{array}{cccccc}
    ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
    A&B&3&8&4&6 \\
    \times &&&&& 4\\ \hline
    6&A&B&3&8&4 \end{array}$



    In column-3: .$\displaystyle 3\!\times\!4 + 3 \:=\:15 \quad\Rightarrow\quad B = 5$

    . . $\displaystyle \begin{array}{cccccc}
    ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
    A&5&3&8&4&6 \\
    \times &&&&& 4\\ \hline
    6&A&5&3&8&4 \end{array}$



    In column-2: .$\displaystyle 5\!\times\!4 + 1 \:=\:21 \quad\Rightarrow\quad A = 1$


    Therefore: . $\displaystyle \begin{array}{cccccc}
    1&5&3&8&4&6 \\
    \times &&&&& 4\\ \hline
    6&1&5&3&8&4 \end{array}$

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  4. #4
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    Soroban, the fact that it is a 6digit number is NOT a given.
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  5. #5
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    Quote Originally Posted by Wilmer
    Soroban, the fact that it is a 6-digit number is NOT a given. . Right!


    I originally set up the problem like this:

    . . $\displaystyle \begin{array}{ccccccccccc}
    A & B & C & D & E & F & G & \hdots & M & N & 6 \\
    \times &&&&&&&&&& 4 \\ \hline
    6 & A & B & C & D & E & F & G & \hdots & M & N
    \end{array}$


    And found that "it came out even" after six digits
    . . which gave me the smallest number: $\displaystyle 153,\!846.$

    The next is the 12-digit number: $\displaystyle 153,\!846,\!153,\!846.$

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  6. #6
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    Fyi, there is a name for these, n-parasitic number, where here n=4 and k=6.
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  7. #7
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    Let n = number of digits

    {20[10^(n-1) - 4] + 78} / 13

    Gives integer solution for n=6k where k = any integer > 0
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