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  1. #1
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    Problem

    The last digit of a six-digit number is 2. If the 2 is moved to the start of the number, the new six-digit number is only a third of the original number. Find the original number.
    Please help, I cant get my head around this.
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by not happy jan View Post
    The last digit of a six-digit number is 2. If the 2 is moved to the start of the number, the new six-digit number is only a third of the original number. Find the original number.
    Please help, I cant get my head around this.
    Digitwise, the number is abcde2

    the new number is 2abcde

    so 3 times the new number equals the old number

    \begin{array}{lllllll}<br />
&a&b&c&d&e&2\\<br />
=&6&3a&3b&3c&3d&3e\\<br />
\end{array}

    3e must have a ones digit of 2. So it could be equal to 2, 12, 22, 32, etc

    But we know it is a digit, so it is less than 10, and since 3*9 = 27, it must equal 2, 12, or 22. And since 12 is the only one of those which is a multiple of 3, it must be 12. then 12/3=4. so e =4

    \begin{array}{lllllll}<br />
&a&b&c&d&4&2\\<br />
=&6&3a&3b&3c&3d&12\\<br />
\end{array}

    we can't have a ones digit of 12, so we carry the one.

    \begin{array}{lllllll}<br />
&a&b&c&d&4&2\\<br />
=&6&3a&3b&3c&3d+1&2\\<br />
\end{array}


    now 3d+1 returns a ones digit of 4. so options are 4, 14, and 24 (because d cannot be greater than 9, so 3d+1 cannot be greater than 28) subtracting one from each of these we get 3d = 3, 13, or 23. and clearly 3 is the only option returning an integer. so 3d=3. thus d = 1

    \begin{array}{lllllll}<br />
&a&b&c&1&4&2\\<br />
=&6&3a&3b&3c&4&2\\<br />
\end{array}

    now we know that 3c has a ones digit of 1. so 3c equals 1, 11, or 21 (again, must be less than 27, because c < 9). 21 is the only multiple of 3. so 3c = 21. therefore c = 7

    \begin{array}{lllllll}<br />
&a&b&7&1&4&2\\<br />
=&6&3a&3b&21&4&2\\<br />
\end{array}

    21 is not a digit, so we carry the 2.

    \begin{array}{lllllll}<br />
&a&b&7&1&4&2\\<br />
=&6&3a&3b+2&1&4&2\\<br />
\end{array}

    now we have that 3b+2 must have a ones digit of 7. so 7, 17, or 27. Subtracting 2 we get that 3b must equal 5, 15, or 25. 15 is the only multiple of 3. so 3b=15. therefore b = 5

    \begin{array}{lllllll}<br />
&a&5&7&1&4&2\\<br />
=&6&3a&17&1&4&2\\<br />
\end{array}

    17 is not a digit so carry the one.

    \begin{array}{lllllll}<br />
&a&5&7&1&4&2\\<br />
=&6&3a+1&7&1&4&2\\<br />
\end{array}

    Now we get that 3a+1 has a ones digit of 5. so 5, 15, or 25. Subtracting one we get that 3a must be 4, 14, or 24. 24 is the only multiple of 3. so 3a=24. therefore a = 8

    \begin{array}{lllllll}<br />
&8&5&7&1&4&2\\<br />
=&6&25&7&1&4&2\\<br />
\end{array}

    25 is not a digit, so carry the 2.

    \begin{array}{lllllll}<br />
&8&5&7&1&4&2\\<br />
=&8&5&7&1&4&2\\<br />
\end{array}

    Now looking back at the original, equation we see that this becomes

    \begin{array}{lllllll}<br />
&a&b&c&d&e&2\\<br />
=&8&5&7&1&4&2\\<br />
\end{array}

    Therefore the number is 857142, and the new number is 285714.
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, not happy jan!

    Lucky for both of us, I'm familiar with this type of problem . . .


    The last digit of a six-digit number is 2. If the 2 is moved to the front,
    the new six-digit number is only a third of the original number.
    Find the original number.
    We have: . ABCDE2

    Then: . 2ABCDE \:=\:\frac{1}{3}\times ABCDE2 \quad\Rightarrow\quad 2ABCDE \times 3 \;=\;ABCDE2

    And we have a multiplication problem . . .

    . . \begin{array}{cccccc}<br />
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\<br />
2 & A & B & C & D & E \\<br />
\times & & & & & 3 \\ \hline<br />
A & B & C & D & E & 2 \end{array}


    In column-6, 3E \text{ ends in 2}\quad\Rightarrow\quad\boxed{ E = 4}\quad\hdots ("carry 1" to column-5)


    . . \begin{array}{cccccc}<br />
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\<br />
2 & A & B & C & D & 4 \\<br />
\times & & & & & 3 \\ \hline<br />
A & B & C & D & 4 & 2 \end{array}


    In column-5, 3D + 1\text{ ends in 4}\quad\Rightarrow\quad\boxed{D = 1}


    . . \begin{array}{cccccc}<br />
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\<br />
2 & A & B & C & 1 & 4 \\<br />
\times & & & & & 3 \\ \hline<br />
A & B & C & 1 & 4 & 2 \end{array}


    In column-4, 3C\text{ ends in 1}\quad\Rightarrow\quad\boxed{C = 7}\quad\hdots ("carry 2" to column-3)


    . . \begin{array}{cccccc}<br />
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\<br />
2 & A & B & 7 & 1 & 4 \\<br />
\times & & & & & 3 \\ \hline<br />
A & B & 7 & 1 & 4 & 2 \end{array}


    In column-3, 3B + 2\text{ ends is 7}\quad\Rightarrow\quad\boxed{B = 5}\quad\hdots ("carry 1" to column -2)


    . . \begin{array}{cccccc}<br />
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\<br />
2 & A & 5 & 7 & 1 & 4 \\<br />
\times & & & & & 3 \\ \hline<br />
A & 5 & 7 & 1 & 4 & 2 \end{array}


    In column-2, 3A + 1\text{ ends in 5}\quad\Rightarrow\quad\boxed{A = 8}


    . . \begin{array}{cccccc}<br />
2 & 8 & 5 & 7 & 1 & 4 \\<br />
\times & & & & & 3 \\ \hline<br />
8 & 5 & 7 & 1 & 4 & 2 \end{array}\qquad\hdots ta-DAA!



    Finally took a look at your solution, angel.white . . .
    Excellent job . . . nice explanations!
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by Soroban View Post
    Finally took a look at your solution, angel.white . . .
    Excellent job . . . nice explanations!
    Thanks ^_^ your way was a bit clearer b/c it didn't rely on unconventional mathematical notations like mine. (I sort of figured it out / made it up as I went).
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