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Math Help - What 4-digit number abcd satisfies this equation?

  1. #1
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    What 4-digit number abcd satisfies this equation?

    What 4-digit number abcd satisfies this equation?
    4 * abcd = dcba
    Last edited by chinchu; June 8th 2010 at 02:45 AM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by chinchu View Post
    What 4-digit number abcd satisfies this equation?
    4 * abcd = dcba
    Since  4 \times 3000 = 12000 > 9999 so 0 < a \leq 2 but if  a=1 from the unit digit of  dcba we conclude that  dcba is odd but it is the multiple of 4 which should be even . Therefore ,  a=2

    Consider the unit digit after multiplying a number among 0,1,2,..,9 by 4 , we have already known it is  2

     4\times 2bcd = dcb2 we have  d=3~,~8 but  d= 3 is impossible because  4 \times 2000 = 8000 > 3999 we conclude  d = 8


     4 \times 2bc8 = 8cb2

    Then consider the actual values of these number :

     2bc8 = 2000 + 100b + 10c + 8

     8cb2 = 8000 + 100c + 10b + 2

    we have

     8000 + 400b + 40c + 32 = 8000 + 100c + 10b + 2

     390b + 30 = 60c

     13b + 1 = 2c

    When  b=1 ~ , 2c = 14 ~, c=7

    When  b=3 ~ , 2c = 40 ~ , c=20 it is impossible , so  b=1 , c = 7 is the unique solution .


    The required number is  abcd = 2178


     13b + 1 = 2c is an indeterminated equation , any number may be the solution of  b provided  c will be an integer :


     c = \frac{1 + 13b}{2} = \frac{12b + (b+1)}{2} = 6b + \frac{b+1}{2} because  \frac{b+1}{2} is an integer ,  b must be a positive odd ,  b=1,3,5,7,... but if  b is very large which makes c to be greater than 9 , we cannot construct a number with a digit c ...
    Last edited by simplependulum; June 19th 2010 at 09:03 PM.
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  3. #3
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    hi

    how did u take b =1 and b = 3??
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  4. #4
    Newbie darknight's Avatar
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    13b + 1 = 2c
    c = (13b + 1)/2
     c = \frac{1 + 13b}{2} = \frac{12b + (b+1)}{2} = 6b + \frac{b+1}{2} because  \frac{b+1}{2} is an integer ,  b must be a positive odd since for (b + 1) to be perfectly divisible by 2 it should be even. And if b+1 is even it follows that b is odd.

    And therefore simplependulum started by trying all the odd integers which start from 1. And quite clearly, the next odd number is 3. Putting b as 3 results in c = 20. Since c lies between 0 and 9, b cannot be 3 or any integer greater than 3. Therefore, b = 1.
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  5. #5
    MHF Contributor undefined's Avatar
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    Initially I misread dcba as dabc and thought there were no solutions. See this Wikipedia article for some interesting related material.
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  6. #6
    Newbie darknight's Avatar
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    That was very interesting. I only knew about 142857 and it's connection to 22/7. Feels good figuring out its intricacies.
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  7. #7
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    Hello, chinchu!

    This is a classic puzzle.
    It can be solved without Algebra, just common sense.


    What 4-digit number ABCD satisfies: . ABCD \times 4 \:=\:DCBA ?

    We have:

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


    In column-1, since D is a digit \le 9,
    . . we see that A = 1 \text{ or }2

    But if A = 1, column-4 says: . 4\! \times\! D ends in 1 . . . impossible.

    Hence: . \boxed{A = 2}

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


    In column-1, we see that: . D \:=\:8\text{ or }9

    In column-4, 4\!\times\!D ends in 2.
    . . Hence: . \boxed{D = 8}

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


    We see that there is no "carry" from column-2.
    . . Hence: . B \:=\:0\text{ or }1


    . . Suppose B = 0

    . . . . \begin{array}{cccc}<br />
^1 & ^2 & ^3 & ^4 \\<br />
2 & 0 & C & 8 \\<br />
\times & & & 4 \\ \hline<br />
8 & C & 0 & 2 \end{array}

    . . In column-3, we have: . 4\!\times\! C + 3 ends in 0 . . . impossible.


    Hence: . \boxed{B = 1}

    . . \begin{array}{cccc}<br />
^1 & ^2 & ^3 & ^4 \\<br />
2 & 1 & C & 8 \\<br />
\times & & & 4 \\ \hline<br />
8 & C & 1 & 2 \end{array}


    In column-3, 4\!\times\!C + 3 ends in 1.
    . . Hence: . C = 2\text{ or }7

    Since A = 2,\,\text{ then }\,\boxed{C = 7}



    Solution:

    . . \boxed{\begin{array}{cccc}<br />
2 & 1 & 7 & 8 \\<br />
\times & & & 4 \\ \hline<br />
8 & 7 & 1 & 2 \end{array}}

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  8. #8
    Junior Member shannu82's Avatar
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    why does it say LaTeX ERROR: Unknown error all over the place.
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  9. #9
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    Quote Originally Posted by shannu82 View Post
    why does it say LaTeX ERROR: Unknown error all over the place.
    Perhaps it fails to convert the codes after the forum upgrade but this problem can be solved , edit the post and save even we don't have any changes on the post . Check my post up there .
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