Results 1 to 9 of 9

Thread: What 4-digit number abcd satisfies this equation?

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    5

    What 4-digit number abcd satisfies this equation?

    What 4-digit number abcd satisfies this equation?
    4 * abcd = dcba
    Last edited by chinchu; Jun 8th 2010 at 02:45 AM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by chinchu View Post
    What 4-digit number abcd satisfies this equation?
    4 * abcd = dcba
    Since $\displaystyle 4 \times 3000 = 12000 > 9999$ so $\displaystyle 0 < a \leq 2$ but if $\displaystyle a=1 $ from the unit digit of $\displaystyle dcba $ we conclude that $\displaystyle dcba $ is odd but it is the multiple of $\displaystyle 4 $ which should be even . Therefore , $\displaystyle a=2 $

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

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


    $\displaystyle 4 \times 2bc8 = 8cb2$

    Then consider the actual values of these number :

    $\displaystyle 2bc8 = 2000 + 100b + 10c + 8 $

    $\displaystyle 8cb2 = 8000 + 100c + 10b + 2 $

    we have

    $\displaystyle 8000 + 400b + 40c + 32 = 8000 + 100c + 10b + 2$

    $\displaystyle 390b + 30 = 60c $

    $\displaystyle 13b + 1 = 2c $

    When $\displaystyle b=1 ~ , 2c = 14 ~, c=7$

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


    The required number is $\displaystyle abcd = 2178$


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


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

  3. #3
    Newbie
    Joined
    Jun 2010
    Posts
    5

    hi

    how did u take b =1 and b = 3??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie darknight's Avatar
    Joined
    Sep 2009
    From
    on the dark side of the moon
    Posts
    16
    13b + 1 = 2c
    c = (13b + 1)/2
    $\displaystyle c = \frac{1 + 13b}{2} = \frac{12b + (b+1)}{2} = 6b + \frac{b+1}{2}$ because $\displaystyle \frac{b+1}{2}$ is an integer , $\displaystyle 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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Initially I misread dcba as dabc and thought there were no solutions. See this Wikipedia article for some interesting related material.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie darknight's Avatar
    Joined
    Sep 2009
    From
    on the dark side of the moon
    Posts
    16
    That was very interesting. I only knew about 142857 and it's connection to 22/7. Feels good figuring out its intricacies.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, chinchu!

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


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

    We have:

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


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

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

    Hence: .$\displaystyle \boxed{A = 2}$

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


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

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

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


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


    . . Suppose $\displaystyle B = 0$

    . . . . $\displaystyle \begin{array}{cccc}
    ^1 & ^2 & ^3 & ^4 \\
    2 & 0 & C & 8 \\
    \times & & & 4 \\ \hline
    8 & C & 0 & 2 \end{array}$

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


    Hence: .$\displaystyle \boxed{B = 1}$

    . . $\displaystyle \begin{array}{cccc}
    ^1 & ^2 & ^3 & ^4 \\
    2 & 1 & C & 8 \\
    \times & & & 4 \\ \hline
    8 & C & 1 & 2 \end{array}$


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

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



    Solution:

    . . $\displaystyle \boxed{\begin{array}{cccc}
    2 & 1 & 7 & 8 \\
    \times & & & 4 \\ \hline
    8 & 7 & 1 & 2 \end{array}}$

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member shannu82's Avatar
    Joined
    May 2010
    Posts
    28
    why does it say LaTeX ERROR: Unknown error all over the place.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Jan 2009
    Posts
    715
    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 .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Digit sum & digit product of number x
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 19th 2011, 08:07 AM
  2. Replies: 7
    Last Post: Jul 24th 2010, 02:21 AM
  3. Replies: 3
    Last Post: Jul 22nd 2010, 07:14 PM
  4. 4 digit PIN number - please help
    Posted in the Statistics Forum
    Replies: 4
    Last Post: Feb 12th 2010, 05:47 AM
  5. 6 digit number
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 10th 2009, 08:10 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum