What 4-digit number abcd satisfies this equation?

4 * abcd = dcba (Worried)

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- Jun 8th 2010, 12:41 AMchinchuWhat 4-digit number abcd satisfies this equation?
What 4-digit number abcd satisfies this equation?

4 * abcd = dcba (Worried) - Jun 8th 2010, 01:55 AMsimplependulum
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 $ ... - Jun 8th 2010, 02:19 AMchinchuhi
how did u take b =1 and b = 3??

- Jun 8th 2010, 03:58 AMdarknight
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. - Jun 8th 2010, 05:36 AMundefined
Initially I misread dcba as dabc and thought there were no solutions. See this Wikipedia article for some interesting related material.

- Jun 8th 2010, 05:57 AMdarknight
That was very interesting. I only knew about 142857 and it's connection to 22/7. Feels good figuring out its intricacies.

- Jun 8th 2010, 06:50 AMSoroban
Hello, chinchu!

This is a classic puzzle.

It can be solved without Algebra, just common sense.

Quote:

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}}$

- Jun 19th 2010, 08:59 PMshannu82
why does it say LaTeX ERROR: Unknown error all over the place.

- Jun 19th 2010, 09:06 PMsimplependulum