What 4-digit number abcd satisfies this equation?

**chinchu** · June 8th 2010, 12:41 AM

What 4-digit number abcd satisfies this equation?
4 * abcd = dcba

**simplependulum** · June 8th 2010, 01:55 AM

Originally Posted by chinchu

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 $c$ ...

**chinchu** · June 8th 2010, 02:19 AM

how did u take b =1 and b = 3??

**darknight** · June 8th 2010, 03:58 AM

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.

**undefined** · June 8th 2010, 05:36 AM

Initially I misread dcba as dabc and thought there were no solutions. See this Wikipedia article for some interesting related material.

**darknight** · June 8th 2010, 05:57 AM

That was very interesting. I only knew about 142857 and it's connection to 22/7. Feels good figuring out its intricacies.

**Soroban** · June 8th 2010, 06:50 AM

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:

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

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: . $\boxed{A = 2}$

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

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

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

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

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

. . Suppose $B = 0$

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

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

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

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

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:

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

**shannu82** · June 19th 2010, 08:59 PM

why does it say LaTeX ERROR: Unknown error all over the place.

**simplependulum** · June 19th 2010, 09:06 PM

Originally Posted by shannu82

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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