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