Corrected in response to Niel de Beaudrap's comment.
Main ideas:
- It is a consequence of the test for divisibility by $11$. The remaider of the division of a number $a$ by $11$ is equal to the remainder of the difference of the sum of the odd ordered digits of $a$ from the sum of the even ordered digits of $a$.
- And the base $10$ property: for some $n$ the power $10^n\equiv 1\pmod {11}$.
- $10\equiv (-(-10))\equiv -1\pmod{11}$
Let
$a=\overline{a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}}=a_{n}10^{n}+a_{n-1}10a^{n-1}+\cdots +a_{2}10^{2}+a_{1}10+a_{0}.$
For $\mod 11$ we have $10^{0}\equiv 1,10^{1}\equiv 10,10^{2}\equiv 1$, and if $n$ even
$10^{n}\equiv 1;$
if $n$ odd $10^{n}\equiv 10.$ Thus
$a\equiv (a_{0}+a_{2}+a_{4}+\cdots )+10(a_{1}+a_{3}+a_{5}+\cdots )\pmod{11}.$
Since $-10\equiv 1\pmod{11}$, we have $10\equiv (-(-10))\equiv -1\pmod{11}$ and
$a\equiv (a_{0}+a_{2}+a_{4}+\cdots )-(a_{1}+a_{3}+a_{5}+\cdots )\pmod{11}.$
Your case is: For $a=A10^{3}+B10^{2}+C10+D$ $a\equiv (D+B)-(C+A)\pmod{11}$ and for $b=D10^{3}+C10^{2}+B10+A$ $b\equiv (A+C)-(B+D)\pmod{11}$ and $a+b\equiv (D+B)-(C+A)+\left( (A+C)-(B+D)\right) \equiv 0\pmod{11}.$
For $a=A10^{4}+B10^{3}+C10^{2}+D10+E$ and $b=E10^{4}+D10^{3}+C10^{2}+B10+A,$ we have $a+b\equiv (A+C+E)-(B+D)+\left( (E+C+A)-(D+B)\right) \not\equiv 0\pmod{11}.$
Can you generalize this?
Corrected in response to Niel de Beaudrap's comment.
I want a generalized proof for the above problem. I want why it is applicable only for 11 and why it is valid for even digit of numbers?
I am not sure in what direction does OP want to generalize his/her result. It is valid only for numbers with an even number of digits. (see above)
- Here is a possible generalization of the proof above (radix $10$) for the radix $r$. Since
$r^{0}\equiv 1\pmod {r+1},r^{1}\equiv r\pmod {r+1},$
$r^{2}\equiv 1\pmod{r+1},r^{3}\equiv r\pmod {r+1},\ldots$
and $ABCD_{r}\equiv (D+B)_{r}-(C+A)_{r}\pmod {r+1},$
$DCBA_{r}\equiv (A+C)_{r}-(B+D)_{r}\pmod {r+1},$
we obtain
$ABCD_{r}+DCBA_{r}\equiv 0\pmod {r+1}.$