Question

A four-digit number $abcd$ is divisible by $11$, if $d+b=$ _______ or _______.

Answer 1

If the difference of the sum of alternative digits of a number is divisible by $11$, then that number is divisible by $11$ completely.

A number is divisible by $11$ if the difference between the sum of its digit at an even place and the sum of its digit at the odd place is $0$ or multiple of $11$.

$(d+b)-(a+c)=0,11,22,33,55,...$

It gives us ,$\left(d+b\right)=(a+c)\mathrm{or}(\mathrm{d}+\mathrm{b})=11,22,33+(\mathrm{a}+\mathrm{c})$

But as $a,b,c,d$are all single-digit numbers

So, the maximum sum of $\left(d+b\right)$ can be $19$

That means $22,33$,… and higher digits are excluded, and only $11+(a+c)$is remaining

Therefore, $\left(d+b\right)=(a+c)\mathrm{or}(\mathrm{d}+\mathrm{b})=11+(\mathrm{a}+\mathrm{c})$