Concatenation sum

Since both $A,B$ are $1729 \times 9 = 15561$ digit number , there is a perfect pairing between their digits in the sum like this:

$1+9=10 \\ 2+8=10 \\ 3+7=10 \\ 4+6=10 \\ 5+5=10 \\ 6+4=10 \\ 7+3=10 \\ 8+2=10 \\ 9+1=10 \\$

So there is a bijection between a $k^{th}$ digit of $A$ and the $k^{th}$ digit of $B$ such that $A_k + B_k = 10$ where $A_i , B_i$ are the $i^{th}$ digits from left in $A$ , $B$ respectively. We observe that $\forall \ i \ 1 \leq i \leq 15561$ , $A_i+B_i=10$ . So $A+B$ can be written in the form:

$\Large{A+B=\displaystyle\sum_{i=1}^{15561} 10^{i-1}(A_{15562-i} + B_{15562-i}) \\ = \displaystyle\sum_{i=1}^{15561} 10^{i} = \underbrace{\overline{111111111\dots111110}}_{\text{15561 } (111111111)\text{'s} , \text{ 1 } 0\text{'s}}}$

Hence sum of digits in $A+B=15561 \times 1 = \huge\boxed{15561}$ .

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Generalization:

If in $A$ , $123456789$ is concatenated $n$ times , in $B$ , $987654321$ is concatenated $n$ times , then sum of digits of $A+B = 9n$ .

For 2 concatenations 123456789123456789 $+$

987654321987654321

1111111111111111110 So you got (18)1's for 2 concatenations same pattern will be followed for 1729 concatenations and you'll get 1729×9 one's and sum of digits will be 1729×9 which 15561

Nice that you used the Hardy-Ramanujan number 1729 too - https://simple.wikipedia.org/wiki/Taxicab_number