I saw a quite vague math riddle like this - given two numbers a = 78404343 and b = 57, derive the third number c.
I think the solution to that might be this:
78404343 + 57 = 78404400
reverse
00440487
440487 + 13 = 440500
reverse
005044
5044 + 56 = 5100
reverse
0015
15 + 85 = 100
c = 57135685
I think it's a quite neat trick because when looking at a and c it is not obvious (not to me at least ) how these numbers are related.
My question is - is there a name for this trick with numbers? I wanted to read more about it, but don't know how to google it .
Also I could have avoided reversing and do dividing by hundred instead or perhaps increate the digits of added numbers each time (not sure which is the right way to do it).
I don't have a name for it, but can explain how it works. I think it is stage magic where you describe the computation in your question, then are given $a$ and are asked to predict the final number mentally. You should only be given $a$, not $b$ as well. I do a lot of mental calculation, but this seems like it would be difficult to me because of the reordering that takes place. $a$ should have a number of digits divisible by $4$ or we have to rethink what happens at the end.
Break $a$ into two digit pieces $$78\ 40\ 43\ 43$$ Now reorder them taking the last, the first, the next to last, the next to first, and so on (if there were more digits). When you take from the front of the number, reverse the digits $$43\ 87\ 43\ 04$$ Each one after the second gets a carry in, so increase it by $1$. The last one, which was just in front of the midpoint, gets a carry in from both sides so increase it by $11$ $$43\ 87\ 44\ 15$$ Finally subtract each pair from $100$ and report the result $$57135685$$
As a longer example, take $$12\ 34\ 56\ 78\ 75\ 31\\ 31\ 21\ 75\ 43\ 78\ 65\\ 31\ 21\ 76\ 44\ 79\ 76\\ 697924562124$$ And following your procedure $$123456787531+69=123456787600\\ 6787654321+79=6787654400\\ 44567876+24=44567900\\ 976544+56=976600\\ 6679+21=6700\\ 76+24=100$$