A little (unsolved) Riddle

This week in math and computer science class we were talking about number riddles, where you can find an explicit solution, which works for every row. For example: 1+452+5123+6214+7?1+4\Rightarrow 5\\2+5\Rightarrow 12\\3+6\Rightarrow 21\\4+7\Rightarrow \boxed { ? } Most of you will get 3232. Because 14+1=51 \cdot 4 +1 = 5; 25+2=122 \cdot 5 + 2 = 12; 36+3=213 \cdot 6 +3 = 21 and finally 47+4=324 \cdot 7 + 4 = 32. Hence, for every interger aa and every integer bb we can say: a+bab+a=a(b+1)a+b\Rightarrow a \cdot b + a = a \cdot \left( b+1 \right) But an important information is, that such a riddle doesn't have a unique solution!

But after a while, we were talking about the following riddle and nobody of us found a solution. Even after searching in the www nobody found any formula, which works row by row. 1+2212+3363+4434+5?1+2\Rightarrow 21\\ 2+3\Rightarrow 36\\ 3+4\Rightarrow 43\\ 4+5\Rightarrow \boxed { ? } As you can easily see, in the first and third row, they only flipped the digits. Anyway, we found a few solutions, but these solutions said, that every second row is an exception. But this is neither a formula, which will work row by row nor a general formula, which will work for any integer aa and for any integer bb.

Do you have any ideas, which will work?

Until now, my best try was: a+bb10+(ab)mod10a+b \Rightarrow b \cdot 10 + (a\cdot b) mod 10 . But my best try will fail at the third row by one.

Note by CodeCrafter 1
1 week, 2 days ago

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Weil, you could suppose the equation is in the form a+bxa+yb+z a + b \rightarrow xa+yb+z . However, this has no solutions for (x,y,z) (x,y,z) .

Then, I tried a+bxab+ya+zb a+b \rightarrow xab+ya+zb and this actually works. We get the solution (x,y,z)=(4,33,2) (x,y,z)=(-4,33,-2) . For 4+5 4+5 this actually results in 42 42 and then decreases even further.

Henry U - 1 week, 1 day ago

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Wow, that's amazing. Thank you!

Funfact: I think it's a divine sign that the answer is 42 ;)

CodeCrafter 1 - 1 week, 1 day ago

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