A smudge in his notebook

Question

Lothar took his notebook and started scribbling some equations:

\begin{align} 2 + 3&= 412 \\ 1+5& = 4321\\ 0+4&=4320\\ 4320-412&=40\\ 4321120-0&=3\\ 80-62&=0\\ 6+6&=\text{■} \end{align}

$$\hspace{10mm}\dots$$

Where the $\text{■} $ represents a smudge on the paper.

Can you restore the missing number? Can you explain these equations?

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Hint 0: Trivia reason was mentioned in the comments and answers posted so far.

Hint 1:

The numbers here, due to their nature, are not necessarily restricted to decimal system notation. For example, valid equation which is consistent with the previous ones:

$ A1120 + A110 = 43215 $

Hint 2:

The solution is quite simple and strongly related to the collatz function.

Hint 3:

More valid equations:

$41B-C2 = 4511342241113111112121132111211212211112322212 $
$43810 - 1 =451134224111311111212113211121121221113522121111220 $
$43810 + 1 =432131412132140 $

Answer 1

It was my first time doing one of these puzzles so it took me quite a while but I think the answer is :

$6 + 6 = 7$

And here is how I arrived to this answer:

Numbers in the equations are constructed by counting the number of even steps between odd steps, in the Collatz evaluation of that number. See this answer for more info.


For example, the Collatz evaluation for $4$ is :

Step 1 $\to4 / 2 = 2$, Started with an EVEN number
Step 2 $\to4 / 2 = 1$, Started with an EVEN number

We have that $4$ terminates after $2$ EVEN steps.
Therefore, when in the equations we see a $2$, it corresponds to a $4$.


Let me explain how $12$ becomes $412$:

Step 1 $\to 12 / 2 = 6$ Started with an EVEN number
Step 2 $\to 6 / 2 = 3$ Started with an EVEN number
Step 3 $\to (3*3) + 1 = 10$ Started with an ODD number
Step 4 $\to 10 / 2 = 5$ Started with an EVEN number
Step 5 $\to (5*3) + 1 = 16$ Started with an ODD number
Step 6 $\to 16 / 2 = 8$ Started with an EVEN number
Step 7 $\to 8 / 2 = 4$ Started with an EVEN number
Step 8 $\to 4 / 2 = 2$ Started with an EVEN number
Step 9 $\to 2 / 2 = 1$ Started with an EVEN number

We use the ODD steps as separators and ignore them:
We have: $2$ EVEN steps + $1$ EVEN step + $4$ EVEN steps = $214$
And then reverse that number = $412$


With this, we can now translate the first equation: $2 + 3 = 412$
Which now actually means: $4 + 8 = 12$

And now we can apply the process in reverse to solve all the equations!

• $1 + 5 = 4321 \to 2 + 32 = 34$

• $0 + 4 = 4320 \to 1 + 16 = 17$

• $4320 - 412 = 40 \to 17- 12 = 5$

• $4321120 - 0 = 3 \to 9 - 1 = 8$

• $80 - 62 = 0 \to 85 - 84 = 1$

• $6 + 6 \to 64 + 64 = 128 → 7$

Answer 2

I haven't gotten the answer yet, but here are some things I found out about:

In response to the OP's comment:

Trivia reason ahead (spoiler/hint): Lothar and his famous unsolved conjecture are related to the construction (values) of numbers in the equations. -Vepir

Here is a little background info:

"Lothar and his famous unsolved conjecture" is the Collatz conjecture, named after Lothar Collatz. It states that given a starting number $n$, and the function $$f(n) = \begin{cases} n/2 & n\equiv 0\pmod2 \\ 3n+1 & n\equiv 1\pmod2 \end{cases}$$ the application of $f(n)$ will reach $1$ after finite operations. $$$$ In other words, consider the function where "if a number is even, divide it by two" and if a number is odd "multiply it by 3 and add 1." After applying this function multiple times, you will always end up with the number one.

Here's a link for some more explanations on this conjecture.

Answer 3

Not sure this brought us any closer to the solution, but here's a table of the numbers mentioned in the equations. $n_0$ is the starting numbers, $N_\text{steps}$ how many "Collatz-iterations" required to reach 1, and $N_\text{odd}$ is how many times an odd number is encountered before 1 is reached ($n_0$ included).

$$ \begin{matrix} n_0 & N_\text{steps} & N_\text{odd}\\ \hline 2 & 1 & 0\\ 3 & 7 & 2\\ 4 & 2 & 0\\ 5 & 5 & 1\\ 6 & 8 & 2\\ 40 & 8 & 1\\ 62 & 107 & 39\\ 80 & 9 & 1\\ 412 & 80 & 9\\ 4320 & 46 & 13\\ 4321 & 170 & 61\\ 4321120 & 92 & 27 \end{matrix} $$