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Why does Kaggle use Log-loss?

Solving a Bar Bet Using at Most Three Different Operations

A bar bet as presented in the YouTube video The HARDEST Puzzle Yet! involves starting with three of the same number from 0 through 9, then adding mathematical operations that result in an evaluation of 6. For example, if we start with three of the number 6, one solution could be \[ 6+6-6=6 .\] I'll demonstrate a method for solving this bar bet puzzle starting with three of any number, say \( N \), which involves using at most three different mathematical operations (although some of these may be used many, many times).

If \( 0 \leq N \leq 2 \) we have the solutions
(0! + 0! + 0!)! &= 6 \\
(1! + 1! + 1!)! &= 6 \\
2+2+2 &= 6.
\end{eqnarray} If \( N\geq 3\), concatenate the three numbers. Repeatedly applying the square-root operation we'll eventually end up with a result \( x \) with \( 3 \leq x < 9\).  If we now take the greatest integer \(\lfloor x \rfloor\) we have an integer \( n \) with \( 3 \leq n \leq 8 \). If we can exhibit solutions for each of these cases that use only square-roots, greatest integers and one other operation, we'll be done. Using factorial for the third operation, some possibilities are
\href{}{3!} &= 6\\
\href{}{\left\lfloor \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{(4!)!}}}}} \right\rfloor!}} \right\rfloor !} &= 6\\
\href{}{\left\lfloor \sqrt{\sqrt{5!}} \right\rfloor !} &= 6\\
\href{}{6} &= 6 \\
\href{}{\left\lfloor \sqrt{\sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{7!}}\right\rfloor !}}} \right\rfloor !} &= 6\\
\href{}{\left\lfloor \sqrt{\sqrt{\sqrt{8!}}} \right\rfloor !} &= 6.
I've added Wolfram Alpha links so you can verify that these do indeed evaluate to 6.

As an illustrative example, when \(N=1337\) we have the solution \[
\href{}{\left\lfloor \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\left(  \left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{133713371337}}}} \right\rfloor !\right)!}}}}} \right\rfloor!}} \right\rfloor !} = 6.\]


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