Denote $33!$ by $n$. It's easy to show that after the $i$th move (when $i\leqslant 2^n-1$), the numbers of balls from box $1$ to $n$ are $a_1,a_2,\dotsc ,a_n$ that satisfy $$i=\overline{a_na_{n-1}\dotsc a_1}_2\text{ and }a_1,a_2,\dotsc ,a_n\in \{ 0,1\}.$$So, the answer is box number $\nu_2(33!)+1=32$, with one ball in it.

You're wrong answer is 36

ThE-dArK-lOrD wrote: Denote $33!$ by $n$. It's easy to show that after the $i$th move (when $i\leqslant 2^n-1$), the numbers of balls from box $1$ to $n$ are $a_1,a_2,\dotsc ,a_n$ that satisfy $$i=\overline{a_na_{n-1}\dotsc a_1}_2\text{ and }a_1,a_2,\dotsc ,a_n\in \{ 0,1\}.$$So, the answer is box number $\nu_2(33!)+1=32$, with one ball in it. Check with smaller numbers and you will see you are mistaken

inmemories wrote: ThE-dArK-lOrD wrote: Denote $33!$ by $n$. It's easy to show that after the $i$th move (when $i\leqslant 2^n-1$), the numbers of balls from box $1$ to $n$ are $a_1,a_2,\dotsc ,a_n$ that satisfy $$i=\overline{a_na_{n-1}\dotsc a_1}_2\text{ and }a_1,a_2,\dotsc ,a_n\in \{ 0,1\}.$$So, the answer is box number $\nu_2(33!)+1=32$, with one ball in it. Check with smaller numbers and you will see you are mistaken İn memories you 're absolutely right bro

Kyolcu22 wrote: inmemories wrote: ThE-dArK-lOrD wrote: Denote $33!$ by $n$. It's easy to show that after the $i$th move (when $i\leqslant 2^n-1$), the numbers of balls from box $1$ to $n$ are $a_1,a_2,\dotsc ,a_n$ that satisfy $$i=\overline{a_na_{n-1}\dotsc a_1}_2\text{ and }a_1,a_2,\dotsc ,a_n\in \{ 0,1\}.$$So, the answer is box number $\nu_2(33!)+1=32$, with one ball in it. Check with smaller numbers and you will see you are mistaken İn memories you 're absolutely right bro You're also absolutely right dude. Cheers.

I get 36, because the period of the kth box is lcm(1,2,...k+1) but I can't find how many balls are there in the 36th box.

Answer: The box labeled with $36$ with $31$ balls in it. Claim.1: When $n-th$ box is filled for the first time, exactly $n$ balls will be placed in it. And after any moves the number of balls in n-th box cannot be greater than n. Proof: It's easy to see when n-th box filled for the first time, all other smaller numbered boxes must contain at least one ball each. And in any move if n-th box isnt empty we cannot increase the number of balls in it. Thus we can easily reach claim.1 . Claim.2: After any move, the total number of balls in boxes with numbers greater than n and n-th box is divisible by n. (For example, after any move the sum of the ball numbers in the $7$th to $33!$-rd box is divisible by $7$.) Proof: If that sum has changed after our move that means we have filled some box with labeled some number which is greater than n. So we increased the sum by exactly $(n-1)+1=n.$ The sum remains $0$ modulo n. Now we are ready to finish. By using $claim.2$ it is easy to show the period of first $35$ box getting empty is $lcm(1,2,...,36)$ which is a divisor of $33!$ but the period of first $36$ box is $lcm(1,2,...,37)$ which is not a divisor of $33!$ $\implies 36-th$ box gonna be the first non-empty box after $33!$ moves. By wilsons theorem, we know that $33!=\frac{1}{6} =31 (mod 37)$. That implies the number of the balls in $36-th$ box must be $31$ by claim.1.