There is number $1$ on the blackboard initially. The first step is to erase $1$ and write two nonnegative reals whose sum is $1$. Call the smaller number of the two $L_2$. For integer $k \ge 2$, the ${k}$ the step is to erase a number on the blackboard arbitrarily and write two nonnegative reals whose sum is the number erased just now. Call the smallest number of the $k+1$ on the blackboard $L_{k+1}$. Find the maximum of $L_2+L_3+\cdots+L_{2024}$.
Problem
Source: 2024 CTST P11
Tags: combinatorics
PRMOisTheHardestExam
12.03.2024 19:12
are numbers written after deleting different??? otherwise... L2 <= 0.5 L3 <= 0.25 L4 <= 0.125 ... ??
joeym2011
12.03.2024 19:25
The maximum sum is taken when all deleted numbers are split in two. We get
$$1\rightarrow\frac12,\frac12\rightarrow\frac14,\frac14,\frac12\rightarrow\frac14,\frac14,\frac14,\frac14\rightarrow\dots.$$We then get that if $2^{n-1}<x\leq2^n$ we get $L_x=\frac1{2^n}$. This means
$$L_{2^{n-1}+1}+L_{2^{n-1}+2}+\dots+L_{2^n}=\frac12,$$$$L_2+L_3+\dots+L_{2024}=\frac92+\frac{1000}{2048}=\boxed{\frac{1277}{256}}.$$
syano
13.03.2024 18:59
@above you are absolutely right. It is easy to find out that we can rearrange our steps to enbiggen our Ls. After splitting 1 into a and b, we use induncion to say that $S_k\le aS_r+bS_s+\min\{a,b\}$ and GAME OVER WE WIN. where $S_k=\sum^{k}{i=2}L_i$ and r and s is the number of steps based on the a branch and the b branch. Although this, but the calculation of the induction might be pretty boring and long. who can tell? for example. k=2024, so r=1024, s=1000 (NOT the only solution to the biggest S), and a=b=0.5.
syano
14.03.2024 09:39
whoa, wait a sec, it should be $\dfrac{10}{2}+\dfrac{1000}{2048}=\dfrac{1405}{256}$, but no big deal
ZussimanSupremacy
14.03.2024 13:10
Actually it is not as abstruse as I thought it would be