There are $n$ children in a room. Each child has at least one piece of candy. In Round $1$, Round $2$, etc., additional pieces of candy are distributed among the children according to the following rule:
In Round $k$, each child whose number of pieces of candy is relatively prime to $k$ receives an additional piece.
Show that after a sufficient number of rounds the children in the room have at most two different numbers of pieces of candy.
(Proposed by Theresia Eisenkölbl)
Claim: After a sufficient number of rounds eventually each child will have either (current round number $-1$) candies or (current round number $+1$) candies.
Proof: We take the set of kids that initially have more than $1$ piece and denote this set by $A$. If after some rounds the difference between a kid from $A$ and the current day number is $x \ge 2$, then if the current day number is divisible by $x$ these numbers will be both divisible by $x$ and, $x$ will decrease until $x=1$, and after that $x$ will not decrease anymore, and after a certain amount of days the difference between number of candies possessed by each kid from $A$ and the current day number will be $1$.
Now take set $B$ in which we include all children that have initially only one candy. After one round they wouldn't get any candy and after each next round they will get $1$ candy so the difference between the current day and their number of candies will be $1$
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