Question 44
Let $a_0$ be an arbitrary positive integer. Let $\{a_n\}$ be an infinite sequence of positive integers such that for every positive integer $n$ the term $a_n$ is the smallest positive integer such that $a_0 + a_1 + \dots + a_n$ is divisible by $n$. Prove that there is a positive integer $N$ such that $a_{n+1} = a_n$ for all $n \ge N$.
Let $a_0$ be an arbitrary positive integer. Consider the infinite sequence $(a_n)_{n \ge 1}$, defined inductively as follows: given $a_0, a_1, \dots, a_{n-1}$ define the term $a_n$ as the smallest positive integer such that $a_0 + a_1 + \dots + a_n$ is divisible by $n$. Prove that there exists a positive integer $M$ such that $a_{n+1} = a_n$ for all $n \ge M$.
Let $a_0$ be an arbitrary positive integer. Consider the infinite sequence $(a_n)_{n \ge 1}$, defined inductively as follows: given $a_0, a_1, \dots, a_{n-1}$ define the term $a_n$ as the smallest positive integer such that $a_0 + a_1 + \dots + a_n$ is divisible by $n$. Prove that there exists a positive integer $M$ such that $a_{n+1} = a_n$ for all $n \ge M$.