Problem 331

Given nonnegative integers $a_1,a_2,\dots, a_{1997}$ satisfying the given inequalities, let $I_n$ be the set of all $x$ such that $a_n=\lfloor nx\rfloor$. Therefore,
$$\text{I n= x ,: ,a n nxk , then by the division algorithm, we can write n=qk+r for nonnegative integers q and r with 0 r ra k . Substituting this into (3), we find [k(a n+1)> kqa k+ra k=(kq+r)a k=na k.}$$
This proves the hypothesis for $n=j>k$.
Suppose that $k=j>n$. Then by the division algorithm, we can write $k=qn+r$ for nonnegative integers $q$ and $r$ with $0\le rn$.
Therefore, if the hypothesis is true for all $k,n<j$, then it must be true for all $k,n\le j$. Hence by induction, it must be true for all positive integers $k$ and $n$. $\text{}$
Now suppose for the sake of contradiction that $I_n$ and $I_k$ are disjoint intervals. Without loss of generality, we may assume that $I_n$ precedes $I_k$ on the number line. Hence the upper bound of $I_n$ is less than or equal to the minimum value of $I_k$, or rather
$$\frac{a_{n}+1}{n}\le \frac{a_k}{k}.$$
This simplifies to $na_k\ge k(a_n+1)$. But this contradicts the statement of Lemma 1. Therefore, $I_k\cap I_n\ne \emptyset$.
Now we claim that $I_1\cap I_2\cap \cdots \cap I_{1997}\ne \emptyset$. We prove this by induction. By the above, we know that $I_1\cap I_2\ne \emptyset$. Now suppose that $I_1\cap I_2\cap\cdots\cap I_k\ne \emptyset$. The intersection of two overlapping intervals of the form $[a,b)$ and $[c,d)$ is an interval of the form $[e,f)$, where $e=\max\{a,c\}$ and $f=\min\{b,d\}$. Therefore, by induction, we know that if the intersection of $k$ overlapping intervals is nonempty, then it must also be an interval, say
$$I_1\cap I_2\cap\cdots\cap I_k=[x_k,y_k).$$
If $I_{k+1}$ does not intersect $[x_k,y_k)$, then as an interval, it must appear either completely before $x_k$ or completely after $y_k$. If $I_{k+1}$ appears completely before $x_k$, then it has a nonempty intersection with each of $I_1, I_2, \dots, I_k$. But we also know that $x_k$ is a lower bound of one of the intervals, hence $I_{k+1}$ cannot intersect that interval, a contradiction. A similar contradiction arises if $I_{k+1}$ appears completely after $y_k$. Therefore,
$$I_1\cap I_2\cap\cdots\cap I_{k+1}\ne \emptyset.$$
Thus by induction, $I_1\cap I_2\cap\cdots\cap I_{1997}\ne \emptyset$. So let $x\in I_1\cap I_2\cap\cdots\cap I_{1997}$. Then by the definition of $I_n$, we know that $a_n=\lfloor nx\rfloor$ for all $1\le n\le 1997$, and we are done.