Track / Stage 6 / 364 of 400 #1364 of 1964

Problem 1364

National olympiad, first round

Number theory Difficulty 6.8 Prove it

For which positive integers $n(>2)$ , do there exist $n$ consecutive positive integers such that the largest number is a divisor of the least common multiple of the other $n-1$ numbers?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

When $n=3$ , for any three consecutive positive integers $a-1, a, a+1$ , if

then
and
so
$\begin{array}{l} a+1 \mid [a-1, a] \\ a+1 \mid a(a-1) \\ (a+1, a)=1 \\ a+1 \mid a-1 \end{array}$

This is a contradiction.
When $n>3$ , if $n$ is even, let $n=2m$ , then among the numbers $2m-1, 2m, \cdots, 2(2m-1)$ , the largest number $2(2m-1)$ is a divisor of the least common multiple of the other $2m-1$ numbers (which include $2m-1$ and $2m$ ); if $n$ is odd, let $n=2m+1$ , then the numbers $2m-2, 2m-1, \cdots, 2(2m-1)$ are $n$ consecutive positive integers (note that $m>1$ is used here), and the largest number among them is a divisor of the least common multiple of the other $n-1$ numbers.

Therefore, when $n>3$ , the positive integer $n$ meets the condition.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.