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I came across this problem:

Suppose we have $$ natural numbers, none of which are greater than 2, such that the least common multiple of any two is greater than 2 . Show that all numbers are greater than 2/3 .

I'm trying to start off with contradiction--that there is some $k_i < 2n/3$. What to do after simply stating this is not at all clear to me. I'm trying to see if there's any $\gcd$ ideas I can incorporate here--we have $\gcd(k_i, k_j) \text{lcm}(k_i, k_j) = k_i k_j$, so $\text{lcm}(k_i, k_j) = \frac{k_i k_j}{\gcd(k_i, k_j)} > 2n$. Since $k_i < \frac{2n}{3}$ we must have $\frac{2n}{3} \frac{k_j}{\gcd(k_i, k_j)} > 2n \implies k_j > 3 \gcd(k_i, k_j)$. This at least narrows what our numbers should be a little but it is not at all clear how to proceed with the problem. Any tips would be appreciated!

Thank you!

mathwarrior
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1 Answers1

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I know this won't pass the criteria of a formal proof but can perhaps help some understand intuitively why the claim in the question is true.

Let's start by taking a close look at numbers in the interval $(2n/3, 2n]$. This range has a size of $4n/3$.

But for the LCM of any two numbers to be greater than $2n$, we must discard $1/3$ of the interval mentioned above. Here's why.

For any number $a \in (2n/3, n]$, its double, $2a$, will lie in the interval $(4n/3, 2n]$. Also, the LCM of $a$ and $2a$ would be $2a$. And that would violate the condition. So, for each number we, pick in the first interval, we must discard another in the second interval.

That means, in the interval $(2n/3, 2n]$, there are exactly $n$ numbers that can make it into our group.

Now, it can be shown that for any number in the interval $[1, 2n/3]$, at least two of its multiples would lie in the interval $(n, 2n]$. So, again, for every number from the first interval we bring into our group, we will lose at least two numbers in the second.

But we can't afford that. We have exactly $n$ permissible numbers in the higher interval to begin with.

I would greatly appreciate if anyone can point out a mistake I am making here. Would love to make a correction or remove the answer, as needed.

Haris
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  • Seems like your proof is good. Let "Erdos set" be defined here as a set in ${1,\ldots, 2n}$ s.t. the LCM of any 2 elements is $>2n$. You show Lemma 1: that the maximal size of an Erdos set in $(2n/3, 2n]$ is $n$. You show Lemma 2: if an Erdos set $E$ has an element $\leq 2n/3$, then you can replace it by 2 of its multiples in $(n,2n]$ (which can not have been in $E$ previously), remaining an Erdos set (replacing by larger multiples can only possibly increase LCM). Lemma 1 and Lemma 2 together show that any Erdos set containing elements $\leq 2n/3$ must have size $<n$. – D.R. Mar 19 '25 at 19:42