Choosing $100$ numbers in which one of the chosen number is divisible by another one

Question

Prove that if $100$ integers are chosen from $1,2, \ldots, 200$, and one of the integers chosen is less than $15$, then there are two chosen numbers such that one of them is divisible by the other.

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It is a classic exercise in combinatorics. For the answer, please refer to Choose 100 numbers from 1~200 (one less than 16) - prove one is divisible by another!. I want to move forward and ask something more. First, let me rephrase the question:

$100$ integers are chosen from $1,2, \ldots, 200$. Let $n$ be the smallest number of these chosen numbers. Prove that if $n=1,2,\ldots, 15$, then two chosen numbers can be found such that one of them is divisible by the other.

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Now my question is:

Find all possible $n$.

This question comes to my mind when I first encounter this exercise. I was curious that why we should have the condition 'less than 16'. What if I change it to 'less than 17'? Does it mean that if $n=16$, it would be possible for me to find a counterexample? Of course, $n$ cannot be too large, as if $n=101$, then the chosen numbers would be $101, 102, \ldots, 200$ and none of them is a multiple of the others. So, I get the sense that some values of $n$ work, while some don't. So I post here to gather your opinions.

By the way, I cannot prove or disprove the statement when $n=16$. I think the idea to prove or disprove this case would help understand more about my ultimate questions.

$100$ integers are chosen from $1,2, \ldots, 200$. Let $n$ be the smallest number of these chosen numbers.

Prove or disprove: If $n=16$, then two chosen numbers can be found such that one of them is divisible by the other.

My attempt to construct a counterexample:

• I noticed that no two numbers in the set $\{101, 102, \ldots, 200\}$ divides each other. So, I started with this set and added in $16$.
• Then I had to remove $112, 128, 144, 160, 176, 192$.
• I needed to add in some new numbers, but I then would need to, again, remove many more existing numbers. So I think it is getting nowhere....
• Think in another direction: Start with an empty set. Add in the smallest number $16$. Then all the multiple of $16$ cannot be in the set.
• Add in another number, say $17, 18, \ldots, 31$. Wait... Then I cannot add in anymore numbers...

I welcome any comments and suggestions.

Answer 1

There is a counterexample for the case where the condition is a number chosen is less than or equal to $16$. It is built on the ideas found in the question link and another similar question:

• Choose 100 numbers from 1~200 (one less than 16) - prove one is divisible by another!
• Each number in a subset $S\subseteq \{1,\ldots,2n\}$ does not divide another one. Then $\max |S|$?

Prove that at most $100$ numbers can be selected

Form descending 2-sequences (these are the pigeonholes) starting with each term in $T=\{101,102,\ldots,199,200\}$, where each sequence is finite and ends on an odd term and the term in any sequence is the previous term divided by $2$. Denote each sequence by its starting term (note that $n=100$), e.g. $$\begin{array}{rll} S_{2n}&=(2n,n,\ldots)&=(200,100,50,25) \\ S_{2n-1}&=(2n-1) &=(199)\\ &\quad\vdots \\ S_{n+2}&=\{n+2,\ldots\}&=(102,51) \\ S_{n+1}&=\{n+1,\ldots\}&=(101) \end{array}$$

Every number in $\{1,2,\ldots,2n\}$ appears in exactly one of the $S_{k}$ because:

1. Surjective. Every number inclusively between $1$ and $n$ has a multiple of a power of two inclusively between $n+1$ and $2n$. If not, $2 \ge \frac{2n+1}{n}$, which is absurd.
2. Injective. No number can appear in more than one sequence. If this was untrue we could find two distinct terms in $\{n+1,n+2,\ldots,2n\}$, one of which is a multiple (of a power of two) of the other.

It is clear that every pair of terms in a given $S_{k}$ has the smaller term dividing the larger one, so we can choose at most one term from each $S_{k}$.
To construct a counterexample, we need to choose exactly one from each, and ensure that $16$ is included.

Sketch of the counteraxample P

When $n$ is odd, $S_n={n}$ has only one element, so this must be chosen. Hence, we need that $P$ contains every odd integer between $100$ and $200$, i.e. $$P_1=\{101,103,105,\ldots,199\}$$

This eliminates all submultiples so for the elements that cannot be chosen: $$Q_1=\{1,3,5,\ldots,65\}$$

For each odd number $n$ in $\{67,69,\ldots,99\}$ the relevant sequence is $S_n=(2n,n)$. It is less restrictive to choose $n$ than $2n$, since all further multiples of $n$ exceed $200$, and the submultiples of $n$ are a subset of those of 2n$. So choose

$$P_2=\{67,69,71,\ldots,99\}$$

This choice eliminates all numbers that are twice these:

$$Q_2=\{134,138,142,\ldots,198\}$$

Then for every large enough $n$ that is a multiple of $3$ in $P_1$ (and $4n>200$), the number $2n$ needs to be chosen (since $n,4n$ and higher multiples are excluded). So then

$$P_3=\{102,106,110,\ldots,130\}$$

The eliminated submultiples are:

$$Q_3=\{2,6,10,14,18,22,26,34,38,42\}$$

Now consider all even numbers at least $100$ that are not in $Q_2$ or $P_3$, a typical case is $196$ for which $S_{196}=(196,98,49)$ i.e. with an odd third number that is in $Q_1$. By an argument similar to that used before, it is preferable to choose the smaller of the two even numbers. This gives

$$P_4=\{50,54,58,62,66,68,70,74,76,78,82,84,86,90,94,98\}$$

which excludes the further multiples

$$Q_4=\{100,108,116,124,132,136,140,148,152,156,164,168,172,180,188,196,200\}$$

and additionally

$$Q_5=\{4,28,30\}$$

We are left with the disjoint 2-sequences (where entries that cannot be selected are removed), where one number must be chosen from each to make up the full quota of $100$:

$$\begin{array}{cllll} S_{128}'&=(128,64,32,16,8) & & S_{192}'&=(192,96,48,24,12) \\ S_{160}'&=(160,80,40,20) & & S_{144}'&=(144,72,36) \\ S_{184}'&=(184,92,46) & & S_{176}'&=(176,88,44) \\ S_{120}'&=(120,60) & & S_{112}'&=(112,56) \\ S_{104}'&=(104,52) \\ \end{array}$$

In addition, the numbers $8,12,20$ cannot be selected without preventing any selection from another 2-sequence. Hence

$$\begin{array}{clcllcll} S_{128}''&=(128,64,32,16) & & S_{192}''&=(192,96,48,24) & & S_{160}''&=(160,80,40) \end{array}$$

So it turns out that the following set is selectable as there are no pairwise multiples:

$$P_5=\{16,24,36,40,44,46,52,56,60\}$$

Resulting counterexample

Putting everything together a possible counterexample is:

$\begin{array}{ll} P&=P_1\cup P_2\cup P_3\cup P_4\cup P_5 \\ &=\{67,69,71,\ldots,199,\color{blue}{16},24,36,40,44,46,\\ &\quad 50,52,\ldots,62,66,68,70,74,76,78,82,84,86,90,94,98, \\ &\quad 102,106,110,\ldots,130 \} \end{array}$

$P$ contains $67$ odd numbers and $33$ even numbers, with the odd numbers being on average larger than the even numbers. This should not be major surprise.