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Prove that there exists a natural number $n$ that has more than $2017$ divisors $d$ satisfying $$\tag 1 \sqrt n \le d < 1.01\sqrt n$$

My reasoning was that $\tag 2 1.01\sqrt n-\sqrt n\ge2019$ must occur. Otherwise, there won't be enough space to fit in enough numbers in $(1)$. $(2)$ gives us that $n\geq 201900^2$. We also have that $201900^2<201900!$ so we can pick any factorial greater than or equal to $201900!$ and that satisfies the desired condition. Is this solution correct?

Jyrki Lahtonen
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somerndguy
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    A problem in your attempt: When you replaced $n=201900^2$ with $n=201900!$ you also changed the interval $[\sqrt n,,1{,}01\sqrt n]$. Why would there be enough divisors of $201900!$ in there? – Jyrki Lahtonen Sep 13 '21 at 02:52
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    Typesetting tip: If you use a comma , as a decimal separator, please put it inside curly braces. Like $1{,}01$ instead of $1,01$. The reason is that MathJax (and most of the English speaking world) treats a comma as a list separator. Therefore a tiny amount of whitespace is added after a comma. The curly braces remove that. – Jyrki Lahtonen Sep 13 '21 at 03:33
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    Having enough space for that many divisors to fit isn't the same as actually having divisors in that space. – Greg Martin Sep 13 '21 at 03:36
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    My quick thoughts, which may not pan out. Can you find such a $d$ with two divisors in the interval? Can you find a number of such $d$s, each with two divisors in its interval? Call them $d_1, d_2, \ldots$ Can you say something about $d_1d_2?$ How about $d_1d_2d_3?$ Maybe you need to make the intervals smaller for the original $d_i$. It wouldn't seem there is anything special about $1.01$ except that it is slightly greater than $1$ – Ross Millikan Sep 13 '21 at 04:58
  • My reasoning that in the interval $[\sqrt n,,1{,}01\sqrt n]$ will be enough divisors for $n\ge201900!$ is that $1,01\sqrt{201900^2}-\sqrt{201900^2} < 1,01\sqrt{201900!}-\sqrt{201900!}$ so the gap between $[\sqrt {201900!},,1{,}01\sqrt {201900!}]$ is large enough to fit in atleast 2018>2017 numbers also every integer in $[1,n]$ dvides $n!$ and for $n\ge2$, $1.01\sqrt{n} < n$. – somerndguy Sep 13 '21 at 10:26
  • Your solution is most likely wrong. It is true that for $ n = 201900!$, we have A) Every integer from 1 to 201900 divides $n$ and B) The gap between $[\sqrt{n}, 1.10\sqrt{n} ]$ has at least $2019$ numbers. However, what fails is that not all of the $2019$ numbers in this gap must be divisors of $n$. EG Some of them could be prime numbers. What is true (which is likely what you're thinking) is that for any $ k < 2019$, there is some number in the gap that is a multiple of $k$. – Calvin Lin Sep 13 '21 at 18:56
  • A problem with your thinking is that none of the integers $1,2,\ldots,201900$ are in the interval $[\sqrt{201900!},\ \dfrac{101}{100}\sqrt{201900!}]$. They are way too small. – Jyrki Lahtonen Sep 13 '21 at 19:16
  • I see the flaw. I'll try to work it out thank you. – somerndguy Sep 13 '21 at 19:29
  • This was posted long ago in this forum. https://math.stackexchange.com/questions/1940584/divisors-problem/2011289#2011289 – Sungjin Kim Sep 14 '21 at 23:44

3 Answers3

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Let $$ n=(10^6-1)^{4032}(10^6+1)^{4032}. $$ Then the numbers $$ s_j=(10^6-1)^j(10^6+1)^{4032-j}, $$ $j=0,1,2,\ldots,2016$, are clearly all factors of $n$. We also have $$s_0>s_1>\cdots >s_{2016}=\sqrt n,$$ so they are distinct and satisfy the lower bound. On the other hand $$ \frac{s_0}{s_{2016}}= \frac{(10^6+1)^{2016}}{(10^6-1)^{2016}}<1.01\qquad(*) $$ so we have $s_j<1.01\sqrt n$ for all $j$ also.

I checked $(*)$ with Mathematica, but elementary estimates will also get the job done. At least if we replace $10^6$ by something bigger.

Jyrki Lahtonen
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  • Use $1+\varepsilon<e^{\varepsilon}$ to prove $(*)$ or its suitable version. – Jyrki Lahtonen Sep 13 '21 at 21:04
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    Nice solution, thank you, could you please tell me how have you come up with a solution? I tried to solve it on my own but I wasn't able find a useful observation. – somerndguy Sep 13 '21 at 21:35
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    Finding the solution was a process, all right. I do have a bit of experience in contest math, but it wasn't obvious how to go about this. A light bulb moment was to realize that if $n$ is product of an even number of roughly equal size factors, then by taking one half of those factors we get a large number of divisors of the size $\approx\sqrt n$. If the factors are really close to each other, so are the respective products, and we should be able to estimate the products and get an upper bound of $1.01\sqrt n$. – Jyrki Lahtonen Sep 14 '21 at 03:47
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    (cont'd) One problem is getting also the lowe bound. Ok, just toss out those products that are too small. My first attempt used products like $$n=(10^k-m)(10^k-m+1)\cdots (10^k-1)(10^k+1)(10^k+2)\cdots(10^k+m)$$ for some $k$ and $m$. If we choose, say, three "small" factors $(10^k-1)(10^k-2)(10^k-3)$ together with $(m-3)$ "large" factors, the resulting $\binom m 3$ products of $m$ numbers will be in the range. However, it is not obvious that those products are all distinct. – Jyrki Lahtonen Sep 14 '21 at 03:52
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    (cont'd) If I knew enough about the distribution of primes, I could prove the existence of a large enough number of prime numbers in a similar "tight" interval, but I don't (I still think it can be done!). That would have solved the problem of the factors being distinct by the uniqueness of prime factorization. Then the final light bulb of using an even larger of number repeated factors was a relatively natural step. – Jyrki Lahtonen Sep 14 '21 at 03:55
  • Everything seems clear, thank you for explanation – somerndguy Sep 14 '21 at 09:49
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    Strengethened Bertrand postulate states that for any $r > 1$, there is some $N$, such that for any $ n > N$, there exists a prime between $n$ and $rn$. EG For $r = 1.5, N = 12$. This is how we can force as many primes as we want between $n$ and $1.01 n$. – Calvin Lin Sep 14 '21 at 16:53
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Generally speaking, I expect that such a contest problem can be done via induction, and we just need to push through.

Define a divisor that satisfies those conditions to be "valid".

Base case: $D = 1$, $n_1=1$ suffices.

Induction step: Suppose it is true for some $D$ with corresponding $n_D$.
Consider $n_{D+1} = p^2 n_{D}$ where $p$ is to be determined below.
Then, for any valid divisor $d_D$ of $n_D$, we have that the corresponding $ pd_D$ satisfies $pd_D \mid n_{D+1}$ and $\sqrt{n_{D+1}} \leq pd_{D} < 1.01 \sqrt{n_{D+1}}$. Hence this gives us (at least) $ D$ valid divisors.

How can we force another distinct valid divisor?
If we have an integer $p \in [ \sqrt{n_D}, 1.01 \sqrt{n_D}) $ that is not a divisor of $n_D$, then we could use $ n_{D+1} = p n_D $.
If not, look at the interval $ [ \sqrt{ 4^k n_D } , 1.01 \sqrt{4^k n_D})$, and show that we will eventually find such a $p$ that is not a divisor of $4^k n_D$. (Try proving without any high power theorem). Then, we can use $n_{D+1} = (2^k p)^2 n_D$.

Calvin Lin
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  • Simple! I haven't thought it through, but I think your method produces a much smaller $n_D$ than the obvious generalization of my approach. – Jyrki Lahtonen Sep 14 '21 at 17:55
  • @JyrkiLahtonen Nah, yours is definitely smaller. Each time I'm (at least) multiplying by $p^2 \approx n_D$ which means i'm squaring the term, while yours is just mulyiplying by a constant $(10^{24} - 1)$. – Calvin Lin Sep 14 '21 at 18:04
  • Oopsie. You're right. Mind you, at some point I need to start using larger numbers instead of $10^6$. This happens approximately when $(1+2\cdot10^{-6})^D$ exceeds $1.01$. – Jyrki Lahtonen Sep 14 '21 at 18:34
  • Hi Calvin, would you be so kind to check my solution here : https://math.stackexchange.com/questions/3033093/coloring-grid-points-with-two-colors/4251342#4251342 – nonuser Sep 16 '21 at 16:35
  • A very unexpected use of induction, I would have never thought of that. I've learnt alot, thank you. – somerndguy Sep 16 '21 at 23:44
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As an alternative to the other excellent answers, let's use the prime number theorem and apply it to ideas similar to those of @JyrkiLahtonen. The only thing we need it for actually is that it assures we can find prime numbers

$$p_1 < p_2 < \cdots p_k < p_{k + 1} < \cdots p_{2k}$$

with

$$\left(\frac{p_{2k}}{p_1}\right)^{k} \le 1.01$$

since it says that the $m$-th prime is of size $P(m)\sim m\log m$.

Then $n = p_1\cdots p_{2k}$ will do for sufficiently large $k$:

  • It has ${2k}\choose{k}$ divisors composed of exactly $k$ primes.

  • Exactly half of them will be greater than $\sqrt n$

  • Let $d$ be such a divisor among the greater 50%. Then

    $$\sqrt n < d < p_{2k}^k < 1.01p_1^k < 1.01\sqrt n.$$

This proves the existence of $n$.

Now let's continue to find an explicit such $n$.

The only thing we need to do is find $k$, and $p_1,\ldots,p_{2k}$.

  • We have ${{2k}\choose{k}} > 2\times 2017 = 4034$ for $k = 8$, so $k = 8$ will do.

  • We want $p_{16}/p_1 \approx \sqrt[8]{1.01}$.

  • To know where to look for such primes, note that the density of primes around $x$ is approximately $1/\log x$. We have 16 primes in the interval from $p_1$ to $p_{16} < \sqrt[8]{1.01}p_1$, so we are looking for $p_1$ around which the density of primes is something like

    $$\frac{16}{\left(\sqrt[8]{1.01} - 1\right)p_1} \sim \frac1{\log p_1}.$$

We can solve for $p_1$ as

$$p_1 = e^{-W_{-1}\left(-\frac{\sqrt[8]{1.01} - 1}{16}\right)} \approx 153520,$$

where $W$ is the Lambert $W$ function (a more approximate solution would have been fine).

The first 16 primes after 153520 satisfy $(p_{16}/p_1)^8 = 1.0111\ldots$, not quite smaller than $1.01$ but almost, and indeed their product, $n = 96114801918559792121496664418537548970202278805505371719773949381515854089880827473$ works, and has $\frac12{{16}\choose8} = 6435$ divisors between $\sqrt n$ and $1.01\sqrt n$.

Note that by just trying rather than estimating, you can find smaller values, e.g. the product of the first 16 primes after 3300 give $n = 248739277096018887021673215137171457841909984636041714233$, which has 2455 divisors between $\sqrt n$ and $1.01\sqrt n$.

doetoe
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  • Nice to see this idea developed! I had a momentary blind spot about why would have of the products be $\ge\sqrt n$ if, say, there is a "relatively" large prime gap near one of the ends. But, the resulting factors form complementary pairs: $d$ and $n/d$, of which surely exactly one is $>\sqrt n$ :-) Also, I thought I would need estimates more accurate than what the prime number theorem gives. You side-stepped that with numerical work. – Jyrki Lahtonen Sep 16 '21 at 03:32
  • Thanks @JyrkiLahtonen! I think that the numerical part is non-essential and that the estimates of the prime number theorem are enough (so that 2017 could be replaced by $N$ and 1.01 by $1 + \varepsilon$): – doetoe Sep 16 '21 at 06:17
  • We have $P(n) \sim n\log n$, and for any fixed $k$ we have $\lim_{n\to\infty}\frac{n\log n}{(n + k)\log(n + k)} = 1$, so also $\lim_{n\to\infty}\frac{P(n + k)}{P(n)} = 1$, so certainly there will be an $n$ so that $\frac{P(n + 2k - 1)}{P(n)} < \sqrt[k]{1 + \varepsilon}$. – doetoe Sep 16 '21 at 06:17
  • Hmm. Armed with only the version $\lim_{n\to\infty}\dfrac{\pi(n)\log n}n=1$ I don't see why it would follow that $\lim_{n\to\infty}\dfrac{P(n+k)}{P(n)}=1$? – Jyrki Lahtonen Sep 16 '21 at 13:24
  • The size $P(n)$ of the $n$-th prime satisfying $\lim_{n\to\infty}\frac{P(n)}{n\log n} \to 1$ is equivalent to $\pi(x) \sim \frac x{\log x}$, $P$ is essentially the inverse of $\pi$ (the actual inverse is the one involving $W_{-1}$ used to compute $p_1$ above, but it is not hard to see that asympotically this is the inverse, just by substituting one in the other). I think that it can then be concluded from $\lim_{n\to\infty}\frac{P(n)}{n\log n} = 1$ and likewise with $P(n+k)$, and the fact that $\lim_{n\to\infty}\frac{n\log n}{(n + k)\log(n + k)} = 1$. – doetoe Sep 16 '21 at 15:50
  • Here we have $\lim_{x\to\infty}\frac{f_1(x)}{g_1(x)} = 1$, $\lim_{x\to\infty}\frac{f_2(x)}{g_2(x)} = 1$ and $\lim_{x\to\infty}\frac{f_1(x)}{f_2(x)} = 1$, and the conclusion is that $\lim_{x\to\infty}\frac{g_1(x)}{g_2(x)} = 1$. – doetoe Sep 16 '21 at 15:51
  • (in my penultimate comment I meant to say that the actual inverse of $x/\log x$ is $e^{-W_{-1}(y)}$, not the inverse of $\pi$ which really is $P$, at least on positive integers.) – doetoe Sep 16 '21 at 16:06