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For natural numbers $n$ and $k$, where $n \ge k$, let $n\underset{k \text{ times}}{\underbrace{!!!\dots!}}=n(n-k)(n-2k)\dots $ , where all the factors are positive integers (excluding zero and negative factors).

What is the greatest integer $n$ for which $n\underset{17 \text{ times}}{\underbrace{!!!\dots!}}<1025\underset{65 \text{ times}}{\underbrace{!!!\dots!}}$

The only idea I got is trial and error. To start by evaluating the right side of the inequality, which is approximately $9.27999 \times 10^{41}$. Then check for each of the following: $n=$RHS, $n(n-17)=$RHS, $n(n-17)(n-34)=$RHS, ... and so on, where $n=17,18,19,\dots$.

But this idea, in my opinion, is not feasible since we will have equations with orders higher than $4$ which have no solution in radicals to general polynomial equations with arbitrary coefficients (Abel–Ruffini theorem).

Can we solve this problem by the help of scientific calculators such as (CASIO fx-991es) but not by computers or websites?

I do not know how to start solving this problem. I am not asking you to solve it, I am asking for (help, hint, useful formulae) to be able to start. Thanks in advance!

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  • $\begingroup$ I don't like that notation. Why not $n!_k$? $\endgroup$
    – marty cohen
    Nov 26, 2019 at 19:40
  • $\begingroup$ @martycohen , Notations are nothing, definitions are everything. Since we define a thing using a notation, we can define the same thing with another notation. Thanks for sharing your opinion. $\endgroup$
    – Hussain-Alqatari
    Nov 26, 2019 at 19:45
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    $\begingroup$ Well, then why don't you use $n!_k := n(n-k)(n-2k) ....$ which is a gazillion times easier to type and to read and to parse? $\endgroup$
    – fleablood
    Nov 26, 2019 at 19:50
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    $\begingroup$ Well hunch: $17 = 2^4 +1$ and $1025 = 2^{10} + 1$ and $65=2^6 + 1$. Don't know if that will help. $\endgroup$
    – fleablood
    Nov 26, 2019 at 20:02
  • $\begingroup$ @fleablood , Good observation, I observed that too, but I just ignored it because as you, I do not know if it is helpful. However, I will give it a try. Thanks. $\endgroup$
    – Hussain-Alqatari
    Nov 26, 2019 at 20:08

1 Answer 1

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Assume $n=17k$: then $s_n=n!\ldots!=17^kk!$. If $k \leq 20$, $s_n \geq (1+1/16)^{20}20!16^{20}\geq 2 \cdot 2^{80}20! \geq 10^{24}10!\cdot 400000 \cdot (14 \cdot 15) \cdot (16 \cdot 17) \cdot (18 \cdot 19) \geq 12 \cdot 10^{35} \cdot 200 \cdot 200 \cdot 300 \geq 144 \cdot 10^{35+6} > RHS$.

Of course, the bound is really crude, but we thus know that we need to search for $n < 17 \cdot 20=340$.

On the other hand, if $n=17k$, $k = 19$, $s_n = 19 \cdot 17^{19}18! \leq 323 \cdot (18 \cdot 17)(15 \cdot 16) (13 \cdot 14)(11 \cdot 12)10! 17^{18}\leq 17^{18} \cdot 323 \cdot 310 \cdot 240 \cdot 190 \cdot 140 \leq 2^{18}10^{22}10! \cdot 24 \cdot 31 \cdot 266 \cdot 323 \leq 768 \cdot 323 \cdot 266 \cdot 2^{20}10^{28} \leq 10^{28}2^{20}66 \cdot 10^6=66 \cdot 1.05 \cdot 10^{28+6+6}<70 \cdot 10^{40} < RHS$.

So we know that you “just” have to look for $323 \leq n < 340$.

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