# $n\underset{17 \text{ times}}{\underbrace{!!!\dots!}}<1025\underset{65 \text{ times}}{\underbrace{!!!\dots!}}$

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!

• I don't like that notation. Why not $n!_k$? – marty cohen Nov 26 '19 at 19:40
• @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. – Hussain-Alqatari Nov 26 '19 at 19:45
• 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? – fleablood Nov 26 '19 at 19:50
• Well hunch: $17 = 2^4 +1$ and $1025 = 2^{10} + 1$ and $65=2^6 + 1$. Don't know if that will help. – fleablood Nov 26 '19 at 20:02
• @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. – Hussain-Alqatari Nov 26 '19 at 20:08

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 $$323 \leq n < 340$$.