If $n\times\lfloor\dfrac{n}{2}\rfloor\times\lfloor\dfrac{n}{3}\rfloor\times\cdots\times\lfloor\dfrac{n}{x}\rfloor=$ 425193449651032277806171880871000564797302545128630885065227836825878902928834560000000000000, where $x=\lfloor\sqrt{n}\rfloor$ and $\lfloor\dfrac{n}{x}\rfloor$ mean floor function, find the value of n.
No calculators allowed. I made this question.
For the sake of concision, we will let $N = $
425193449651032277806171880871000564797302545128630885065227836825878902928834560000000000000
and define $f(n,k) = \lfloor \frac{n}{k} \rfloor$ and $F(n) = \prod_{x=1}^{\lfloor \sqrt n \rfloor} f(n,x)$ (the original function).
We can take a reasonable guess at what $n$ should be by noting that this number has 93 digits. We can assume an upper bound of $n<3600$ as $3600^{60} \gg N$ and $60^{60} = 10^{60} \cdot 6^{60} = 10^{60} \cdot (6^4)^{15} > 10^{60} \cdot (10^3)^{15} = 10^{105} > N$.
We then assume a conservative lower bound of $n>1000$ as $1000^{30}<N$ and $30^{30} \ll 1000^{30}<N$.
We now know that $31 < \lfloor \sqrt{n} \rfloor < 60$ and $1000<n<3600$.
We now refine our bounds by refining our estimation function from simply $n^{\lfloor \sqrt{n} \rfloor}$ and $\left(\lfloor \sqrt{n} \rfloor\right)^{\lfloor \sqrt{n} \rfloor}$ to one that more closely matches the real function.
For $n=2000$, we break it up into $2000\cdot 1000$ for the first two terms. The terms up until $k=20$ will have three digits. Assume a conservative estimate of $200$ for the quotients of these 18 terms. All other terms are below 100, going down to $k=44$ which results in a value of 45. We approximate these 24 terms all having a value of 50. The estimate now becomes $2\cdot 10^6 \cdot 200^{18} \cdot 50^{24} = 10^{42}\cdot 2^{19} \cdot 50^{24} = 100^{19} \cdot 10^{42} \cdot 50^5 = 3.125\cdot10^{88} < N$.
Our lower bound is now $2000$ (the actual value of $F(2000)$ should be greater than the estimate of $F(2000)$). Because our estimate is relatively close to the relatively actual value of $N$, $n$ should be close to 2000. At $n=2025$, a 45th term is added to the product, effectively increasing the product by roughly 2 digits (and the estimate by the same factor).
Function Definition
We note that we can approximate $F(n)$ as $G(n) = n \cdot \frac{n}{2} \cdot \frac{n}{3} \cdots \frac{n}{{\lfloor \sqrt{n} \rfloor}} = \frac{n^{\lfloor \sqrt{n} \rfloor}}{{\lfloor \sqrt{n} \rfloor}!}$ and $g(n,k) = \frac{n}{k}$. We expect this to be an overestimate as the individual numbers being multiplied are not being floored. This should give much better bounds than the rudimentary calculations previously shown.
Because this is always an overestimate, we should figure out how much error this new $G(n)$ function gives us. The maximum possible error for each number is 1. We will now use error propagation to figure out the margin of error of the estimate: $$\sum_{k=1}^{\lfloor \sqrt n \rfloor} \frac{1}{g(n,k)} = \sum_{k=1}^{\lfloor \sqrt n \rfloor} \frac{k}{n} = \frac{1}{n} \sum_{k=1}^{\lfloor \sqrt n \rfloor} k = \frac{\left(\lfloor \sqrt n \rfloor\right)\left(\lfloor \sqrt n \rfloor + 1\right)}{2n}$$ For sufficiently large $n$, $\lfloor \sqrt n \rfloor \approx \sqrt{n}$ and therefore we can take the limit of the error function to be $\frac{1}{2}$. This is much, much better than what we had before!
Essentially, the bounds can be made much tighter once we find $n$ such that $\frac{N}{2} < G(n) < 2N$.
The baseline
As a baseline, we will now try to estimate $G(2000) = \frac{2000^{44}}{44!}$. It is trivial to find that $44! = 2^{41}\cdot3^{19}\cdot5^9\cdot7^6\cdot11^4\cdot13^3\cdot17^2\cdot19^2\cdot23\cdot29\cdot31\cdot37\cdot41\cdot43$. Since $2000 = 2^4 \cdot 5^3$ we can immediately simplify the fraction to $$\frac{2000^{41}}{2^{29}\cdot3^{19}\cdot7^6\cdot11^4\cdot13^3\cdot17^2\cdot19^2\cdot23\cdot29\cdot31\cdot37\cdot41\cdot43}$$ We factor out $2^{41}$ to cancel the $2^{29}$ on the bottom to get $$\frac{2^{12} \cdot 1000^{41}}{3^{19}\cdot7^6\cdot11^4\cdot13^3\cdot17^2\cdot19^2\cdot23\cdot29\cdot31\cdot37\cdot41\cdot43}$$ We now note that $23+43=29+37=66$ so we can reasonably approximate $23\cdot29\cdot37\cdot43 \approx 33^4$ and we are now left with $$\frac{2^{12}\cdot1000^{41}}{3^{19}\cdot7^{6}\cdot11^{4}\cdot13^{3}\cdot17^{2}\cdot19^{2}\cdot33^{4}\cdot31\cdot41}$$ To simplify the $3^{19}$, we use $\left(\frac{n-1}{n}\right)^n \approx \frac{1}{e}$ to get $3^{19} = \frac{1}{3} \cdot 9^{10} \approx \frac{1}{3} \cdot 10^9 \cdot \frac{1}{e} \approx \frac{10^9}{9}$. Putting this back into the fraction gives us $$\frac{9\cdot2^{12}\cdot10^{113}}{7^{6}\cdot11^{4}\cdot13^{3}\cdot17^{2}\cdot19^{2}\cdot33^{4}\cdot31\cdot41}$$. We employ another approximation of $11^2 \cdot 17 = 2057 \approx 2048 = 2^{11}$ to get rid of these. Since we only have $2^{12}$ on top and $11^4 \cdot 17^2$ on the bottom, we need to borrow some 2's out of the $10^{113}$. This leaves us with $$\frac{9\cdot5^{10}\cdot10^{103}}{7^{6}\cdot13^{3}\cdot19^{2}\cdot33^{4}\cdot31\cdot41}$$ To cancel out the $7^6$ we employ some bashing to find $7^6 = 117649$. Dividing this by $10^6$ results in a value very close to $8.5 = \frac{17}{2}$, which is very convenient: $$\frac{17\cdot9\cdot5^{10}\cdot10^{97}}{2\cdot13^{3}\cdot19^{2}\cdot33^{4}\cdot31\cdot41} = \frac{17\cdot9\cdot5^{11}\cdot10^{96}}{13^{3}\cdot19^{2}\cdot33^{4}\cdot31\cdot41}$$ We do a little more bashing and find $5^{11} = 48828125 \approx 48622761 = 33^4\cdot41$ so the whole expression now becomes $$\frac{17\cdot9\cdot10^{96}}{13^{3}\cdot19^{2}\cdot31}$$ We now can just go ahead and make the rough approximation of $\frac{17\cdot9}{13^{3}\cdot19^{2}\cdot31} \approx \frac{150}{2000\cdot400\cdot30} = 6.25\cdot10^{-6}$ and thus $G(2000) \approx 6.25 \cdot 10^{90}$.
This is still off by two orders of magnitude, so we conclude that $2025 \leq n \leq 46^2 - 1 = 2115$ as we found in the first rough estimate that adding another factor will increase the number by roughly 2 digits (as we have 45 and an unknown error factor from 1 to 2 as found when we defined $G$).
Better than before
This time, instead of doing over 50+ painful long divisions and spending a few hours going through papers, I was able to reduce the number of divisions needed to 4.
The actual work
We begin by checking the primes 101 and 103 from $f(n,10)$ and $f(n,20)$. If $2025\leq n\leq 2039$, then 101 must be a factor of this number as dividing by either 10 or 20 will result in the floored value being 202 or 101, respectively. Similarly, for 103, if $2060 \leq n \leq 2079$, then 103 must be a factor of this number. Divisibility checking for 101 is simple as $101 = 10^2 + 1$ which means that we can check using the paired alternating sum (similar to 11 but with two-digit numbers). Checking this on $N$ results in $303$ which shows that 101 is a factor. Therefore, $2025\leq n \leq 2039$. We can do long division and check again to determine which half of this range $n$ belongs to if necessary. The first division of $\frac{N}{101}$ results in (with the trailing zeros removed as they will not affect divisibility)
420983613515873542382348396901980757225052024879832559470522610718691983097856We now check this one more time and get a sum of $101$, showing that $101^2$ is a factor of $N$ and thus $2025 \leq n \leq 2029$.
Another check we can now do is the $k=6$ check. We know that $f(n,6)$ is either equal to 337 or 338. If $f(n,6) = 338$ we should expect a factor of $13^2$. We divide the quotient in the previous step by 169 and get a remainder of 156. Therefore, we must have a factor of 337 and thus $n$ is one of $2025, 2026, 2027$. We will not bother to check dividing because long division with such large numbers is very time-consuming.
Our final check to determine $n$ will be checking for divisibility by 1013. If the running quotient is not divisible, $n=2025$ as $f(2025,2) = 1012 = 2^2 \cdot 11 \cdot 23$. If it is, then $n=2026$ or $n=2027$. If $n=2026 = 2\cdot1013$, we would expect that 1013 goes into the running quotient twice. If $n=2027$ then it will only go in once.
The first division results in415581059739263121798961892302054054516339609950476366703378687777583398912eliminating $n=2025$. The next division results in410247837847248886277356260910221179186909782774409049065526838872244224eliminating $n=2027$. Therefore, $n=\boxed{2026}$.
I haven't used the trailing zeros or final non-zero digit yet. We know N = $2^{4m} 5^{13} Q$ for some $m,Q$.
The exponent of 5 cannot increase when $n \bmod 5 \neq 0$. With $n=2025$ trivial solutions of $f(2025,k) \bmod 5 = 0$ gives $k = 1, 3, 5, 9, 15, 27, 45$ as $2025 = 3^4 \cdot 5^2$. However, finding nontrivial $k$ will be difficult without an exhaustive search, and will be especially unfeasible when testing all $2025 \leq n \leq 2039$.
Of course, long division by 101 once already has told us that $2025 \leq n \leq 2029$ which would reduce the amount of calculations significantly, but I'm trying to avoid doing such long division entirely.
