Solving a crazy iterated floor equation.

We present a solution to an equation which involves a certain iteration of the floor function.

[blackpenredpen]

Man! This is super crazy!!
@20:47, I will try to see if I can solve any of that. If I can, I will be super hyped!

[bsharpmajorscale]

"Comment what the future is like"

It sucks, don't come here. :P

[friedrichotto]

At the example: You know 3<x<4, thus floor(x)=3. Since x*floor(x)=10; x=10/3

[cycklist]

What a great puzzle! It would be great to see more of these.

[samuelmeyer]

Question 2:
For the negative values:
floor(x) = -7
=> x*floor(x*floor(-7x)) = 2020
floor(x) <= x and floor(x) >= x-1
=> floor(-7x) <= -7x
=>x*floor(-7x) >= -7x^2
>=-7x^2-1
=>x*floor(x*floor(-7x))) =2020<= -7x^3-x
=> x <= -6.6

floor(-7x) >= -7x-1
x*floor(-7x) <= -7x^2-x
floor(x*floor(-7x)) <= x*floor(-7x)<= -7x^2-x
x*floor(x*floor(-7x)) = 2020 >= -7x^3-x^2
x >=-6.65

[josir]

This "solving" just doesn't feel good. I worked on it by substituting each floor function by a subtraction of a small number between 0 and 1, so the thing becomes (n+a){(n+a)[(n+a)n-b]-c}=2020, where 0<a, b, c<1 and n is integer. After showing no positive solution, substitute n=-7 and get something like 1029a+...=381. By consider the leading term, a~0.37 or x~-6.63, check the two nearby m in x*m=2020 and it is easy to conclude that x=-2020/305.
And this method generalize well to either solving n times floor or the alternating floor ceiling problem

[laszloliptak]

A small correction at 18:44: Strict inequality means that you can't have equality, and you want to say the opposite: taking the ceiling (or floor) will now make the inequality not strict.

[niekteeuwen]

I think I figured a way to tighten the bounds on x. When we fill in x=∜(2020) into the equation we will get an answer smaller than 2020, since floor(∜(2020)) < ∜(2020). This means ∜(2020) is a lower bound for the positive solution, so ∜(2020) < x < 7. In a similar way we can argue that the bounds for the negative solutions can be tightened to -∜(2020) < x < -6.

[jroemling]

Wow, this was a random recommendation that was super-interesting with a good production quality and pleasant to listen to. And you have been churning those out many times a week for 9 months and gotten like 70 views per video, but you kept going and now lately some of them blew up a bit. Thanks for sticking to it, I will definitely subscribe to this one!

[AlephThree]

This is brilliant. I started off by observing that 6<x<7, but then noting the expression is strictly less than 7^3 * 6, since the rightmost floor of x must be 6. But 7^3 * 6 <2020 so there must be no solution. I then made a cup of coffee to see how Michael would spend 22 mins telling me there is no solution, only to realise I’d overlooked the negative possibilities. Doh!

[TheOneThreeSeven]

I am totally floored with how cool this is!

[williamsean]

The left alternating floor ceiling has a solution 7 - 17/291, the right alternating floor ceiling has a solution 7 - 23/61. And wrapping another gives a range of solutions from -2021/305 to -2020/305 including -2020/305

[dzspdref]

5:19 : reciprocating the inequality should have been written 1/4 < m/10 < 1/3 ; oopsed on the 1/3 by writing 3.

[SwapnilAnand]

When this guy starts to present solutions, there are moments when I actually pause the video and clap for the ingenious mathematical tools this guy provides us with. 🙌🙌🙌👏👏👏

[zoonvanmichiel]

For the fourth extra question: 2020/291 for the first equation and 2020/305 for the second equation. No other solutions exist. I will highlight my approach below:
assume x in (1, inf)
then floor(x) in (x-1, x] isct [1, inf)
then x floor(x) in (x^2-x, x^2] isct [x, inf) = (x^2-x, x^2] isct (1, inf)
then ceil(x floor(x)) in (x^2-1, x^2+1) isct [2, inf)
...
We use these ranges to put bounds on x
for which x does x^4-x^3-x < 2020 < x^4+x^2 hold?
As I don't know how to solve 4-th order polynomials I use Wolfram Alpha to find that approximately 6.66687 < x < 6.97453
we convert these bounds on x to bounds on m and check the functions on all possible 2020/m
A similar argument for x < 1. For x in [-1, 1] such floor/ceiling functions remain in [-1, 1] and therefore can't be a solution.

[Flammewar]

Guys, I‘m from 2025. It’s crazy to see an old video of Michael before he proved the Riemann Hypothesis.

[goseigentwitch]

The trick for me was realizing that the answer must be some integer of the form x = 2020/k for some integer k
also for 6 < x < 7, floor(x floor( x)) is at most 41, so floor(x floor(x floor(x))) is at most 286, so our original equation is at most 2001
therefore -7 < x < -6
some small calculator works shows -6.7 < x < -6.6, so m can only be (negative) 304, 305, 306, or 307
(this just shrinks your exhaustive search, so i didn't need anything beyond a calculator to solve it in a few minutes)
plug into calculator looking for your starting number to show up as your answer to floor(x floor(x floor(x)))
voila, m is -305, so x = 2020/-305
check the answer, and it checks out :)

[ordanguric]

Dude, your videos are just amazing, I learn from them a lot, thank you!!!

[hymanimy]

You present such interesting questions and have great production value. Keep it up 👍

[hwxyz]

Nice video! Just a little remark: at 12:05 m should technically be greater than or equal to floor(2020/7)+1 and less than or equal to ceil(2020/6)-1. In this case it doesn't matter but if, for example, instead of 2020 it were 2022, which is a multible of 6, then m being greater than 2022/6=337 doesn't imply that m is greater than or equal to ceiling(2022/6)=337.