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!

Bold to assume there will be a future

Nobody: Michael Penn: okay great

"Comment what the future is like" It sucks, don't come here. :P

At the example: You know 3

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

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

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.

This is brilliant. I started off by observing that 6

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

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!

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

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.

I am totally floored with how cool this is!

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. 🙌🙌🙌👏👏👏

This is amazing!! I hope for more videos like this.

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

I love this channel. Thanks for the content!

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

This is a great channel, wish I found it sooner!

Awesome problem! I got super hyped with this.

In 12:25 you probably want a write ceil(2020/7) = 289, not 2020/6.

This is exactly what I needed for quarantine

With everything floored you get at least extra solutions that slightly increase your value of x, since the multiplication of all of it will increase it by less than one. The goal is to be so small no intermediate floor increases (in absolute values due to sign). You can tell that gives you a half open/closed line with as least value your solution (below your value the original product is less than 2020 so the floor is 2019) Also happy to have this popped up in my recommendations, subscribed.

Hello, your second question and the end is in my opinion a very interesting and exciting question. "Can we tighten the bounds of the intervalls?" I haven´t prepared anything, but I can imagine, that following steps could lead to a solution. First we saw: The given equation has no solution for positive x because k = 2020 is bigger than the maximum possible value as shown, let´s say M. One could think about a solveble range of values M for this intervall. The goal is: Find a maximum epsilon e > 0, s. t. the existent solution x is in the open intervall (a + e ; b - e). And my feeling tells me, that possibly the relation between the range of possible M and maximum epsilon is something involving squareroot and the difference of M and K.

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

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

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.

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.

Subscribed because I can tell good things are going to come from this channel. :D

Thank you for the video!

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 :)

When I solved this I worked from the inside out from the beginning. You get the same kind of thing. I concluded there was no solution, but I forgot to consider negative values.

As far as tightening bonds goes, you've used floor( n ^ (1 / no_of_iterations)) < x < ceil(n ^ (1 / no_of_iterations)), but clearly if |x| < n ^ ( 1 / no_of_iterations), then this floor multiplication will yield results less than n.

Probably be a good idea not to say "two twenty" for "2020" since "two twenty" would be 220 instead. Just saying.

The search is reduced to 6 cases if you express x = -6.7 + d, where 0 < d < 0.1. Plus you only need to check 3 of those cases this way. Like you did, work your way out, the third floor operation gives you that range of 6 integers to explore (remember d0.1 so no checks needed. Of the other three, two of them don’t work: There’s no need to check the whole thing, just that the inner [-308.2 + 46d] doesn’t match the integer that produced that “d” Quite happy about this!

if this is from a competition, there should be a way how to solve the equation without using the computer or checking the possible solutions (even though you don't need to check them all, since the function is monotone, so you can find it by binary search and that might be manageable)

You are one of the world's greatest gifts.

Very interesting. Thank you.

Great suggested problems at the end.

I tried to replace x with 6+a for positive x and with -7+a for negative x, with 0

Great video, but i‘m watching it in the past so I can’t comment how the future is.

2009 is also far fetched. Extending you logic for finding (x(x)) I found that maximum it can take is 2002. Still. Loved it.

For the positive solution there is a very easy way to prove it is impossible. You know that 6 < x < 7. We also know that LHS is a non decreasing function of x. Just assume x = 6.999999....., which is the largest possible value still less than 7 LHS = x * floor(x * floor(x * floor(x))) LHS = x * floor(x * floor(x * 6)) LHS = x * floor(x * (7 * 6 - 1)) LHS = x * (7 * (7 * 6 - 1) - 1) LHS = x * (7 * 41 - 1) LHS = x * 286 LHS < 7 * 286 LHS < 2002 Therefore LHS cannot ever equal 2020 and hence there are no positive solutions. For the tighter bound, you had asked in followup questions: Solving for x^4 = 2020 gives x = 2020 ^ 0.25 = 6.704059..... Hence if you use floor, the x value will be 6.704059...

Thank you professor for this video 👌👌👌👌👍👍👍

Interesting problem. In the simpler solution, after finding the boundaries for x isn't it directly clear that integer of x has to be three without the way around m? I know, you need it for the other problem but you should mention more?

Gran canal, buen video.

there's probably a more efficient way to search through all the possibilieis e.g. guess halfway, depending on whether it's and over- or under-estimate, you guess halfway of the appropriate half. something like a binary search algorithm.

Nifty! Thank you!

From the fact that x is multiplied by an integer to yield 2020, x must be a rational number of the form 2020/y. The absolute value of x falls between the floor and ceiling of the positive real fourth root of 2020, which restricts x to two ranges, either -7 to -6 or +6 to +7, which means that y is either from -336 to -289 or from 289 to 336. Using divide-and-conquer reveals that y = -305, giving a reduced value of -404/61 for x.

If -7

What chalk and board are you using man, i want that clear look

i bet someone already solved this but i found that n needs to be less or equal to 10 so that the equation has positive solutions ( if x is less than 1 then it's trivial and there are no solutions , if x is 1 then there are no solutions either , if x is bigger than 1 the only solution starts at x =2 because if x=1.9 then x[x] =x as the definition of the floor function , we find that the first possible solution is for x=2 , and we notice that for n>10.98 the equation has no solutions because then root of 2020 becomes smaller than 2 , thus since n is an integer n=10 is the maximal n for which there are positive solutions , I will try to do the negative ones because they don't work by simple analogy and need some writing haha , great video , loving them !

Woohoo! What a beautiful problem! ❤️ I'm so excited I got this answer -404/61, took me more than half an hour, using the same approach and didn't use a computer (though used calculator for few checks).... I was almost about to give up at the end of 30 mins as I found no solutions, but then suddenly it struck me that there could be negative solutions.... I saw at the start that x had to lie between 6 and 7 and tried to reduce the bounds of the interval as you suggested at the end.... I quickly saw that the expression in question f(x) kept increasing with x.... But due to the discontinuous nature of function, there was a jump near 7.... If we approach 7 from the right we have 7*7*7*7, but if we approach from left, x = 6.999999........ x times fl.x = 41.999999....... x times fl of fl of x = 286.999999..... So final expression is 2001.999999..... = 2002 < 2020, so no positive solutions But for negative solutions, I started with x as fourth root of 2020 = -6.7041 and kept increasing x by 0.01 till x reached -6.2, then all the inside terms come to 305, making x = -2020/305 The thing to note as that for the second floor operation inside, the number is positive so smaller number should be taken for floor value.... ☺️

Great video, but why the long explanation for finding m at the 4:58 minute mark? If m is floor(x), and x is between 3 and 4 (not including 4), isn't that enough to confidently state that m=3?

Setting constants as something relatable such as calendar dates makes math more relatable somehow.

At 5:00, note: We already know that m is floor(x) where x is between 3 and 4. The floor of any number between 3 and 4 (not inclusive) is 3. Done! We don't need all that other stuff up to 5:45 to get to m = 3.

I loved it!

Well, it seams, that binary search require less math. 289 and 336 average is 312, so check if 2020/312 is too small or too big.

The graph of x*floor(x) in Desmos is pretty interesting.

I tried a condition that was a bit too strong (x itself is an integer multiple of a certain integer n, as in kn, plus an integer multiple of 1/n, as in j/n, where 0

when you know the interval of integers, I would do a binary search if I had to do it on paper.

Ok, some boundary on the 3rd question at 20:50 As soon as the number of floors exceeds log_2(2020)=10.9... (i.e. the number of floors is at least 11), we can't have any positive solution. Proof: If x is a solution, then x=2^11=2048>2020. But then floor(x)=1, so x floor(x)=x, hence we get floor(x floor(...)))=1. Negative values for x seem to be more subtle, since the rounding of the absolute value changes between floor and ceiling.

@3:51 but the floor function is increasing for negative values too. As you said, it behaves like the square function.

Acually doing numbers with theory is kinda stupid but a nice way to test someone anyways thanks for makeing things like this available for the public.

2020 is almost over, that's roughly all there is left to be grateful for.

Jeees, You lost me on "solve x" xD

This is pretty dope tbh. Never trained too much in number or graph theory but this was easy to follow.

Noo, here I was working out the problem and watching the video to check whether there was a nicer/cleaner way to do this only to see you did the same thing. Oh well. One thing to note is I went a bit further to conclude that x=-6.6... so to make the gap and amount of things to check only about 5 values.

In the first equation, how did you select 10 as the value. If you selected a diff value, say 100 what would happen?

what about for complex values of x? with floor(a+bi) = floor(a) + floor(b)i

Future is good as I found this channel.

Again, this makes Math alive guys!!! More fun, more excitement...I will try the last alternate one. Thank you for the gift.

The floor function over the negative integers mirrors the ceiling function over the positive ones. So the equation x*ceiling(x*ceiling(x*ceiling(x))) = 2020 has a positive but no negative solution and it has the same absolute value as the solution here.

18:52 Where did that arrow come from?

(positive no-solutions) Take x=7-a for a on (0,1) b = x*floor(x) = x*6 = 42-6a, so floor(b) is at most 41 c = x*floor(b) = x*41 = 287-41a, so floor(c) is at most 286. d = x*floor(c) = x*286 = 2002-286a. So for tight bounds, 1296

For x > 0, 6 < x < 7 is too wide a range. We know that for any non-integer x, x^4 > ⌊x ⌊x ⌊x⌋⌋⌋, so 2020^.25 < x < 7 is a better range, then we have ⌈2020/7⌉ ≤ m ≤ ⌊2020^.75⌋, or 289 ≤ m ≤ 301. Similarly, for x < 0, we have -(2020^.25) < x < -6 as a better range, and ⌈-2020^.75⌉ ≤ m ≤ ⌊-2020/7⌋, or -301 ≤ m ≤ -289.

Regarding the third question: I quickly wrote a little Python script that gives me five solutions for n. I only checked for n

The future’s bright!

"okay I think this is a good place to breathe"

12:29 that should be 2020/7, not 2020/6 19:00 should be -336≤m≤-228, not -336

At 19:11, shouldn't the bounds have the same absolute value as for the case that the solution is positive, 336 and 289 respectively? Taking the elevator down in the positive case should mirror taking the elevator up in the negative one, shouldn't it? Edit: Also, around 19:43 you incidentally go further negative from -2020/288, the next element in the set should be -2020/289 . But whatever, just a slip-up.

Amazing!!!!

I’m still confused but great explanation!

5:00 that's weird. If you already said 3

hmm. there are probably a couple complex solutions to this too but finding them wouldn't be trivial I think....in fact I think it'd end up looking a lot like the alternating floor/ceiling problem you proposed at the end

Nice video

It feels like you should check for existence in the interval first before jumping through all the hoops, but hey, you do you (by the method you used to prove non-existence, I mean)

When you say 3

What about the other two Fourth roots of 2020.?

this dude lifts

I don't really understand why we can say that the floor funtion behaves like x^n. I mean is obvious for the upper value, but not so much for the lower one.

I missed the reasoning behind the transformation at about 12:50. What theorem or part of the vid says we can do that…?

0:18 _Ooooooh boy......_ honestly, I doubt that even in your wildest dreams you could’ve imagined quite how f-ing crazy things have turned for us over the past 2.5yrs

Why all the rigamarole? If 3

12:30 I didn't understand why ceil(2020/6) is 289 instead of 337. What am i doing wrong?

Can you do a Mathematica tutorial

The fact that you found no negative in the first case ( x floor(x) = 10) does not mean no solution exist, but that the approximation x floor(x) = x^2 may be bad (it is a simplification of the true inequality x-1 < floor(x) 10, it is possible that there is a solution in (-3,-2)

I tried it myself, but forgot about the negative x, so I got no solutions😕

I come from 1 month in the future of this video, and I can safely say that we're still in lockdown XD