International Math Olympiad | 1998 Q6.

[@MichaelPennMath]

Ok guys, I was out climbing today and saw the comments about my sketchiness with the inequality argument. Here is a fix.... Around 15:28 I prove that s

[@caesar_cipher]

Hello Michael, thanks for clarifying. On a separate note, in IMO papers the set N refers to "positive integers". This is usually always explicitly stated in every question that the "N" is used, including in this problem if u look at the original paper. Please stop using the term "Natural numbers" as it meaninglessly triggers people who feel a compulsive duty to comment whether 0 is natural or not - imo its irrelevant. In your last Q&A, u explicitly addressed this but its understandable if not everyone has seen that. Also in this video u mention "positive integers" also to clarify. Anyway my suggestion would be to use "N" as set of "positive integers" as IMO does and not needlessly say the term "natural" numbers - thanks for reading

[@oussamaennafii7233]

How about f(n) = 0, \forall n it solves the problem.

[@MrRyanroberson1]

yeah i agree. you could have instead shown that f(a)^n -> f(1)^(n-1) f(a^(n-1)) and then argue (n-1)s

[@megauser8512]

. . . and you can prove that ns=1. Then: Induction step: f(a)^(k+1)=f(a)*f(a)^k=f(a)*f(1)^(k-1)*f(a^k)=f(1)^(k-1)*f(1)*f(a*a^k)=f(1)^k*f(a^(k+1))

[@megauser8512]

Also, see the first comment of the thread below: By @UC3CiApPbkPn6BXYPKySvCcA5 days ago: "I think, there is a mistake at 6:00 , if you´re using equation 1 and want to get the equation as f(...), the answer follows with n=a and m=f(b^2f(1)): f(a)^2f(b^2f(1))=f(a^2f(f(b^2f(1))))"

[@hfix307]

Goodness, this is a brutal question. Props to people who solved this in test conditions.

[@DerZerberster1]

I think, there is a mistake at 6:00 , if you´re using equation 1 and want to get the equation as f(...), the answer follows with n=a and m=f(b^2f(1)): f(a)^2f(b^2f(1))=f(a^2f(f(b^2f(1))))

[@avnertishby]

Exactly.

[@cr1216]

Is there a fix for this?

[@andreben6224]

It's actually a benign error: indeed we need to use eq.(1) with n=a and m=f(b²f(1)), which gives us f(a)² f(b² f(1)) = f( a² f(f( b²f(1) )) ) But the argument in the second factor is by use of eq.(2) with m=b²f(1) f(f( b²f(1) )) = b²f(1) f(1)² = b²f(1)^3 So at the end f(a)² f(b² f(1)) = f( a²b² f(1)^3 ) which is the same as what Michael ended up with two lines later, so n harm done :)

[@wesleydeng71]

25:40 No, you cannot permute those two. g is self-inverting, g(g(3)) = 3. Because g(3)=2 then g(2) must be equal to 3. Hence, g(3)=2, g(2)=3, g(37)=5 is the only solution.

[@adityaprakashsingh2675]

I think there is an error at ~16:10. For example, take s = 8 and r = 6. It satisfies both s

[@danielva9309]

yes and no, he arrived at that conclusion the wrong way. however, if u keep powering, you will get (n-1)*sinf u'll get s

[@IanXMiller]

Yeh that step confused me too. You can't subtract inequalities like that. Add them yes, not not subtract. He could have generalised to f(a)^n and/or used induction to get n*s

[@leif1075]

Does the notation in the equality on the right mean f of n squared just the input n squared or the whole function f of n squared..poor notation..wasn't anyine else confused by this? I'd be surprisednif people weren't.

[@anonymoususer9837]

Agreed, I was looking for this. As Daniel VA pointed out, you should indeed be able to take a limit as [the power of f(a) on the left hand side] approaches infinity, which would yield s

[@andreivila7607]

Finally! I’ve been waiting for one of those for a long long time. Nice problem, and I hope you will do more of them. Have a fantastic day!

[@Monolith-yb6yl]

There is a mistake: first using of equasion 1 to prove "almost multiplicative" property. Have to be n =a, m=f(b^2 f(1))

[@HagenvonEitzen]

At the end, you forgot to mention that every involution of the set of primes gives rise to a valid g. Otherwise, it might happen that 120 cannot be achieved

[@Czeckie]

is it clear that g defined as g(n)=f(n)/f(1) even satisfies the equation (1)? EDIT: I'm sure it does, but it seems to me we need another tricky computation to see that. EDIT2: I've written the computations here, it's not annotated or commented, but I hope it's clear. I'm curious if this is needlessly complicated and I am missing more direct route www.desmos.com/calculator/v40v2hwi3a

[@benkelly2024]

@@Czeckie Any multiplicative self-inverting function satisfies equation (1). g is multiplicative and self-inverting. In full: let g be any such function. Then g(n^2.g(m)) = g(n)^2.g(g(m)) = m.g(n)^2 as required.

[@Czeckie]

@@benkelly2024 but if you don't know g satisfies (1), you don't know g is an involution (self-inverting)

[@benkelly2024]

@@Czeckie Yes, I agree that I was assuming that. There is a bit of calculation skipped over in the video needed to show it, but it looks like it follows fairly easily from the definitions (unless I've made a mistake, which is certainly possible!) g(g(a)) = g(f(a) / f(1)) => g(f(1)) . g(g(a)) = g(f(1)) . g(f(a) / f(1)) = g(f(1)f(a) / f(1)) = g(f(a)) = f(f(a)) / f(1) = a.f(1) g(f(1)) = f(f(1)) / f(1) = 1.f(1)^2 / f(1) = f(1) => g(g(a)) = a. Not sure that's any simpler that yours tbh, though it does also directly establish that g is a bijection. Edit: A rather simpler way in this comment below: kzbin.info/www/bejne/joLdnYuXgbKSgKc&lc=UgyrBLKJHyukLJnkY7p4AaABAg.9Ey8kMgDInf9EyHWiXKpUC

[@Czeckie]

@@benkelly2024 cool! both your method and in the comment. I think that settles my question: it's not obvious to see, it's not hard in any way, but you need to do some ad hoc computation.

[@HagenvonEitzen]

25:15 actually, ">="

[@jimallysonnevado3973]

Didnt get at 6:45 if we are using equation 3 shouldn’t it become f(a^2f(b)^2) instead of the one written on the board

[@saroshadenwalla398]

When he applied the first equation it should have been with n=f(a) and m=f(b^2*f(1)) and he would have got what he gets at 6:45 as the answer

[@Mr5nan]

@@saroshadenwalla398 where is f(a) that you mean?

[@saroshadenwalla398]

@@Mr5nan Sorry meant n=a

[@Mr5nan]

@@saroshadenwalla398if so, then it would be f(b^2*f(1))*f(a)^2 which is the step back

[@saroshadenwalla398]

@@Mr5nan No you go from f(a)^2 f(b^2 f(1)) to f(a^2 f(f(b^2 f(1)))) using equation 1 taking n=a and m=f(b^2 f(1)). You're going from the RHS of equation 1 to the LHS.

[@andreben6224]

Wow this one was brutal. There were some mistakes along the way too, but in the end, it all holds together, even though I feel one should show that the minimum possible value can be attained by some function f (or g, since the g function will also verify (1)). Question 6 is always pretty colossal, thanks for sticking with it and thank you for making these videos. They really motivate to do more day to day math and I'm really grateful for it.

[@gustavowadaslopes2479]

16:10 So, this is wrong. a

[@caesar_cipher]

Illogical comment

[@gustavowadaslopes2479]

@@caesar_cipher How is it illogical? All of it's parts are logical, even if not written on the best maner.

[@caesar_cipher]

@@gustavowadaslopes2479 I shouldn't have used "illogical" for this comment - I take it back.

[@antonryzhov]

You cannot subtract inequalities in such way like at 16:20. Consider 4

[@timothyrosenvall1496]

So ranking math channels, you have channels like Numberphile, Veritasium, and Eddy Woo which produce absolutely beautiful content explained well and designed for perhaps the less mathematically practiced. Then you have Mathologer, 3B1B for those with a more rudimentary understanding of. Maybe throw some of Mind Your Decisions in here too. Then I'd probably stick BPRP and Dr. Peyam here for those who have taken some college math classes. Then there's this guy. I can mostly follow what he's doing, but theres some jargon and some leaps of logic that I can't quite make yet. Hope to keep learning and practicing!

[@zhiruo7675]

Oh my god I am following all of them. Guess it makes me a big nerd.

[@jkid1134]

Pleb list tbh, but you did kinda already out yourself as a pleb, so I shouldn't just pile on

[@rishabbomma9361]

And then next are like Fields medalists doing Commutative Algebra(Nod at Richard B)

[@Sam-xt1zk]

For sure. I like to watch math KZbin videos as a poor to average math student (never went past pre-calc formally), just to see how much I can understand. I watch a fair amount of channels, with BPRP and Math Sorcerer as my top two. Michael's videos are just about always above my head, however I keep watching anyway.

[@whiteboytft8604]

Blackpenredpen big sad..

[@adityamohan7366]

I didn't quite understand the 6:06 jump. It seems the values of n and m as in [f(a)^2 f(b^2 f(1))] are the arguments of the overarching function in equation 1. And so shouldn't it be equivalent the inverse function over the RHS in equation 1. What you did was instead substituting n with just 'a' rather than 'f(a)'. Or am I getting this wrong?

[@mathijs1752]

I don't understand it as well. You cannot apply an inverse there as you do not know at that point whether an inverse does exist

[@mathijs1752]

He got the wrong n and m. When you pick n=a and m=f(b^2*f(1)), you go from RHS to LHS to get m*f(n)^2=f(n^2*f(m))=f(a^2*f(f(b^2*f(1)))). Now use (2) to get f(f(b^2*f(1)))=f(b^2*f(1)^3). All in all this gives f(a)^2*f(b^2*f(1))=f(a^2*f(f(b^2*f(1))))=f(a^2*f(b^2*f(1)^3)). This is where I get with properly using (1). I do not know how to get rid of f(1)^2 to arrive at f(a^2*f(b^2*f(1)))

[@adityamohan7366]

Okay so I got what he was actually intending to do. There are couple of mistakes. - The value of n is just *a* instead of f(a). {see 5:49} - The value of m is actually *f(b^2 f(1))* instead of just b^2 f(1). {see 5:53 } - These values are the *RHS* of equation 1 and not the 'arguments' the LHS function of equation 1. {see 5:58} Solving for those you get the actual equivalent which is *f[a^2 f(f(b^2 f(1)))]* instead of just f[a^2 f(b^2 f(1) ) ] To give him the credit, Michael indeed corrects it in the next step {see 6:56}.

[@mathijs1752]

@@adityamohan7366 I do not get it yet. Because in the next step from f(a^2*f(b^2*f(1))) with (3), you should get f(a^2*f(b)), but somehow he gets f(a^2*f(f(b^2*f(1))))

[@mathijs1752]

Wait, if you get my last f(a^2*f(b^2*f(1)^3)) and apply (2), then you get f(a^2*f(f(b^2*f(1))) as in the video. Now I get it! I feel that he tried to 'save' a mistake by bluffing that then applying (3) would lead to the correct answer. I did not see this being corrected in the video. If I have one critique to Michael, then it is that if you state that a step is difficult, please take your time to properly and correctly explain it. Even so in a video of half an hour.

[@helloitsme7553]

6:06 anyone being lost at what he does in this step? im so confused

[@yusrizalakbar3983]

Me too lol

[@avnertishby]

It's a mistake. Just skip over that expression. Instead use a=n and use f(b^2 f(1))=m to get directly to the expression after it.

[@juanixzx]

I didn't find f(a)^2 f(b)^2, but f(f(a)^2 f(b)^2) initial -> f(f(a)^2 f(b)^2) (3) w/ n = b -> f(f(a)^2 f(b^2 f(1))) (1) w/ n = f(a), m = b^2 f(1) -> b^2 f(1) f(f(a))^2 (2) w/ m = a -> b^2 f(1) (af(1))^2 -> = b^2 f(1) (af(1))^2 = a^2 b^2 f(1)^5 (2) w/ m = a^2 b^2 f(1)^3 -> f(f(a^2 b^2 f(1)^3)) so I got f(f(a)^2 f(b)^2) = f(f(a^2 b^2 f(1)^3)), then, applying inverse function: -> f(a)^2 f(b)^2 = f(a^2 b^2 f(1)^3) = f((abf(1))^2 f(1)) (3) w/ n = abf(1) -> f(abf(1))^2 at last, I got the same result Michael got in 9:15

[@goodplacetostop2973]

26:34 Any NFL fan in the comments? Anyway, here’s the daily. We have a cube of side length 8. At each corner of the cube, we remove the corner by making a cut through the midpoints of the three adjacent sides. Determine the increase or decrease in total surface area as a result of slicing the eight corners off the original cube.

[@pbj4184]

Is it a decrease of 64(3-√3)?

[@ambassador-of-misogyny]

83.2 unit^2 decrease in TSA. We are removing 8 pyramids with equilateral triangles as their base and adding the equilateral base area to the remaining solid. So, TSA(Cube) - 8×CSA(pyramid){a pyramid is not curved but get the picture} + 8×area of the equilateral triangle[a=4sqrt(2)].

[@err954]

26:34

[@pbj4184]

@@ambassador-of-misogyny I got a decrease of 81.14 using a calculator. Algebraically, it was a^2(3-√3). Did you calculate the value by hand?

[@malignusvonbottershnike563]

Well, you'd end up with 6 squares of side 4sqrt(2), as well as 8 equilateral triangles of side 4sqrt(2). Adding these surface areas together, you get 32*6 + 8sqrt(3)*8, which is 192 + 64sqrt(3). You can factorise that to get 64(3+sqrt(3)). Back to the cube, 8^2 * 6 = 64*6, which can be factorised to get 64(6). Just by observing the factorised forms, 3+sqrt(3) is less than 6, so we know that the surface area decreases overall.

[@thephysicistcuber175]

16:04 what you just did is illegal. You can't subtract inequalities like this.

[@zafarb4219]

SInce s≦3/2r, it automatically satisfies s≦r anyway, so it doesn't mater.

[@a_llama]

i agree, doesn't seem directly possible

[@thephysicistcuber175]

@@zafarb4219 s=8 r=6 is a counterexample as someone else said.

[@boristerbeek319]

@@zafarb4219 Except it does not. Consider the following: do all numbers x satisfying x ≤ 3/2 also immediately satisfy x ≤ 1?

[@chhabisarkar9057]

@@boristerbeek319 s is natural bruh so it doesn't even matter

[@benjaminbrat3922]

Great video as always. What threw me off a little is that the indication you write at the third step of the quasi-multiplicativity derivation is wrong (notes reading, I am sure): at 5:58 it is n=a and m=f(b^2*f(1)), so the first f() switched the result gives directly the next step of the equation, with no need to use eq (3) in the other direction. I tried with the indication you wrote, but nothing fruitful. See you tomorrow =)

[@lucassandleris4486]

You can't just subtract inequalities like that... A way of proving r≥s is to show by induction that f(a)ⁿ=f(1)^(n-1).f(aⁿ), then nr≥(n-1)s for all n, so r/s≥(n-1)/n, but letting n approach infinity we get r/s≥1 and therefore r≥s.

[@benkelly2024]

Or slightly simpler, nr≥(n-1)s => s≥n(s-r) for all n => s-r≤0.

[@wesleydeng71]

16:04 This is obviously wrong - you cannot substract them. To prove s= s-s/n for any n. Given any ts/(s-t). Then we have r > s-s/(s/s-t)) = t. So r is greater than any t

[@HagenvonEitzen]

The problem is much simpler for those considering 0 a natural number ;)

[@caesar_cipher]

It DOESN'T MATTER if 0 is natural or not, nothing in mathematics fundamentally changes or any great problem solved / unsolved based on that definition. The domain and co-domain are positive integers as Michael says out loud in the problem

[@arnavdeep8396]

Chill, why so serious?

[@richardfarrer5616]

A couple of minor points. in g(p) = A.B you technically need to show that both A and B can't be 1. Easy by injection. At the end the mapping happened to work because g(2) = 3 and 2 = g(g(2) = g(3). If, say. you had g(2) = 3, g(3)=5, g(37)=2 then the product wuld be the same but it would be invalid but g would not be an involution, which you showed it had to be.

[@theartisticactuary]

Not only were there errors here (like the subtraction of inequalities) but this seems to be at about double your normal feed, making it really difficult to follow. One minute we're solving quadratic equations very slowly using a formula, the next minute I'm having to keep pausing the video.

[@ethancheung1676]

11:11 how?? I failed to see how to sub it back and pull out the f(1)

[@mackenziekelly1148]

In the second line he substituted ab for a, making the second line turn into f(ab)f(1)=f(abf(1)). f(abf(1)) was also the result from the first line, so he set them equal, making f(a)f(b)=f(ab)f(1).

[@Blabla0124]

typo at 25:44: smaller or equal should be bigger or equal

[@wesleydeng71]

For those who want a concrete instance of function g, here is one. g(1)=1, g(2)=3, g(3)=2, g(5)=37, g(37)=5, then arbitrarily pair the rest of prime numbers together such that g(p)=q and g(q)=p, and for composite numbers apply the multiplicative rule. One can verify this function g() satisfies the conditions given by the problem.

[@jitzukinanaya4626]

at 19:43, I don't think it is that obvious to have g(g(a))=a here, while your only proposition is f(a)f(b)=f(1)f(ab), by which you define the g(a)=f(a)/f(1). then you still have to do some steps to get g(g(a))=a first. and only if you already have g(g(a))=a, can you verify that g(.) as well satisfies the equation (1), hence g(.) is one of that function family.

[@yurenchu]

Actually, we can show from f(a)f(b) = f(1)f(ab) and the definition g(n) = f(n)/f(1) that g(.) does satisfy equation [1] ; and from there, we can then easily derive the self-inverting property g(g(a)) = a . So yeah, I agree that he should have presented some steps first before mentioning g(g(a)) = a . Below are the necessary steps: - - - - The equation that Michael eventually found is f(a) f(b) = f(1) f(ab) This equation also holds for g , which can easily be seen by dividing both sides by f(1)² : f(a)/f(1) * f(b)/f(1) = f(1)/f(1) * f(ab)/f(1) g(a) * g(b) = g(1) * g(ab) which, with g(1) = f(1)/f(1) = 1, reduces to the multiplicative identity g(a) * g(b) = g(ab) Now we can show that the original equation [1] also holds for g , as folllows: f(n² f(m)) = m f(n)² [1] Multiply LHS by 1/f(1) and RHS by f(1)/f(1)² : f(n² f(m))/f(1) = m f(1) f(n)²/f(1)² g(n² f(m)) = m f(1) g(n)² Substitute f(m) = f(1) f(m)/f(1) = f(1) g(m) in LHS : g(n² f(1) g(m)) = m f(1) g(n)² Apply multiplicative identity on LHS, with a = f(1) and b = n² g(m) : g(f(1)) * g(n² g(m)) = m f(1) g(n)² Now what is g(f(1)) ? g(f(1)) = f(f(1)) / f(1) , and according to [1] with n=1 and m=1 we have f(f(1)) = 1*f(1)² , hence g(f(1)) = f(1)²/f(1) = f(1) . Back into the equation where we left it: f(1) * g(n² g(m)) = m f(1) g(n)² Divide both sides by f(1) and voilá : g(n² g(m)) = m g(n)² So indeed g satisfies equation [1]. With n=1 and g(1) = 1, we can then derive: g(1² g(m)) = m g(1)² g(g(m)) = m * 1² g(g(m)) = m which shows the self-inversion property of g. I hope that helps.

[@thibaultlabatide_alanore9456]

I haven’t looked at the answer but does N not include 0 for Americans? Otherwise f:N-> {0} would be the trivial answer right ?

[@moshadj]

He used the N symbol for natural numbers, but said positive integers multiple times (one commonly accepted "definition" of the natural numbers).

[@goodplacetostop2973]

Usually Michael doesn’t consider 0 as a natural number in the videos... sometimes yes. More about his views about 0 in N or not : kzbin.info/www/bejne/nnO1p2ikj9-esJYm55s

[@caesar_cipher]

It DOESN'T MATTER if 0 is natural or not, nothing in mathematics fundamentally changes or any great problem solved / unsolved based on that definition. The domain and co-domain are positive integers as Michael says out loud in the problem

[@gustavowadaslopes2479]

@@caesar_cipher "It DOESN'T MATTER if 0 is natural or not, nothing in mathematics fundamentally changes or any great problem solved / unsolved based on that definition." It literally does, because depending on which you consider, a problem writen as Michael wrote would have a whole new number that would change the entire result.

[@caesar_cipher]

@@gustavowadaslopes2479 Please understand the difference between GENERALIZING and TRIVIALIZING. If someone generalizes the problem to say expand the domian/ co-domain to all integers or rationals or reals, then its commendable. Adding a ZERO to "natural numbers" TRIVIALIZES the solution. It adds nothing to the problem - also it would then not be a problem for any math contest, leave alone IMO P6. All it does is spam the comment section

[@moshadj]

For the very last line wouldn't g(1998) be greater than or equal to the final expression (as those were its smallest values). Equality still holds when it is equal, but is it trivial that such a satisfying function exists?

[@konraddapper7764]

Yes but you can explicitly construct a function h with h(1998) =120 h is the function that swaps all prim factors( 2 and 3) and (37 and 5) of its input h(1998) =120 And it satisfies equation 1

[@leickrobinson5186]

To expand on what Konrad said, you can easily show that a function that starts with a prime factorization of the input and swaps prime factors 2 and 3, and also swaps prime factors 5 and 37, *and leaves all other prime factors alone*, will satisfy the original equation and evaluate to 120 at 1998. Whew!! That was a tough problem!

[@tomatrix7525]

Wow. Great problem but I could’ve never known to try these specific things

[@adandap]

I have more chance of running a sub 10 second 100m in the "other Olympics" than I have of solving that problem under test conditions!

[@accountname1047]

You get the same result but in the initial calculation I think taking n=a and m=f(b^2f(1)) is more straightforward to see when using equation (1), since you're working over N.

[@ВасилийТёркин-к8х]

so any function satisfying the equation is a natural number multiplied by g(x) which is multiplacable and self-inversively permutates prime numbers

[@egillandersson1780]

I don't understand at all at 6:10 ! Even reading the comments below ... Some say you make a mistake, some other seem understand ... Can you explain please ? It would be sad to get stuck at the beginning of this long video. Thank you in advance.

[@keysh618]

I think the correct usage in the equation should be n=a, m=f(b^2 f(1)).

[@avnertishby]

It's a mistake. Skip that expression and get directly to the next one using the substitution others have mentioned.

[@yurenchu]

He makes some sloppy writing mistakes and takes erroneous steps. Here is the proper derivation: f(a)²⋅f(b)² = ... apply [3] with n = b ... = f(a)²⋅f(b² f(1)) ... apply [1] with n = a and m = f(b²⋅f(1)) ... = f( a²⋅f(f(b²⋅f(1))) ) ... apply [2] with m = b²⋅f(1) ... = f( a²⋅b²⋅f(1)⋅f(1)² ) = f( [a⋅b⋅f(1)]²⋅f(1) ) ... apply [3] with n = a⋅b⋅f(1) ... = [ f( a⋅b⋅f(1) ) ]² ⇒ f(a)⋅f(b) = f( a⋅b⋅f(1) ) I hope that helps.

[@AW0348]

At 6:45 I don't understand how equation (3) gives you: f(a^2*f(b^2*f(1)) = f(a^2*f(f(b^2*f(1))) It triggered me because I don't see how to get b out of the f

[@yosid1702]

ye i also wonder about 6:44, i dont understandddd

[@shahidafridi528]

I think that is a mistake.

[@yurenchu]

He makes some sloppy writing mistakes and takes erroneous steps. Here is the proper derivation: f(a)²⋅f(b)² = ... apply [3] with n = b ... = f(a)²⋅f(b² f(1)) ... apply [1] with n = a and m = f(b²⋅f(1)) ... = f( a²⋅f(f(b²⋅f(1))) ) ... apply [2] with m = b²⋅f(1) ... = f( a²⋅b²⋅f(1)⋅f(1)² ) = f( [a⋅b⋅f(1)]²⋅f(1) ) ... apply [3] with n = a⋅b⋅f(1) ... = [ f( a⋅b⋅f(1) ) ]² ⇒ f(a)⋅f(b) = f( a⋅b⋅f(1) ) I hope that helps.

[@byronwatkins2565]

At 6:17 I don't see how you can turn n^2 f(m) into anything except [inverse f(m f(n)^2)].

[@valerysmague9628]

I just have one question - except g(a)=a, are there other functions that satisfies g(a)g(b)=g(ab) and g(1)=1 ? Only solutions in R are exponentials and it looks to me that there are none others than Identity in N as well - any clue ?

[@daniellosh8341]

6:50, why f(f(b^2f(1))) = f(b^2f(1))?

[@movax20h]

Nice shirt. But the usage of equation 1 at 6:00 is weird. I don't see this substitution.

[@DarkOne713]

18:18 How do you prove g satisfies (1)? g(n²g(m)) = f(n²f(m)/f(1))/f(1) = ..... ????? ...... = m.g(n)²

[@yurenchu]

If g is a multiplicative function (which means g(ab) = g(a)g(b) ) and also self-inverting (which means that g(a) = b implies g(b) = a, or in other words g(g(m)) = m ), then g satisfies equation [1]. This can be easily seen by applying the multiplicative identity onto the LHS g(n² g(m)) and showing it to be equal to g(n)² m . g(n² g(m)) = ... apply multiplicative property g(ab) = g(a)g(b), with a = n² and b = g(m) ... = g(n²) g(g(m)) = g(n)*g(n) * g(g(m)) ... apply self-inverting property g(g(m)) = m ... = g(n)² m

[@Aramil4]

In the end, don’t we need to show such a function g yielding 120 even exists? How do we know? I feel we’ve only proven that IF it exists then for sure it’s the minimizing one.

[@benkelly2024]

g is a self-inverting multiplicative function. It follows that g(n^2.g(m)) = g(n)^2.g(g(m)) = g(n)^2.m as required (using first the multiplicative and then the self-inverting properties of g). So g is one of the family of functions being considered.

[@karlbernhard4656]

What is about f(x)=0? I wasn't able to follow the whole proof and can't see my mistake.

[@shafikbarah9273]

Is 0 in lN ?

[@jerrymouse3420]

Hey Michael,i solved it in a very tricky and different way At first,i utilized the given condition to prove that f is an injective map.Then i derived the condition that f(1)=1...(A) Using these two results, i proved that f•f (f composed with itself) is just the identity map on the set of natural numbers N.....(B) After that, it was very easy to derive the condition f(mn)=f(m)*f(n) for all natural numbers m and n.....(C) Then, using conditions (A), (B) and (C), i derived the condition that a natural number a divides another natural number b if and only if f(a) divides f(b)......(D) Thus,1998= 2*37*3^3 So, number of positive divisors of 1998 = tau(1998)= 16 ( using the known formula to compute tau function) Now,using the condition (D), we can safely conclude that f(1998) is the least positive integer with 16 divisors and this turns out to be 120,which is the final answer.

[@alexeydrobyshevsky3375]

You have proved that for all g satisfying (1) g(1998) is greater than or equal to 120. But you did not prove that there in fact exists a function g that satisfies (1) AND such that g(2)=3, g(3)=2, g(37)=5. So you can't claim to have found the actual minimum.

[@gustavowadaslopes2479]

I'm afraid he didn't. At least not if 0 is considered a natural. If zero is a natural number, the answer is 0

[@zprmscorner1769]

You have to realize that a g function is completely defined by its value at prime numbers. So you can pick a g function with g(2) = 3 and g(3) = 2 and g(37) = 5, it must also have g(5) = 37 for involution, all other values for primes are free, with the constraint that g(prime) is prime', and g(prime') is prime, but that has no impact on g(1998)

[@exopsykearyvous436]

@@zprmscorner1769 I agree with this argument, but would like to be a bit more precise in how it plays out to others that are curious. Notice how the last two yellow properties given at 19:46 imply that the function g must satisfy (1) (this is left as an exercise to the reader). In addition, notice how the existence of g and f are linked (if one exists, then so does the other). With this in mind, it should be easy to define a g such that the two yellow properties are satisfied, so the function g as given does certainly exist. Consider the following piece wise function of g. Let g(a) = a when a does not have prime factors of 2, 3, 5, or 37. If a does have these factors, with w, x, y, z as their exponents respectively, then let a = 2^w*3^x*5^y*37^z*k. Let g(a)=3^w*2^x*37^y*5^z*k under those conditions. It should be easy to see that this example g satisfies those 2 properties.

[@gustavowadaslopes2479]

He did prove g always satisfices f, he just didn't explain it to us.

[@alexeydrobyshevsky3375]

@@zprmscorner1769 The facts that (a) g is completely defined by its values at primes and (b) those values can indeed be chosen arbitrarily must be proven, or at the very least, explicitly mentioned, if we're to call this a complete solution.

[@Nick-kg7sk]

Why is there a g such that g(3) = 2, g(2) = 3, and g(37) = 5?

[@lemousquetaire]

i think you mean at the end that the min of g() is less or equal To ..and not g

[@natepolidoro4565]

6:00 yea n should be equal to a instead of f(a)

[@clilhuseynov1364]

What do you think, Karabagh is Azerbaijan or Armenia?

[@iamrepairmanman]

If 0 is considered a natural number, why can't f(x) just be 0 or x-x?

[@HVH_49]

Very nice, the sixth question is usually the hardest on the test

[@nicolascalvisi3359]

Ans if se juste use the 0 function (f(n)=0 for all n in N).

[@JalebJay]

I didn't consider changing the prime values of f(p) being anything equal to anything but p. Seems a little weird.

[@DubioserKerl]

If 0 was included in the domain and codomain of f, f(x) = 0 would have been a valid answer, since f(n^2 * f(m)) = f(n^2*0) = f(0)= 0 = m * 0^2 = m * f(n)^2

[@gustavowadaslopes2479]

Is zero considered a natural? Because if that's the case, the minimun value would be 0

[@gustavowadaslopes2479]

@@angelmendez-rivera351 Thanks for the polite answer. I guess I should check this Q&A to avoid other possible misunderstandings.

[@CommanderdMtllca]

Quick question: on the RHS is the power as f(n^(2)) or (f(n))^2? Maybe I missed it in the video

[@wasserstein5110]

Hello, I dont follow when you give your last argument that we can pick the values of g to be any primes. Do we know that then equation (1) is still satisfied necessarily?

[@GooglerPraveen]

Why can't we have f(x)=0 for all x?

[@reeeeeplease1178]

Depends on the definition of natural numbers but defining 0 to be a natural number would defeat the purpose of the question

[@natepolidoro4565]

6:38 This makes no sense to me

[@chandramoulikamanchi8520]

Is g it's own inverse?. g is 1-1 from the definition and the proof still goes through

[@saroshadenwalla398]

I don't think g is its own inverse At least not from what he showed but how do you show it's 1-1?

[@chandramoulikamanchi8520]

@@saroshadenwalla398 g is 1-1 iff f is 1-1. It is straight forward to show this from identity (2) or (3) in the video .

[@MrRyanroberson1]

wait a second. you find the condition f(abf(1)) = f(1) f(ab), is this not already sufficient to show that f(1) = 1? though i suppose that's only guaranteed if f(x) = kx

[@HagenvonEitzen]

16:17 Whaat?? Suppose s = 10 and r=9. Then s = 10 = 1+c, so c >= 1.

[@matthiasbergner8911]

There is an argument missing right at the end of your proof. You assume g(3)=2, g(2)=3 and g(37)=5. But you still need to show that such a function g:N to N with these values and the required property exists!

[@benkelly2024]

Any self-inverting multiplicative function trivially satisfies the required property.

[@gustavowadaslopes2479]

6:20 is very wrong.

[@randomcubing7106]

omg this comment session is filled to the brim with math intellectuals

[@KioillOstapenko]

Why is the function g involutive

[@exopsykearyvous436]

If one is able to accept that g satisfies (1) and that g(1)=1, an easier way of seeing that is by setting n=1 and m=a in the version of (1) with g instead of f.

[@dimy931]

@@exopsykearyvous436 but proving it satisfies (1) becomes the problem. With this proof we can use the fact that it is involution to prove it satisfies (1)

[@ozzydozzy4116]

1. Let's start by factoring 1998. 2. Wait..

[@dimitrisstam7596]

Excellent content

[@brettaspivey]

This one was pretty botched, better try again

[@rijubhatt8366]

Hi Michael! I request you solve IMO 2001 Question 3. 21 Boys and 21 girls.

[@buxeessingh2571]

Surf Arrakis? I hope you have some Spice handy.

[@movax20h]

:) noticed too. Just as i am rereading the book.

[@rafael7696]

Hard and beautiful problem

[@minh9545]

Finally an IMO. Ive been waiting.

[@nagmeldinal-farabi7450]

Yes

[@hellosquirrel7271]

Hey, you have a typo in your video title, *International*. Edit: ^_^

[@bravobessa3684]

Error on the subtraction 16.10

[@ClashofWizards-hh2wu]

8:49 look at the board. who would try to do this

[@rafael7696]

At the end, the inequality is g >=

[@ТарасЛисецький-м5к]

why g(g(a))=a?

[@E1Luch]

g(a) satisfies the original equation (1) which is g(n^2g(m)) = mg(n)^2, and evaluated at n = 1 it gives g(g(m)) = mg(1)^2 = m. He didn't prove that g satisfies (1) but it's relatively easy to do knowing that f(a)f(b) = f(1)f(ab)

[@E1Luch]

​@@angelmendez-rivera351 No, the idea is to use already proven "almost multiplicativity" for f to show that g, which is defined in terms of f, satisfies (1)

[@E1Luch]

@@angelmendez-rivera351 I saw direct proof of involution somewhere in the comments, but you can use practically the same steps to prove what I am saying. I can write my proof here, I hope I didnt fuck it up lol

[@E1Luch]

​@@angelmendez-rivera351 g(n^2*g(m)) = = f(n^2*g(m)) / f(1) --- by definition of g = f(n^2*g(m))*f(1) / f(1)^2 --- multiplying numerator and denominator by f(1) = f(n^2*g(m)*f(1)) / f(1)^2 --- using f(a)*f(b) = f(a*b*f(1)) with a = n^2*g(m), b = 1 = f(n^2*f(m)) / f(1)^2 --- by definition of g: f(m) = g(m)*f(1) = m*f(n)^2 / f(1)^2 --- using eq. (1) = m*g(n)^2 --- by definition of g. edit: even shorter proof

[@E1Luch]

​@@angelmendez-rivera351 So if I am correct its not exactly trivial but not so brutal either. But yeah, I wish he didnt skip these steps in the video

[@arvindsrinivasan424]

My god... this problem... is brutal

[@gustavowadaslopes2479]

If 0 is considered natural: f(n².f(m))=m.f(n)² f(a)=0 => f(0)=m.f(0)=0

[@caesar_cipher]

Here we go again with the redundancy of an argument about 0 being natural or not. It DOESN'T MATTER if 0 is natural or not, nothing in mathematics fundamentally changes or any great problem solved / unsolved based on that definition. The domain and co-domain are positive integers as Michael says out loud in the problem

[@gustavowadaslopes2479]

@@caesar_cipher When he keeps saying natural numbers in different points, it matters, as people often miss that.

[@caesar_cipher]

@@gustavowadaslopes2479 It really doesn't as he mentions "positive integers" explicitly at the start of the video. But anyway to repeat a comment which I already replied to u in another post (since u repeated the comment) : "Please understand the difference between GENERALIZING and TRIVIALIZING. If someone generalizes the problem to say expand the domian/ co-domain to all integers or rationals or reals, then its commendable. Adding a ZERO to "natural numbers" TRIVIALIZES the solution. It adds nothing to the problem - also it would then not be a problem for any math contest, leave alone IMO P6. All it does is spam the comment section"

[@njsh2008]

But more importantly, is this the first appearance of Dune?

[@raystinger6261]

Huh... All those years I thought 0 was a natural number. That was what I learned back in junior high (that's not how we call it in my country but it's somewhat equivalent) and nobody touched on that subject even during my entire Math graduation. S***, I gave classes saying that 0 was a natural number! One mistake by one teacher almost 20 years ago was only corrected now...

[@lilellia]

It’s really just a choice of convention, and it depends on who you ask (or what textbook you use). Personally, I also take N to be {0, 1, 2, ...}, and Dr. Penn uses {1, 2, 3, ...}. Which is totally fine. Because of this confusion, the original problem probably said “positive integers” instead of natural numbers.

[@grinpisu]

Very interesting! I think "0" can't be a natural number, because it was "invented" just for the position of the digits in a number (4, 40, 400, etc.), thus in the nature, 0 something means there isn't such a something. On the other hand, in every point of the nature there are an infinite number of 0 entities, meaning no entities. At the end it's just a convention, like in the above answer.

[@richardfarrer5616]

If you go by the original approach, then natural numbers started from one. No point counting zero sheep in your flock - you just don't have a flock. If you go by modern convention with them underlaid by set theory then you equate 0 to the empty set and work from there, so 0 is included.

[@djvalentedochp]

rough question I'd say... good video

[@roboto12345]

I thnik you cant subtract inequalities...

[@gustavowadaslopes2479]

There is a case where you can. a

[@gustavowadaslopes2479]

Mind you, the way he did it on the video is wrong

[@caesar_cipher]

@@gustavowadaslopes2479 Wrong example to counter a wrong example. By your "logic" u should try to prove a-c < b-d, because thats what Michael did wrong in the video and which u are saying is possible in some cases. Its obviously possible in some cases but u messed up the logic, as also your other comments in this section

[@gustavowadaslopes2479]

@@caesar_cipher I was answering a comment saying "I think you can't subtract inequalities", showcasing there is a case in which it happens. Nowhere in the comment I answered there were expecifications or restrictions to only how it was done in the video. Following that, I pointed in another comment that the way he did in the video is wrong, as you have noticed. There was nothing wrong with my first comment for you to be calling it out.

[@dodokgp]

Lots of ffffff....ing in this problem :D

[@tomatrix7525]

Mother of God this one was long and tedious

[@ericlabrique]

Am I the only one that doesn't find obvious that g satisfy (1). In fact, I can't prove it.

[@annebouillon8036]

It took me a while to solve it. Here it is: 1 (easy): show that g is multiplicative. 2 (easy again): show that g(f(1))= f(1). 3. Then compute g(n^2*f(m))= 1/f(1)*f(n^2*f(m))= 1/f(1)*m*(f(n))^2=f(1)*m*(g(n))^2. But g(n^2*f(m))=g(n^2*f(1)*g(m))= (multiplicative) g(f(1))*g(n^2*g(m))=f(1)*g(n^2*g(m)). You divide by f(1) and then have the result.

[@ImaginaryMdA]

Oh, I messed up, I somehow proved that f(1)=1... which isn't true... Damn it.

[@goodplacetostop2973]

More and more comments about 0 being a natural number or not so I think it’s a good place to post Michael views on that matter : kzbin.info/www/bejne/nnO1p2ikj9-esJYm55s

[@MTd2]

Michael, can you do all International Math Olympiads? There's a channel that does it already, Osman Nal kzbin.info . But it would be awesome to see your take on them, because each person has a different way of solve.

[@bollyfan1330]

s

[@bollyfan1330]

In fact if s

[@thephysicistcuber175]

The answer is zero. Next please.

[@JoKer-vu1zh]

Is he American?

[@elaadt]

You lost me at the first hint

[@clilhuseynov1364]

İt is not suitable with video. But I want everybody be about this(sorry for my english)

[@clilhuseynov1364]

For know true answer, you must be history. I think everybody know about Armenia terrorizm.

[@daniellosh8341]

6:50, why f(f(b^2f(1))) = f(b^2f(1))?

[@yurenchu]

That step appears to be in error, and must be simply skipped/ignored. If you look closely, the previous step and the next step (with m = b^2 f(1) ) follow directly into eachother.