International Math Olympiad | 1998 Q6. 

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

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

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

. . . 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))

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))))"

Exactly.

Is there a fix for this?

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 :)

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

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

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.

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

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

@@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.

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

@@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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-XQzlXaLM_JE.html&lc=UgyrBLKJHyukLJnkY7p4AaABAg.9Ey8kMgDInf9EyHWiXKpUC

@@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.

Illogical comment

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

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

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

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

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

For sure. I like to watch math RU-vid 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.

Blackpenredpen big sad..

Is it a decrease of 64(3-√3)?

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)].

26:34

@@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?

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.

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

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

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

@@Mr5nan Sorry meant n=a

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

@@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.

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

i agree, doesn't seem directly possible

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

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

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

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

Me too lol

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.

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

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

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)))

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}.

@@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))))

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.

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.

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).

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

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 : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-hBRv5nZzkz4.htmlm55s

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

@@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.

@@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

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.

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

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!

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

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

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

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.

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?

@@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 .

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

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)

@@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.

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

@@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.

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

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

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

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

I think that is a mistake.

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.

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

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

@@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"

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

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.

@@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)

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.

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.

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.

There is a case where you can. a

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

@@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

@@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.

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.

In fact if s

it depends, 0 is often included in N

@@angelmendez-rivera351 thanks, good to know

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)

​@@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)

@@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

​@@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

​@@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