1995 British Mathematics Olympiad problem

[@michaelkotthaus7120]

Thank you for this great episode. This problem is really awesome! And I tell you why. Being also interested in music theory and tuning systems, I found a strong relation to this domain. Translated into the language of music, this equation asks: How can I divide an octave (which is the frequency ratio of 2 to 1) into three simple intervals where the nominator is one more than the denominator. In just intonation, such intervals are: octave = 2/1, fifth = 3/2, fourth = 4/3, major third = 5/4, minor third = 6/5, major second = 9/8 or 10/9, minor second = 16/15 or 17/16. For example, the solution a=4, b=5 and c=3 leads to the factors 5/4, 6/5 and 4/3, which are a major third, a minor third and a fourth. If I start at C3, the following notes are E3, G3 and C4. This is a major chord. Great. Another example, the solution a=2, b=4 and c=15 maps to the intervals of a fifth, a major third and a minor second. Starting at C3 the following notes are G3, B3 and C4. For me, there is a heavenly connection between mathematics and music.

[@PrimeNewtons]

I love this!

[@toveirenestrand3547]

Thank you for this comment! I love the connection between STEM (science, technology, engineering, mathematics) and music.

[@gustavoheras557]

Why can't a be 1 ?

[@justcommentator]

This is futile. If you start to calculate intervals like that, you´ll never finish one single piece of music.

[@Ranoake]

@@gustavoheras557 Because that would cause the first term to be 2 and so the other terms would have to be 1 total, and the only way to do that is if b and c were infinity, which is not a real number.

[@quack8550]

"I don't want to write it", proceed to write out more than previously needed.

[@sh0gun570]

absolute god i love the videos

[@4m0d]

''I am gonna write it 😈''

[@jamesnewcomer5963]

There was absolutely no reason to write out all the permutations...

[@ИринаРзаева-ф2с]

А и Б - точные или натуральные единицы, а С - это кислота от кислорода и снова хим реакция и 9.Надеюсь , понял как мои отстающие одноклассники, а меня ругали, что издеваюсь над преподом. Мне нужно обдумать, что сказал этот из Бауманки.

[@cykkm]

​@@jamesnewcomer5963 “… no reason to write out all the permutations” - If you're returning a solution to the Olympiad problem, you should: you won't get the full score otherwise because it would look like you had forgotten your WLOG assumption. You may try writing in words next to the solution that any permutation is also a solution, but IRL I wouldn't take the risk.

[@jlmassir]

I like very much the way you teach. You experiment with possibilities, instead of being smarty and immediately presenting the solution. In this way, the student feels like participating in the solution instead of passively watching it. You're a great tracher.

[@Paddy-ip7qk]

De acuerdo con el comentario. Es odioso leer a tantos otros que al parecer preferirían videos sobre problemas con este grado de dificultad siendo resueltos en dos minutos y usando solo la mitad de la pizarra, como si se tratara de un concurso. Dejen a los matemáticos hacer verdadera matemática, por favor.

[@syamantagogoi]

Awesome reasoning towards getting these solutions.Very much captivating... please continue discussion of such problems.

[@salloom1949]

Great analysis and solution. I have to admit that I viewed this video three time to thoroughly understand your solution. Thank you!!

[@RichardDurn1818]

It's faster to ask an AI for Python code. Lol

[@MaximusU76]

One of the finest math explanation I've ever seen on KZbin. Excelent job. Man you are a real Teacher.

[@borisjaulmes5773]

Never stop learning. I'm 34, out of school since a while, and I loved this video. Thank you ❤

[@cykkm]

I'm 56.

[@Isaac-Playlists]

@@cykkm I am 67 and a mathematician by education since 1981 (but never worked as a mathematician) . I found it eyes opening. Thought provoking task and a great teacher. Thank you very much, Teacher!

[@nishan375]

Your presentation and demeanor is just co calming and engaging. Love from Sri Lanka.

[@TheBoeingCompany-h9z]

(1+1/3)(1+1/4)(1+1/5) is just beautiful

[@londonalicante]

4/3 x 5/4 x 6/5. The frequency ratios between pairs of notes in the sequence G C E G (C major with an extra G in the bass.) The major chord has the frequency ratio 4:5:6 (or 20:25:30) The minor chord has the frequency ratio 1/6:1/5:1/4 (or 20:24:30) (this is based on pure mathematical, "just" chords. To be able to play in all keys without retuning, modern fixed-pitch instruments are tuned in equal temperament, which means all semitones are exactly 2^(1/12) so 4:6 is very well approximated by 2^(7/12) and 4:5 is rather crudely approximated by 2^(4/12)=2^(1/3)

[@OLAFBONDD]

(1+1/1)...(1+1/5)=6 :)

[@Larsbutb4d]

​@@londonalicante i js realized the answer is two (6x4x5)/(3x4x5) cancel 4x5 =6/3 =2

[@plazma5343]

Im sure you have been told, but your voice, rythm and handwriting make for a highly enjoyable and almost relaxing video :) Thank you. Subed

[@Hussain-px3fc]

Love these type of questions and how you explain them

[@patomatma]

The most charismatic teacher that I have ever seen.

[@dougaugustine4075]

That was a fun one. I'm trying to take your advice and to never stop learning so that I don't die.

[@oli111222]

i love your videos and the high quality explenations! Keep up the good work 😊🎉

[@xJetbrains]

The problem becomes much easier when you check that 1.25 ^ 3 < 2, which means one of the numbers must be 2 or 3.

[@tobinpayne3276]

First thing I saw when I tried this problem. Try Q3 from this paper (BMO2 1995): it’s not easy but if you know stuff about roots of equations and basic calculus it’s wonderful.

[@tekbox7909]

Yeah usually a good idea when you see a problem like that is to start plugging in some numbers to see if you can figure out patterns or boundaries like that.

[@pjmoran42]

Fantastic presentation: the mystery, excitement, and guidance are motivational and entertaining.

[@ClarkPotter]

Your handwriting is so good!

[@lindor941]

You are a great teacher and i love your enthusiasm! Here's how i did it: If you set a=b=c=x and solve, you get smth around x=3.8. To balance it, if you make one number larger, you have to make another smaller. One can use this principle to get upper bounds for the smallest of the numbers. Here it means that all solutions must contain at least one 1, 2 or 3. It can't contain a 1 since that's already 2 for one factor. And it can't be (2, 2) or (2, 3) (or (3, 2), to make that clear) for two factors since that's already larger than 2. So now we have lower bounds restrictions for b depending on a. If you set a=2 and b=c=x, you get smth around x=6.5. Repeating the balancing principle and using the restrictions we figured out, you can now set b∈{4,5,6} and solve for c for each case. You get the solutions: (2, 4, 15), (2, 5, 9), (2, 6, 7). Now to set a=3 and b=c=x, you get smth around x=4.4. Using the balancing principle and restrictions again, we can set b∈{3,4} and get the solutions (3, 3, 8), (3, 4, 5) So we have five distinct sets of natural numbers as solutions and we know all other solutions are permutations of those. Figuring out the permutations is kinda trivial.

[@lukamiler5824]

I was going through it a few times confused about how could you possibly just not write out the balancing description. And then I realized you just counted down to A. Nice proof :D

[@dannybarrs]

Excellent. Well done, well presented and just about the right pace to maintain interest.

[@paulpedersen2713]

Hey. Very entertaining problem. Thx! I suggest a quick path: consider [(a+1)/a] ⋅ [(b+1)/b] ⋅ [(c+1)/c], and then use an _ansatz_ where the denominator of the middle term cancels the numerators of the first and third terms (something like that is bound to happen). So try b = (a+1)(c+1), then the problem is to find a, c such that (a+1)(c+1)+1 = 2ac, (i.e.) ac + a + c + 2 = 2ac, or ac - (a + c) = 2. Can we find two positive integers where the product equals the sum plus 2. Sure. Small cases gives you a = 2, c = 4 immediately, and b = 15. We get: (1+1/2)(1+1/15)(1+1/4) = (3/2)(16/15)(5/4) = 16/8 = 2.

[@victorpetrenko2736]

Thank you. Interesting problem for logical thinking. Engaging and even entertaining. Well done!

[@EneldoSancocho]

There's a great tool related to this kind of problems. Whenever a_n is a sequence of real numbers, we consider Sn to be the sum of the terms of the sequence to the n'th power, for example S2=a_0^2+a_1^2... We consider Pn to be a product that generalizes this: P2= a_0*a_1+a_0*a_2.... +a_1*a_2.... Pn is the sum of the different products of n different terms of the sequence. Finally we have this relation: n*Pn=Σ(-1)^(k+1)*P(n-k)*Sk for k between 1 and n, where P0=1

[@EneldoSancocho]

Also, Pn is proportional to the n'th derivative of the polynomial c*(1-x/a_0)*(1-x/a_1)... Which sometimes lets you find infinte sums of powers, if you know the derivatives of the function in question

[@EneldoSancocho]

Pd. You can prove that Σ 1/k^3 = pi^3/32 for k= 1,-3,5,-7,9.... ( if i remember correctly) using this tool, although to be rigorous you would use something like the Weierstrass factoring theorem for the function cos(pi*(1-x)/4)

[@زكريا_حسناوي]

معالجة مفصلة للمسألة، شكرا على مجهودك

[@savonliquide7677]

Hello I did not know your Chanel man you have such a great charisma and your explanations are so smooth and didactic, I wish I could speak like this without any stop or cuts congratulation!!! I have a nice geometric interpretation of this, feel free to present it in your lovely style in a video : let's change the inputs by (x,y,z)=(a-1,b-1,c-1). We get after some little basic operations xyz=2(x+y+z)+6. If we have a prism that sides are x,y,z the left member of the equality is the volume while we see that the write member can unbed in the union f six little cubes plus the union of two triplet of sides, each intersecting in say two opposites vertices (each one on the extremity of the big diagonal) these two triplets do not intersect each other but each edge in a triplet is intersecting another in one little 1x1 cube, so the cardinality of the cube in each triplet is sum of each side minus 2. We have also six more edges and IF we have enough place to put one cube on it plus the 2+2 extras cube that we discussed, we can be sure that 2(x+y+z)+6 will always be smaller than the total volume. That mean we just have to consider the very few cases that you did!

[@CorneliusXI]

Dude, outstanding penmanship! I've tried to improve my own, which would definitely enhance my impact as a teacher, but it seems I've got my limits.

[@jonathanburros6762]

Equivalent to 1 +a+b+c+ a b+a c +b c - a b c = 0 Let a=p,b=p+1,c = p+2 Substitute and by some algebra 6+ 9p +3 p^2 -p(p^2 +3 p+2) = 0 3(p^2 + 3p +2)-p(p^2 +3p +2) = 0 (3-p)(p +1)(p +2) = 0 For natural number solutions p must be positive. p = 3 Then a = 3,b= 4,c = 5 There are five more permutations of a,b,c

[@l_a_h797]

Good, but that only gives you cases where a, b, c are consecutive. What about the others?

[@KermitCyrus]

Multiplying by abc => (a+1)(b+1)(c+1) = 2abc. (Note that a, b, & c can't all be even, and if a = b = c /in R, then a ~= 4 as 5^3 = 125 ~= 128 = 2*4^3. So we're looking for small whole numbers whose product, abc, is somewhat near 64, e.g. a=b=c=~4 and a4, assuming a

[@paulsander4930]

Multiplying with abc gives (a+1)(b+1)(c+1)=2abc, which is also very interesting. Now the question is "Name a product of three numbers which is doubled if the numbers are increased by 1."

[@francescomacheda3721]

Other solution: (a+1)(b+1)(c+1)=2abc. For decomposition, it needs to be for example {(a+1)=b,(b+1)=c,(c+1)=2a} this set of equations leads to a=3 b=4 c=5. There's some other way to factorize, which gives negative or non integer results. The other factorization which gives result is a=b and {(a+1)^2=c,c+1=2a^2}, which leads to a=b=3,c=8.

[@krishnanadityan2017]

Brilliant. No words to describe..

[@uwanttono4012]

AWESOME SH*T!! I just LOVE your teaching style. I learn something new EVERY time!!

[@charlziedouglas-mo7uc]

Now I want to see the calculus version 😁

[@alauddinalam1854]

Me too

[@mr.nicolas4367]

Calculs operates with real numbers in which case there are infinite solutions

[@Lightseeker1-j5p]

The only problem, though, is that you may not know an approximate value of cube root of 2 since you mustn't use a calculator in olympiads (btw approximations aren't exact values or constraints so may sometimes lead to the loss of solutions). Instead, I'd recommend to assume that a>=4. Then 4

[@картонобзоры]

You can estimate it manually

[@brownmold]

I do not understand why writing out the permutations is needed. Multiplication is communicative, and said a

[@arungosavi5698]

Aparantly simple looking problem becomes quite complex.your logic assumptions leads to great solution.explained so nicely that I understood every thing.Great teaching art.

[@Christian_Martel]

Ok people, let’s start over by replacing = 2 by = 3 in the initial proposition.

[@alohamark3025]

Great presentation, coach. It gives me hope that the young people who compete in the Math Olympiad will save the planet.

[@stuartmatthews9108]

Or all go to work at hedge funds and make things worse

[@christopherphelps2326]

I really enjoyed the organized and clear explanation of this problem! Broadening the solution space to all integers, notice that two or three of a, b, and c cannot be negative. Otherwise, if all three of them were negative, we'd have a product of three zero or positive fractions each less than 1, which could not multiply to 2, and if two of them were negative, 1 + 1/a would have to equal or be greater than 8, also impossible. Nevertheless, without loss of generality, we could modify the assumption of a being smallest to a being less than or equal to the other positive integer b, retaining 1 + 1/a as the largest factor in the product, leading to the same bound on a. Then, with the same technique, not including permutations, we can find three new solutions: (a, b, c) = (1, 2, -3), (2, 2, -9), and (1, 3, -4).

[@someperson188]

@3:37: cubr(2) is irrational and 1+1/a is rational. Further, eliminating the case "a = b = c" is actually not necessary. @4:57: As the denominator of a fraction with a "positive" numerator gets bigger and stays the same sign, the fraction gets smaller. @8:43: A correct argument is: cubr(2) > 1.25, since (1.25)^3 = 1.953125. Therefore, a = b - 3 >= a - 3 = -1. Hence, if c - 3 and b - 3 are negative, they both must be -1, which doesn't work. @19:02: 6 could have two negative integer factors. But c - 2 >= b - 2 >= a - 2 = 1.

[@CharlesAbernathy-u6r]

We are waiting for your full course on calculus. The Lenard online courses come to mind. Viewing his calculus lessons could explain what we need. Understanding what you are doing is more important than just being able to perform an operation. People only remember so much. Knowing how to get to solutions solves the memory problem. We are talking about starting with the fundamental theorem of calculus and moving through multivariable calculus. Perhaps I should speak for myself. I could greatly benefit from a review of the basics and learning all the rest of calculus. Thank you for your assistance.

[@cykkm]

Have you looked at the "Essence of Calculus" playlist by 3blue1brown?

[@TurquoizeGoldscraper]

Lovely solution. It was fun to watch you.

[@renforna]

Beautiful problem and beautiful solution!

[@giovannilorenzi8104]

Pure joy hearing you... it all sounds easy

[@BRUBRUETNONO]

Brilliant explaination ! Thanks and greetings !

[@samcopner8661]

Another way to find the upper limit is to consider 2

[@dazedheart9006]

16:40 “I don’t want to write it…” “I am going to write it.” *sudden cut to board being completely filled*

[@herbertsusmann986]

You sir are a genius. This is quite an unusual problem indeed!

[@brandonk9299]

I agree that this is a very interesting problem. While I enjoyed your presentation, it nagged at me that there must be a more number theoretic solution. This is what I found. (1+1/a)(1+1/b)(1+1/c) = (a+1)(b+1)(c+1)/abc = 2 We know that for c>1, it does not divide c+1. So we can safely assume that a, b and c divide the alternate terms. So that: (a+1)(b+1) = ck; (a+1)(c+1) = bm; and (b+1)(c+1) = an. It can be a little difficult to see at first. In the first equation, we know that c doesn't divide c+1 but the factors of c must divide a+1 and/or b+1. The 'k' represents the amount left over. Multiplying: [(a+1)(b+1)(c+1)]^2 = abc * kmn. Since we know [(a+1)(b+1)(c+1)]/abc = 2, if we divide both sides by (abc)^2 kmn/abc= [2]^2=4 so kmn = 4abc Interesting, but how does that help us? Looking back at the equations in , we can find the ratios (a+1)(b+1)/(a+1)(c+1) = ck/bm = (b+1)/(c+1) so kc(c+1) = mb(b+1) and similarly, kc(c+1) = na(a+1) and mb(b+1) = na(a+1) This implies t = kc(c+1) = na(a+1) = mb(b+1) All the values are related. Can we find t? What if we multiply? t^3= kmn*abc*(a+1)(b+1)(c+1) = [kmn/abc] * (abc)^3 * [(a+1)(b+1)(c+1)/abc] and substitute '2' using and then t^3 = 4 * (abc)^3 * 2 = 8 * (abc)^3 = (2abc)^3 so that t = 2abc Now we are very close since can look at the relations in na(a+1) = t = 2abc so n(a+1) = 2bc = na + n and from n + an = n + (b+1)(c+1) = n + bc + b + c + 1 = 2bc so n = bc - b - c - 1 But how do we use n? Going back to an(a+1)/abc = 2 changes to a = 2bc/n - 1 = 2bc/(bc - b - c - 1) -1 since a>0. Further this rotates for all the terms so b = 2ac/(ac - a - c - 1) -1 and c = 2ab/(ab - a - b - 1) -1 This now defines all the solution sets. We can test it using a known solution (3,4,5) = (a, b, c) a =3 so b = 2*3*5/(3*5-3-5-1) - 1 = 30/(15-9) - 1 = 5-1 =4 and c= 2*3*4/(3*4-3-4-1)-1 = 24/(12-8) - 1 = 6 - 1 =5 For a = 1, the values are negative but still works. (1, b, -b-1) => 2*(b+1)/b*[-b/-(b+1)] = 2 For a = 2, b = 4c/(2c-2-c-1)-1 = 4c/(c-3) -1 here c=5 or 7 gives b = 9 and 6 3/2*6/5*10/9 = 2 and 3/2*8/7*7/6 = 2 This doesn't guarantee that every number 'a' has a solution. a=13 b = 26c/(13c-13-c-1) - 1 = 26c/(12c-12)-1 there is no value for c that would make this work. Every solution set should satisfy .

[@emanuellandeholm5657]

I tried the simpler problem with just two variables. Solns are 2, 3 and 3, 2. Before watching the video, I expect a factorization like (a - 1)(b - 1)(c - 1) = something

[@kateknowles8055]

This is an excellent tip: Try a simpler situation first. Also works outside the maths class sometimes. Builds skills and useful confidence.

[@alexgarcia77]

I like very much how you solved the problem except for the square root of two, which requires a calculator. Instead you could have tried calculating small numbers' cubes and see that (1+1/4)^32.

[@usikpa]

If a=2, we have b+1/b = 4c/3(c+1), which works out to c = (3b+3)/(b-3) = 3+12/(b-3). Given the constraints, easy to see that b

[@peterreali3950]

Excellent Problem, I got the 3,4,5 but missed the 3,3,8 solution but was suspicious that there might be other solutions but was spending too much time on it. I wonder was this problem given under a test scenerio where time was limited? It is very challenging but a great exercise in logic. Unlike some of these other Olimpiad videos which are much less interesting.

[@pojuantsalo3475]

My first try to solve this was to multiply both side by abc: (a+1)(b+1)(c+1) = 2abc, but this didn't lead nicely to anywhere. Then I did exactly the same thing that was done in this video to solve for the possible values of a = 2 or 3 (given a ≤ b ≤ c). This gives (1+1/b)(1+1/c) = 4/3 when a = 2 (1+1/b)(1+1/c) = 3/2 when a = 3 From there my solution was different from this video. I solved the equations above as mini versions of the original problem: Since b ≤ c, (1+1/b) ≥ √k, where k = 4/3 when a = 2 and k = 3/2 when a = 3. This gives the possible values for b and the corresponding values of c are calculated from b: c = 3*(b+1)/(b-3) when a = 2, b = 4, 5, 6 c = 2*(b+1)/(b-2) when a = 3, b = 3, 4 This gives the solutions when a ≤ b ≤ c. All the solutions are all the permutations.

[@DrDeuteron]

So what’s the answer?

[@nanamacapagal8342]

What I did is similar to what you did, but instead of solving for c afterwards, I already had the ranges of b ready and just tried them all out. So I still had (1 + 1/b)(1 + 1/c) = 4/3 if a = 2 and 3/2 if a = 3 I checked and if a = 2 then b = 2, 3, 4, 5, or 6. If a = 3 then b = 3 or 4 Then I just tried every setting of a and b available and solved for c instead of coming up with a general form of c So the solutions are: (2, 4, 15) (2, 5, 9) (2, 6, 7) (3, 3, 8) (3, 4, 5)

[@pojuantsalo3475]

​@@DrDeuteron (2, 4, 15) (2, 5, 9) (2, 6, 7) (3, 3, 8) (3, 4, 5) (4, 2, 15) (5, 2, 9) (6, 2, 7) (4, 3, 5) (15, 4, 2) (9, 5, 2) (7, 6, 1) (8, 3, 3) (5, 4, 3) (2, 15, 4) (2, 9, 5) (2, 7, 6) (3, 8, 3) (3, 5, 4) (15, 2, 4) (9, 2, 5) (7, 2, 6) (5, 3, 4) (4, 15, 2) (5, 9, 2) (6, 7, 1) (4, 5, 3)

[@Paddy-ip7qk]

​@@DrDeuteronThere are 27 answers.

[@Jarlaxe111]

I did the same thing and it actually led to the solutions quite nicely. Because you realize that (a+1)(b+1)(c+1) = 2*a*b*c means that adding one to the numbers just introduces another 2 in prime factorization. So at least one of the numbers must be uneven (or else the left side of the equation is uneven), and obv. none of the numbers can be 1. Checking for a = 3 gives 2*(b+1)*(c+1) = 3*b*c. It is again easy to see that neither b nor c can be equal to two. Checking b = 3 gives c = 8 and b = 4 gives c = 5. It remains to prove that there are no further solutions however

[@ronginzler6600]

Very good presentation! With some simple arithmetic I was able to show a and b could not both be 2, and not all three of a, b, c could be greater than 4. Then by writing out the possible answers I found the 3,4,5 solution. I would have found the other solutions eventually, but your way is faster.

[@adw1z]

The only comment I want to make is, people might not know the cube root of 2. Another way to plugging in is to use binomial, we get that cbrt(2) = (1+1)^(1/3)

[@Narennmallya]

For your hardwork and amazing teaching style the least I can do is subscribe 😄

[@angyuekai2747]

How about this perspective. By turning them into improper fraction we get a form like (a+1)/a … , then observe that when the denominator and numerator can cancel each out, that is when a+1/c=2, a=b+1, c=b+1,then we have our answer, by solving1 this linear equation system, we can get our desired answer (3,4,5)

[@wannabeactuary01]

Wow, wow and wow! Super clear explanation.

[@mathewmunro3770]

I think that you can use a < sign rather than a

[@AubreyForever]

I would like to see a way to do this with Calculus

[@maaikevreugdemaker9210]

Wow I would never have thought it would have so many solutions (even wothout the permutations)

[@rahulmmpl]

Beautiful question. Nicely explained

[@EfiXa]

If a=1 then the other two factors should multiply to 1. Then by reducing the equation b+c=-1 which gives as much combinations of b and c as you want...

[@shashankgangwar7687]

Imagine the denominator of the previous term cancels the numerator of the next term. Then the equation can be written as (K+1)/K * (K+2)/(K+1) * (K+3)/(K+2) = 2. which gives 3,4,5 Similarly We can have (K+1)/K * (K+1)/K * (K*K)/(K*K - 1) = 2 This gives 3,3,8

[@razendra2011]

Fantastic proof.Super like and appreciate 👍

[@kateknowles8055]

Good teaching which is fulsome , very clear and easily listened to. We did not say " not equal to" , so (1+1/3)(1+1/3)(1+1/8) = (4/3)*(4/3)*(9/8)= (16/9)*(9/8) =2 also gives us a possible solution if a=b is allowed. a=3, b=3, c=8. This came at 20.00 However a=b=c is covered at 3.00. I missed 3 sets of solutions out of the 5 sets . i become more humble! 9/27 is 33+1/3% aaw!

[@suelingsusu1339]

There are only 5 solutions.... the permutation stuff is not necessary since a, b, c are arbitrary naming we can assign the three numbers in each of the solutions to a, b, c so that a is

[@cantkeepitin]

Treating each solution as a set (which is the task) you have to treat the permutations

[@suelingsusu1339]

@@cantkeepitin ... which results in many sets that have the same elements thus the same sets.

[@okonol]

Sometimes, examiners ask certain questions just to see if the student is stupid enough to waste time on it.

[@d.m.7096]

Awesome solution! Love it.

[@ericknutson8310]

Interesting how there is 3^3 solutions. I wonder if a similar problem can be constructed to have 4^4 solutions. What do you set it equal to? (1+1/a)product through d =?

[@ericknutson8310]

So. You can expand this to 4 variables in the same kind of product and symmetrical structure. To save everyone time you bound a to be 5,4,3,2,1 and b the same but a=\= b so b can be 4,3,2,1. I’m counting out how many combos this is

[@ericknutson8310]

I find 72 solutions for then it goes out to d. Maybe somebody better can stumble into it

[@PrimeNewtons]

This is an interesting question.

[@holyshit922]

2

[@PrimeNewtons]

How did you figure this out? I didn't see it. If I could see this without doing all that work, I'd just brute force it.

[@holyshit922]

@@PrimeNewtons Also brute force but with Python code

[@lindor941]

It is possible to see. If you want one number to be as large as possible, you need to make the others as small as possible. If you figure out the lower bounds for the first two numbers, you get the upper bound for the third. (1, n) doesn't work for any n (2, 2) doesn't work (2, 3) doesn't work (2, 4) could work (3, 3) could work Any other pair would be larger on average. Trying the two possible solutions, you get 15 and 8 as possible upper bounds, meaning that the true upper bound is the larger of those values, 15. I still wouldn't brute force from here. There would be 14!/(3!*(14-3)!) =(14*13*12)/(3*2) =7*13*4 =364 possible ways to choose 3 numbers out of 14 possibilities ignoring the order.

[@ChristianMartins]

That playful smile in the beggining of every video is funny 😆

[@apnakaamkrelala]

20:54 nice quote man 🫡

[@babatundeajibola2589]

This is so sweet to watch

[@abrahamgizaw9313]

the way I did it before seeing yours is like first making the equation (a+1)(b+2)(c+1)/abc=2 and then this will be true if the numerators can be divided by denominator that means a+1=b, b+1=c and c+1=2a and then solve for equation from this three equations for three variables we have and we get 3, 4, 5 as solution for this.

[@patrikwihlke4170]

I just remembered this question yesterdat which I got as homework as a kid… and here it was, suggested to me on youtube 😄

[@SGuerra]

Uau! Que questão linda! A solução dada foi mais linda ainda. Parabéns! Brazil - Julho 2024. Wow! What a beautiful question! The solution given was even more beautiful. Congratulations! Brazil - July 2024.

[@holyshit922]

If you dont know cube root of two you can calculate it with pencil and paper method 2^(1/3)= 1.25... 1 1000| 304 (Triple square of current approximation and append square of last digit of next approximation) 3 * 2 (Triple current approximation shifted one position to the left and multiply by last digit of next approximation) 364*2 (add them up and multiply by last digit of next approximation) 1000 728 (then subtract from remainder) 272000|43225 36 * 5 43225 180 45025 272000|45025*5 225125 46875000

[@wannabeactuary01]

loved this. The bounds were the highlight

[@TheAlavini]

Congrat! Nice teaching class,

[@pacolibre5411]

Is there standard notation for writing an unordered triple, vs an ordered triple?

[@ashutoshpendse4273]

Loved it. Clear and understandable.

[@Kantharr]

That was an amazing way to explain how to solve it. Would it be possible to implement the same techniques for the following question or how would you solve this: x + 1/x + y + 1/y - z - 1/z = 2 where x > y > z

[@laurv8370]

Very nice explanation and video. However, a bit too long/complicate. You could just amplify and use prime decomposition to make all combinations.

[@chrisl.9750]

Came here to learn calculus. Fell asleep peacefully instead. Thank you!

[@daboffey]

The question should be: “find all solutions of …” as otherwise just giving one solution would solve the equation. Is it really necessary to write out all the permutations? This seems somewhat OTT. Surely, giving just one permutation for each solution and then sufficing the phrase, “and all permutations” should be sufficient, or even something like {a, b, c} = {2, 4, 15}, etc.

[@ubk-qm4rz]

In mathematics "Solve the equation" means to give all solutions.

[@smaari]

Great work, thank you.

[@bizikimiz6003]

2 books for my fellow "I like not to involve calculus" problem solvers: Ivan Niven - Maxima and Minima Without Calculus; and Michael Steele: The Cauchy-Schwartz Master Class.

[@PrimeNewtons]

Thanks for the suggestion

[@flipwithin]

Can you show how to do it through calculus

[@gamingshaming3357]

Sir please also make questions of IOQM and RMO and INMO......They so high level problems

[@grazianovenanzoni9858]

beautiful!🤎

[@025437359]

But if we want to calculate solutions with machines,we need a long algorithmic process. Can we do this ?!

[@alauddinalam1854]

We want to see the Calculas version please

[@niravgandhi7357]

Fantastic!

[@PhilFogle]

since the solutions are symmetric, why do you count the permutations? Are they not just permutations of labels?

[@pankaj8ri]

Is there any shortcut solution to this problem ?

[@Arkapravo]

How could you have used Calculus?

[@pinoes31416]

I started writing the equation on a easy way. (a+1)(b+1)(c+1)=2abc then I thought... It would be "cool" if we could simplify some factors, so... I suposed b=c-1 (a+1)(c-1+1)(c+1)=2a(c-1)c (a+1)(c+1)=2a(c-1) ac+a+c+1=2ac-2a 3a+c+1=ac a and c "must" be small. so... try and try... a=3, c=5 works... (3,4,5) but I cannot probe that those are the unic solutions and I do not know why I started with b=c-1 and not a=c-1

[@thomaswan4956]

This question tests the application of equations. It is known that (1+(1/a))(1+(1/b))(1+(1/c))=2. Then (a+1)/a×(b+1)/b×(c+1)/c=2. That is (a+1)(b+1)(c+1)=2abc. Expand it: abc+ab+ac+bc+a+b+c+1=2abc. Therefore, ab+ac+bc+a+b+c+1=abc. Since a, b, and c are all positive integers, Therefore, at least one of a, b, and c is greater than or equal to 2. Assume a≥2, b≥1, c≥1, Then, ab≥2, ac≥2, bc≥1, a≥2, b≥1, c≥1. Therefore, ab+ac+bc+a+b+c+1≥2+2+1+2+1+1=9. And abc≥2×1×1=2. Therefore, ab+ac+bc+a+b+c+1>abc. This is contradictory to ab+ac+bc+a+b+c+1=abc. Therefore, it is impossible for a, b, and c to all be greater than or equal to 2. Then, only one of a, b, and c can be 2, and the other two are both 1. Let's set a=2, b=1, c=1. At this time, (a+1)(b+1)(c+1)=3×2×2=12, 2abc=2×2×1×1=4. Satisfies (a+1)(b+1)(c+1)=2abc. In summary, a=2, b=1, c=1 is the solution of the equation (1+(1/a))(1+(1/b))(1+(1/c))=2.

[@vinceturner3863]

But if you put your solution values into the LHS it equals 6 which is not 2 so a=2. b=1. c-1 is not a solution.

[@thomaswan4956]

@@vinceturner3863 let's re-examine this step-by-step: The original equation is: a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) The given values are: a = 2, b = 1, c = -1 Plugging these values into the left-hand side (LHS): LHS = a ÷ (b + c) = 2 ÷ (1 + (-1)) = 2 ÷ 0 = undefined Plugging the values into the right-hand side (RHS): RHS = (a ÷ b) + (a ÷ c) = (2 ÷ 1) + (2 ÷ (-1)) = 2 + (-2) = 0 Comparing the LHS and RHS: LHS = undefined RHS = 0 Therefore, the original equation is not satisfied, as the LHS is undefined and the RHS is 0. So you are correct - the given values of a = 2, b = 1, c = -1 do not satisfy the original equation, as the LHS evaluates to an undefined value, not 6 as stated. The solution values do not make the LHS equal 6 instead of 2. Thank you for catching my mistake!

[@kianushmaleki]

"... just don't get to infinity... " ❤️😄

[@delalinamamegago5198]

Great man, I salute you