these questions can actually sometimes be found in highschool entrance exams. crazy how we’ve learned all this.

A more concrete answer would be 1/393793903937577106479301743414185865462150020261045958348688864012331731304516123198845761433324488438219593808570961852064612014780620624672175445445623382895917883727710633016494081643796401279080059029404279015592123445442355194726188606869002467332145960773819828320146098500719543708842828909672937331352454838564658760130686616706686521039026472870725835633911635732397771739426937534540098162234792530262277400952960260275149179804966675094176984508713046317788443292024044223680238386586329406803944254718916260871815935375111696289754445942035497936545125868564313977047665770022714319377213978160622697269487278873119845809917406721550071458221311050676531396693901681843064092887169936333415826855426013543139044970705738911781672691422364080471999492019809282504071796204154517495882790498393867199563541419863517221565959261739686501178959417759309568516456104708920229992395230228903953063833065986161681444648022574918434659891914833112053428143196925629879411752075401355532544017417476788308017152966790564260548928897478386312037770428673468613582763168614277948609278870006347055324532210630733323281118380011252615931646701004690182075032123767325437226004873702559515067964070718560797194154998855092813980238968034648305431279385781862205048865127192139055595268009315185126800025295469023394604199061086359254433212349799318521040427532666989236907747701171802346780593669488132594120937585143570489494597412015258973495937532371308552027673665862452462496996340353575649988424249535811860120802497285563561335755572798634944409538335137253766015557723887443982767984207525075962600447534996987404419602791575303516970031347030489168581235574850237385515774445711182351283425539377557090153111789115365503761779089069170411660638323033599411217957676143335096505791910934533669664902763885344507187253341645042322357702417368822373002645986021656797833144415930392123612634034531334528128279264226280389919144186148073433090356229629574487001609945588996700009857931190244582787781981877820785356558637277173725635202063621762415552708860330373067798892396051070149086816524225341526084130866750847634080515977399218344052153044129811988865625708895172247970818244717511475389102792431715060679839217756732955249425619632592548104964162574028939416200191589900040276284996423183580269537829301856916783456589435696218814439711066029936512328512051652967106102752720230021501058568859576165644361305619324532969927181691887792263541733558947644497455771801727310607803753021989435677497179468397332031842874581636660316250318856415094613696232537028140098672438652578876667706317339702327770587307597082164866745269924314810790854857409922099931888208876920474910109408204469258308185663826674345827477443684328744712685776923850147437077008409008570138828158346805197980277364705951140387226569743797152383910274271531809109053141501420405008364306215728862473281280982903067496264707023676950228041397641708182387456387192125261922049660728334893434721711554690113649751953762093719649544106259506282234544368417844344916848902517657155685153361558834810291149212214391780357471530032013708218274165571182212666485358769490050426743923310358555482637184144991176745266924632888615386855771520624921929239357960686134080168753708870543782562454540905771134388971688342011522148286951383041405906792468715329987262184997340193718566532878070026244749756460332336440365671247018928365095159372452799133428334342520366099906761182526083567108201141288588724104232238357914686143195372407715037434209977040415079022412854911606517805279357734165915680428312491934938472712791319580956496831212171879918029774311776051369728473165255068483483066931996186347817761123025174218896915546274705502111123668819605957959687155582900219489982983060785692794814528133977551149181329523978870287909448589934972634913413696109945669390906497298702110302827788158757732929471952188976645780904858003262302119529067867009734725419656542228580307526339149805610503254400913020517140751943563039426792010722261237757210999688897482051826499691470761972612659923515867214526010081917608017819635760125117598408088791374527767798380362721651923772243995626328857066137238047500869683279361993297990285234019363341025302107780021780342462141320104632035676056177623067294477786045233140202827641212364911735918289045162519723315587505292207851118406684560762538232991887799490330991049955318927640825822145387807221109421203765448093540126843802557973601665027839307238700114797369398404860298797664371224359632764205544939799250187322420566038334231005428039078708951716390191762973818257167203166588555881429752420253511610115654895377758332885282660436756527741558431433851666754892799862780042186668220619278028300460620808670210170226049422822267856889193656382801651197151367708914970663241547420235743423123679241376599515853093965541673470565769082214045565719402743347871856207715349252073940433454197876400318235192833670976078217422087070910648435845672844035095784780021928512973373711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I started trying to solve this algebraically, but once I realized that one of the three variables has to be negative the solution becomes fairly intuitive.

When the trees start speaking algebruh

These old days, I have learnt from Vietnamese teacher, but it seems your explanation gets more impressive, I can get all the ideas easily. Thanks a lot. The way of education is inspirational.

In Vietnam, these basic solutions you will encounter very often in high school, but they are not as scary as geometry.

The only Imo question i have managed to solve. It is quiet straightforward that 1/a+ 1/b+1/c = 1/(a+b+c) only when two of the letters are opposite numbers.

Ah yes. Vietnamese here. Problems like this made me hate math. I had to solve stuff like this when I was in 11th or 12th grade. My math teachers kept telling us these aren't that hard. That they always have some sort of trick (a cheat trick) to solving them. By the time I was having finals, I had a shit ton of tricks to solve different problems shoved into my head, but understood extremely little about why anything is solved in that certain way. At present, my teachers' impressions remain still and I don't know if this sort of math problem is normal for highschoolers around the world or not. I never truly understood anything. All I received were fast like the wind explanations which I took as scooping the surface and threw it onto my plate. Asking anymore didn't help. None of my teachers helped but I had to pretend to understand. Otherwise, they'd never get off my back. All throughout middle school and highschool, all I was was a machine using formulas without much thinking. That's it. I envy those who do have passionate and understanding teachers who actually teach.

Good solution there is one more way to solve this from second equation we can get (ab+bc+ca) /abc = (1/2022). Let's assume ab+bc+ca=k, hence abc= 2022k. Hence we can form a cubic equation as below. x3- 2022x2+kx-2022k=0 whose roots would be a, b and c. Solving above quation we get x= 2022(a),+i√k(b), -i√k(c) Now putting the value we can cancel out the conjugate terms as i^2023 will be -i and (-i) ^2023 is +i and k^2023 remains common factor hence we will be left with (1/2022) ^2023

Everybody gansgta until you need to calculate the numeric value of 1/2022^2023 by hand.

glad to see some representation for us in Vietnam 🙏🙏🙏🙏

Thanks for posting this fun one!

Ah yes, the easiest math question in my country's primary school entrance exam. I did not answer it correctly and brought shame to my family. All jokes aside, this was one hell of an interesting problem, for all of its intricacy. Thank you Papa Flammy!

Father flammable mathematics I express my sincerest gratitude towards you for this here marvellous solution

Although I had known how to solve this math problem, I am impressed by your method of addressing it. Keep it up sir

Ich hab dich so lang gefolgt und jetzt kam endlich ein Video über Mathe aus meiner Heimat. Yayyy.

I got stuck at factoring when trying this, I sometimes forget that you can even factor expressions like b^2+ab+bc+ca

I did it without solving for a b c specifically(I still found S first by setting a b c to some specific numbers since it's much easier to solve if you know the desired equality). This is probably the most straightforward solution I've seen so far. Multiplying the first 2 equation give (a+b+c)[(ab+bc+ac)/abc]=1 Then multiply out ab(a+b)+bc(b+c)+ac(a+c)=-2abc =>3abc+ab(a+b)+bc(b+c)+ac(a+c)=abc =>(ab+bc+ac)(a+b+c)=abc Raise the first equation to the 2023 then multiply with third equation give S×2022^2023=[(a+b+c)(ab+bc+ac)/abc]^2023 Substitute the equality above and done

I wasn't going to post a comment on Math Olympiad, but personally these, I think are perfect for Robotics or extracurricular activities and G/T classes where students have mastered the States' objectives and need something to do. My experiences as a teacher I found out that when I brought these students who could perform and solve these problems to sit the State Exam, they failed miserably since they overthink the concepts. I run into these problems with NCTM magazines, but I usually read and looked for activities and word problems to be used in class, rather than these types of problems. Just personal pref.

Thanks for the nice problem and such a beautiful solution!

met this type of question in my friend's notebook when he was in 10th grade in a vietnamese gifted school in math major, but it might be suitable for high school entrance exam for the mark-10 question

Very interesting question! Thank your teaching!

The fact this can be a last question for an exam for grade 7th in Vietnamese is funny

I couldn't prove that S is constant by myself, but from the expression of this problem I can tell it's a constant value, so I just found one set of a, b, c that satisfies the conditions (a=2022, b=1, c=-1) and got the answer.

Awesome! Keep it up lad!

Here's another way to solve the problem. Since b+c=2022-a, and 2022(ab+bc+ca)=abc, we can solve simultaneously to obtain b=-2022a/c. Then, we sub this expression for b into b+c=2022-a, and solve the resulting quadratic equation, giving us a=-c or c=2022. But we can easily check that both solutions lead to the same value of S=1/2022^2023.

In fact, mathematicians or even just mathematics lecturers in Viet Nam were truly unsung geniuses. Considering the field of Maths, attention drawn to them must be intensified. Maybe just because of the wars that deteriorate the improvement of education in Vietnam, many people in our countries couldn't stand a chance to reach further contemporary knowledge and invention. Otherwise, the world would probably have had a different view towards the Vietnamese. I really hope that the contributions and dedication of us to Maths will derive respect and reconsideration from the rest of the world.

Thanks for the content this is nice !

I bookmarked this video. Love it!

Flammy, do you by chance have an estimate on the release date of Wuck? 🧐 I want to spend a few weeks wucking with my fellow wuckers.

Ah yes The usage of Algebra-chan Great one, Papa Flammy!

I'm a Vietnamese, I'm proud when I can't solve this problem

I managed to get into Asia Pacific Mathematical Olympiad and that was the day I decided not to major in mathematics. Still love maths though

That chalk has a great sound on that board

pretty interesting sum with a surprisingly easy answer honestly. liked the sum a lot

Also possible to solve with recurrence

(a+b)(b+c)(a+c)=0 means a+b=0 or a+c=0 or b+c=0. which means a = -b or a = -c or b = -c. then we can get that c = 2022 or b = 2022 or c = 2022. and in all, the result still 1/2022^2023.

Can we do it by taking natural log on both sides of the eqn….?

A less rigorous way to do this is to just say that since there are two equations and three unknowns, just assume one of the unknowns is a free variable. (1) a + b + c = 2022 and (2)1/a + 1/b + 1/c = 1/2022 S = 1/a^2023 + 1/b^2023 + 1/c^2023 Lcd(2): (bc + ac + ab)/abc = 1/2022 (c(b + a) + ab)/abc = 1/2022 Let b = free variable = -a (2) becomes: (ab)/abc = 1/2022 1/c = 1/2022 c = 2022 substitute b = -a and c = 2022 to (1) a - a + 2022 = 2022 2022 = 2022 (Tautology) (1/c)^2023 = (1/2022)^2023 = 1/(2022)^2023 and a = -b implies 1/a = -1/b 1/a^2023 = -1/b^2023 Substituting the above to S: S = 1/a^2023 - 1/a^2023 + 1/2022^2023 S = 1/2022^2023

im a vietnamese and damn these problems hit close to home

I remember this when studying for entrance exam into high school for the gifted in Vietnam

3:08 giving us b + c divided by the set of complex numbers ;)

Ah yes, the kind of question frequently given out on the end of Vietnamese junior high school math exams, taking up 5% of the score to gate off students without a math orientation from achieving a perfect 100%.

Great video

I got it through a bit more thought in a manner that made it much faster and easier to solve. I first noted that for 1/a+1/b+1/c=1/2022 if a,b,c are all positive then a>2022,b>2022, and c>2022 which makes the a+b+c>2022 thus I concluded that at least 1 number must be negative. And once I realized that it needed negative numbers I realized that additive inverses work for multiplicative inverses as well thus I got the solution of a=2022,b=2022,c=-2022 which satisfies both initial equations thus since they are raised to a odd power the additive inverses still apply thus 1/(2022)^2023 = S.

I find the product of sums together that adds b + c to be zero to be somewhat suspicious

How can i find the solution to that Gauss trick question in the intro of brilliant please?

Hi, Thank you, It was cool 👌🙏🌹. please make more videos like this 🙏🌹. good luck 🖐

Flammy what is your opinion on umbral calculus? Would you ever make a video explaining these edgy math bois?

Really I have always been terrible at math and this is certainly not helping however I do find these videos fascinating. These problems seem to always be either 0 or 1? So great job I think and continued success. Physics is on a level that alludes me also. And I can't stop watching those videos either.

that's a wucking awesome problem

I'd solved this kind of question few years ago. So it was a piece of cake for me 🔥

I really like this exam so much !

where is the wuck ad smh. also i hope the cold gets better. i got covid last week and i finally recovered lol.

Very nice job. One point to note: it is a symmetric equation, and once you get - (b+c)/(a(a+b+c))=(b+c)/(bc) a solution is b+c=0 and a=2022. Being symmetric, it follows that b+a=0 and c=2022 and that a+c=0 and b=2022.

AM >= GM implies that only one term can be there since (a+b+c)(1/a+1/b+1/c) >=9. Hence either a=-b or b = -c or a = -c.

you can also use vieta's formula for this. For the sake of brevity, replace a with 1/a(and vice versa), do it also with other variables. We have (ab+bc+ac)/(abc)=1/a+1/b+1/c=2022 ab+bc+ac=2022abc Let f(x)=x³-αx²+βx-γ be a polynomial with roots a,b,c. Then, α=a+b+c=¹/₂₀₂₂ γ=abc β=ab+bc+ac=2022abc=2022γ Let χ be any of the roots, then 0=f(χ)=χ³-¹/₂₀₂₂χ²+2022γ(χ-¹/₂₀₂₂) (χ-¹/₂₀₂₂)(χ²+2022γ)=0 Without loss of generality, a=¹/₂₀₂₂ Let b and c be the solutions of χ²+2022γ=0. Either γ0, we have b=-c ⇒ b²⁰²³+c²⁰²³=0 a²⁰²³+b²⁰²³+c²⁰²³ = (¹/₂₀₂₂)²⁰²³

These are the formulas for the series and parallel association of resistors (capacitors) :P

"das is cool" the german coming back with exitement

In olympiades there are often shortcuts you can take. For example here, the question implies that for every a,b,c you get the same value. So just try a simple a,b and c. For example 1, -1, and 2022. Then 1/1-1/1+1/2022=1/2022 So 1/1²⁰²³-1/1²⁰²³+1/2022²⁰²³= 1/2022²⁰²³

Ha, this kind of exam usually appears in the last question of a Maths Test(secondary school) in Vietnam, it's 0.5/10 pts 😅

IMO 2022 is not hosted in Vietnam? Which contest does this problem come from?

Need more Vietnamese IMO questions 🇻🇳🇻🇳🇻🇳

Ahhh, i remember i did this math 7 years ago when i was in high school. Still dont know how to do this and wonder how this help my life

You deserve to get many more geeky subscribers!

One out of a,b,c has to be negative so just consider: a = 2022 b = -2022 c = 2022 and it should work because its an equation and any random number that fits the equations should give you the answer.

i just saw that you divide the first equation by 1^2022 to get the second equation so i divided 1/2022 (2022^-1) by that to get 2022^-2023 which is 1/2022^2023???

Yaaaas competition maths!

As a Vietnamese, looking at this already made me writhed in pain

I used the following 2 homogeneous symmetric identities a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc = (a+b)(b+c)(c+a) and a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc = (a+b++c)(ab+bc+ca) 🙂

Just how do they make these questions...

When you realize this is the question on a test paper of a grade 8 student in Vietnam

Somebody busts this one out almost every year I feel

wha does Vietnamese IMO mean? This year the IMO is in Oslo and it hasn't happened yet

Papa, this is one that Wolfram Alpha didn't solve. At least not on the free version. It said I had to go Pro. I guess this proves you are Pro! 😝🤪😏😎🤓

I still remember this exact question in my 7th grade end of term exam lol

in equation min 4:19 we can say (b+c) = 0 so b = - c , then a= 2022 ; thiway you can go to minute 8:22 and save precious time .... anyway nice and good job

this question can be solved by the "wouldn't be nice" theorem of blackpenredpen

I solved this one like this. After some mumbo jumbo, I got 2 sets of equations 1) a=-b 2) c^2 = -(ab+bc+ac) Now, I had assumed that as a+b+c = 2022 , this the amount of positives must exceed the amount of negatives. This, c^2 comes out to be negative (not allowed). Thus, a=-b, and finally I too got the expression as 2022^-2023. Is this right??

Lol i'm from Vietnam and I can confirm we see this kind of question for bonus mark at primary school and it appear more often around grade 7-9

I have a question: When youve already factored (b+c)[that monster] = 0, couldnt you assume b+c =0 from there already? Or do we still need to factorize the monster to make sure (b+c) wont be cancelled out? Sorry if this is a dumb question 🤣

Would it be incorrect to just use a triple of numbers that obviously satisfies the two equations? Two of the variables set to zero and the 3rd set to 2022 seems obvious and can be checked with two equations to see it’s correct. Since it satisfies those two why would the sum be any other value?

in the part of a+b=0 or b+c=0 or c+a=0, you must use [ instead of {

I definitely didnt get a general solution but found the trivial solution of a=2022 b=-1 c =1 which i guess could have easily been extrapolated for the general but the factor method was much cooler

Yes i have to solve this crap when im 5th grader and a different way of solving, i also have to do many of this type and the number literally dont matter anymore.

im vietnamese though i don't remember having seen this q, it's not a challenging one ngl

Ah yes. I remember this question in my highschool entrance exam. Damm, I didn't think it was this high level mathematics !

3:19: Why not just divide through by (b+c)? Ummnmnnnmm, what if b= -c? Nevermind.

As a vietnamese i suffered from this thing TOO MUCH

Let's put, u=1/a, w=1/b, v=1/c. Denote uw+uv+wv=K. Consider polynomial W(x)= x^3 -1/2022x^2+K*x-K/2022. u,w,v are roots of this polynomial, but also W(x)=(x^2 +K)(x-1/2022). Therefore (u,w,v)=(sqrt(-K),-sqrt(-K),1/2022) or permutations. This immediately gives us u^2023+w^2023+v^2023=(1/2022)^2023

You should consider doing a Vietnamese geometry problem.

Nice maht problem papa 😘

Oh, that math problem in my good student math test grade 8

I'm still crying on how I continuously skipped this kind of questions during my whole school life until now ( I'm a 10th grader btw ) as a Vietnamese *Now I gotta question my ability*

1/2022^2022 is smaller than 1/2000^2000 which is 0.00..... (at least 6600 zeros, and then some). I'd say 0 is close enough as the answer.

at the second step, just put 1, -1, 2022 into the equation and you got the answer.

can't belive that i had to solve this since grade 9

I'm Vietnamese, for me math is easy and I enjoy solving math. But since they switched the form of essay exam to multiple choice, that passion was disappeared. I hate multiple choice.

What's the reference of your blackboard please ?

in Vietnam always give the time value in each year.