Almost an IMO Problem | IMO Shortlist 2019 N2

[@mcwulf25]

I always start these kinds of questions with a=b=c. In this case we get a = cube root of 3. As all three variables can't be higher than this, at least one is lower, which means c=1. Repeat and we get b=1 or 2. Try both to get (3,2,1) and its combos.

[@typo691]

Can you explain why all three variables cant be higher than cbrt(3)? I dont quite get the implication

[@mcwulf25]

@@typo691 Now I look again I see that it's not a given. If the RHS is a constant, yes. But actually we need to demonstrate that it's true (or not).

[@raytheboss4650]

@@typo691 using the AM GM inequality

[@fede9003]

@@raytheboss4650Hi! I know it’s from a while ago, but how would you do this with AM-GM?

[@jyotishka]

@@typo691 cbrt(3) is the upper bound of LHS. Also can we say that abc is the volume of a cuboid which is maximum when a=b=c?

[@AllanKobelansky]

Well done. Well presented. You are in Michael Penn territory. The road to 10k subs looks easily attainable.

[@clarinetowl7554]

there will never be another michael penn. He is far too powerful.

[@browhat6935]

​@@clarinetowl7554the one piece is real

[@tonyhaddad1394]

For an easy way if a=b a^3(a-2)=1 then we do not have a positive integer solution for (a) There for (a) can not equal (b)

[@letsthinkcritically]

Yes, that’s exactly what I meant.

[@moonlightcocktail]

My solution was similar up to 3:00. The equation is symmetric. Wlog sps a >= b >= c. a + (b3 + c3)/a2 = b2c2. this is an integer, so a2 | b3 + c3. Since a >= b >= c, b3 + c3

[@Kettwiesel25]

b3+c3 isn't actually maximized if b=c=√√(3a)though. I see that bc=√(3a), but b=c is actually the minimum over all such solutions.

[@sa0o923]

2:29 how do you conclude a2

[@lucasmcguire1554]

Because a^2 divides b^3 + c^3. To see why this is just look at the original equation to find that a^2 clearly divides a^3 + b^3 + c^3 then get rid of the a^3

[@theuserings]

3:40 why is b >= 2 when c = 2?

[@ericlorentzen9089]

Because of the initial inequality a >= b >= c.

[@jaysy2980]

Why not just remove common factor in 7:44? We can divide them out since a is a positive integer and therefore the expression a^2-a+1 is never 0.

[@MgMG-ig4qg]

I don’t get how we get that a^2

[@reeeeeplease1178]

Just a line above, he got that a^2 is a divisor of b^3 + c^3 and since all of them are positive, a^2 has to be less or equal to b^3 + c^3

[@MgMG-ig4qg]

@@reeeeeplease1178 thanks. What does he mean when he says that a^2 “is divisor” of …. and why can he say it?

[@MgMG-ig4qg]

Applying it to the solution, we that 9 (a^2) isa divisor of 27 (a^3)+ 8 (b^3) +1 which is right, but how do we know? And knocking off a^3 how do we know the inequality still holds?

[@reeeeeplease1178]

@@MgMG-ig4qg well from the original equation we know that a^2 is a factor of (abc)^2 and since (abc)^2 = a^3 + b^3 + c^3, we know that a^2 has to be a factor of a^3 + b^3 + c^3 This implies that a^2 is a factor of b^3 + c^3 (because a^2 divides a^3)

[@MgMG-ig4qg]

@@reeeeeplease1178 riiiiiiight :). Thank you so much….

[@tvvt005]

4:02 wait how did we Ho from c^4 b =32??

[@suhail9511]

c⁴b =2 Then c⁴b>=32 but it must be less than or equal to 18 Hence c cannot be 2 or more

[@adamzoltan1685]

Or (abc)**2 = 36 thus abc = 6 so a and b and and c is less than 6 and you can only divide 6 by 1,2,3,6 you cam just check for them

[@absolutezero9874]

Hi Why is it at 5:07, a² - b | a³ + 1 - a(a² - b) ? Thank you

[@absolutezero9874]

Did you see my question?

[@jm2270]

Because we established a² - b | a³ + 1 and naturally a² - b | a(a² - b)

[@absolutezero9874]

@@jm2270 Oh I see. Thank you This KZbinr doesn’t reply at all

[@Emre67511]

​@@jm2270what does a^2 - b | a^3 + 1 mean ? I don't know if they use different symbols in my country but I don't understand that operation at all ?

[@jm2270]

@@Emre67511 x | y means x divides y.

[@smashliek5086]

3:27 i dont get it, why can RHS exceed 2 ?

[@icebeargt5142]

i think because C > B hence it (c/b) will be less than 1

[@joehollander1810]

@@icebeargt5142*C

[@mumtrz]

Here's a homework: a^b + b^c + c^a = (abc)^abc

[@AyushKumar-ot3zr]

do it urself

[@mumtrz]

@@AyushKumar-ot3zr you think I'd give homework without knowing the answer myself? 🤣

[@KaranSingh-qw7cn]

How to solve it ? Give me a hint

[@amgalanbayarshagdar9435]

This equation has only 3 solutions that [a; b; c]= [2; 1; 1], [1; 2; 1], [1; 1; 2]

[@amgalanbayarshagdar9435]

​@@KaranSingh-qw7cn use equality that a=>b=>c

[@ericlorentzen9089]

How did you determine that (abc)^2

[@DoReMeDesign]

Look again at the question statement and the previous line ;)

[@ericlorentzen9089]

@@DoReMeDesignAh. I got it. Thanks for the reply!

[@rounaksambhwani2623]

man pls can you recomend me any book of maths which starts from basic to this level pls

[@letsthinkcritically]

I learn problem solving tricks by browsing content in Art of Problem Solving (AoPS: artofproblemsolving.com). I usually recommend online resources rather than books nowadays because online resources are better organised and you can find them very easily.

[@rounaksambhwani2623]

@@letsthinkcritically thanks !!!!

[@krishgupta6518]

You can try pathfinder for prmo by prashant jain

[@TechToppers]

@@krishgupta6518 That book just has a ton of problems without solutions. If somehow you get stuck on a problem, game over. Plus it does not have advanced level problems.

[@ShefsofProblemSolving]

Depends on the field. Here are some good books: Number Theory: ->104 Problems in Number Theory by Titu Andreescu Geometry: ->EGMO by Evan Chen Combinatorics: -> Problem Solving Strategies by Engel (old, but has some good explanations of basic techniques) -> Olympiad Combinatorics by Pranav Sriram (though this is advanced already) Competitions to look at (in order of difficulty): -> Junior Balkan Math Olympiad -> Tournament of Towns (especially for combinatorics) -> JUSAMO -> European Girls Math Olympiad -> National Olympiads of countries like Russia, Bulgaria, Brazil, Vietnam, South Korea, China, USA

[@anupamrawat2581]

Amazing solution Hope your channel grows soon Your problems are great

[@turtlez2776]

At 5:07, how do we know a < b^2? Isn't writing that a^3 + 1 - a(a^2 - b) is positive an assumption?

[@fasebingterfe6354]

--- a < b^2 --- We know that the inequality a < b^2 is true , because for an equation M-N , where M and N are natural numbers , If M-N is negative , N-M is positive and vice-versa. This also applies when one of the numbers is squared . You can try this with any two natural numbers that you want. --- a^3 + 1 - a(a^2 - b ) --- According to the inequality a^2 - b < 0 , If we multiply both sides by a , we get a ( a^2 - b ) < a ( 0 ) , which doesn't change what it holds . Coincidentally , a^2 - b | a^3 + 1 - a( a^2 - b ) has the same effect that a^2 - b | a^3 + 1 - ( a^2 - b ) has.

[@cantcommute]

a(a^2-b)

[@turtlez2776]

@@cantcommute understand.

[@Emre67511]

Where does it even say a < b^2 ? Or where does it come into play ?

[@bagasramadhan1677]

why (c/b)^3 Is Equal To 1 ?

[@sender1496]

I'm guessing that since c and b are integers, and c < b, c/b has to be less than 1 and hence bounded by 1

[@Navodayaclass_2211]

By intuition i found abc triples -1,1,1 ordered

[@tonyhaddad1394]

5:45 mistake If (a)=1 then a^3(a-2) is = -1 which not 1

[@letsthinkcritically]

That’s my point, I was trying to say that this would lead to a contradiction.

[@tonyhaddad1394]

@@letsthinkcritically thank u for your reply and yes now i get it 💓💓

[@katagagyi3523]

I feel really dumb but I don't understand how comes in 1+ab>=a²-b😅😅. And also above that part a²-b| a²+1 means that a²-b divides a²+1 or does it mean subtraction?

[@fix5072]

No youre not dumb, he just speaks really fast! The step follows because he earlyier showed that (a^2)-b divides a*b+1 and if an integer divides another integer, the dividing integer is less then or equal than the divided integer. Dont let your head hang down if u dont hear/understand one Thing!

[@fix5072]

Sometimes i also cant catch up in These yt Videos, because its way above the Things we do in class and we dont do any number theory in class as well

[@katagagyi3523]

@@fix5072 thank you, i get it now. :::)

[@katagagyi3523]

@@fix5072 and one more question: if c²b

[@fix5072]

@@katagagyi3523 if youre talking about (c^4)*b =c thus b>=2 and the entire lhs woul be atleast (2^4)*2 which is bigger than 18

[@mra4167]

If negative was possible then, a=1,b=-1 and c=1, can be a solution.

[@boomer1248]

how can we say that a^2 | a^3+b^3+c^3?

[@Jan-vz5ge]

because from (abc)^2

[@boomer1248]

Oh thanks lol@@Jan-vz5ge

[@abirliouk8155]

Im sorry but why 1+ab>a(b+1)-b 6:26

[@marcotosini7156]

because a^2 -b | 1 + ab then 1 + ab >= a^2 -b from a >= b+1 follow that a^2 -b >= a * (b+1) - b (it's simple substitution of a with b in a^2) then 1+ab >= a(b+1)-b

[@prithujsarkar2010]

Can you check this problem out from 1992 AIME Problems/Problem 15 ? The problem states that Define a positive integer $n$ to be a factorial tail if there is some positive integer $m$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails? One of my fav AIME problems :)

[@letsthinkcritically]

Sure thing!

[@mr.kaiden7159]

@@letsthinkcritically when c=1 a³+b³+1=(ab)² (a+b)(a²-ab+b²)=(ab-1)(ab+1) Two condition a+b=ab+1 and a²-ab+b²=ab-1 (no solutions because (a-b) ²=-1) a+b=ab-1 and a²-ab+b²=ab+1 (from this system we get b=2 and a=3) How to find c=1 is insane solutions 🙏

[@dionisis1917]

Please make a platform which we can write our favourite problems

[@letsthinkcritically]

Feel free to let me know here! I will make videos on problems that you suggest.

[@ryanclothier6853]

Why does it become 3a^3 in the beginning and not a^6?

[@ryanclothier6853]

Never mind I’m stupid

[@aditiapermana5074]

@@ryanclothier6853 ayoo broo 🥺

[@randombanana640]

😅

[@theimmux3034]

I woulda not been able to pull of any of those calculations but I just happened to know that the sum of first n cubes is the sum of the first n integers squared and six happens to be equal to the sum of its factors. Does my reasoning also mean that (a,b,c) = (1,2,3) is the only solution?

[@mishania6678]

we don’t know whether a,b,c are consecutive so this formula does not actually works here

[@이름-k8v6v]

The most entertaining math video I've ever watched

[@Wu-Li]

You're a genius

[@mukaddastaj5223]

Thanks for these vids u make! U really do inspire)

[@Mrpallekuling]

A nice problem. I like!

[@syedamashukaashrafi2755]

Have you tried plugging a=-1, b=1, c=0,1?

[@yuthikasenaratne7250]

0 is not an inetiger brother

[@sejr8053]

Its 1, 2 and 3.

[@sparkaks-gr8647]

Wow nice problem and soln too!

[@letsthinkcritically]

Thank you!!

[@alexkonopatski429]

Nice solution! How long did it take you to solve that?

[@Emre67511]

He didn't find it himself. He just writes it down. Why do you think he doesn't explain anything at all except the most obvious things ?

[@andrea-mj9ce]

​@@Emre67511How do you know 3:19 this. He did explain the whole reasoning in the video

[@chrismcgowan3938]

a=1,b=2,c=3

[@fansuli427]

Good

[@mathdetectivej9764]

very interesting problem!

[@HemantPandey123]

Using AM>=GM we get (abc)^2 is < = 3 (Applying Am>=Gm on LHS). Hence all bounds found without any lengthy steps.

[@alanjoelj1591]

Wrong. AM-GM gives 3abc

[@HemantPandey123]

@@alanjoelj1591 But if (abc)^2 is >= 3abc then isn't it contradicts original solution?

[@alanjoelj1591]

@@HemantPandey123 Where is the contradiction?

[@HemantPandey123]

@@alanjoelj1591 Solution assumes (abc)^2

[@alanjoelj1591]

(only) solutions to the equation are (1,2,3) and permutations. (abc)^2 is clearly 36 which is > 3.

[@mathsislove197]

I didn’t get anything after 2.00 min, can someone pls explain me how he got the other inequality at 3.35mins

[@DaveyJonesLocka]

He multiplied the previous result by 9.

[@ZIN24031980]

Thanks, I liked your way of solution, it's very nice.

[@awildscrub]

Why does this guy sound like Uncle Roger

[@muhammadafifuddin9353]

Nice to learn it.

[@fansuli427]

I really like your video. Thank you

[@letsthinkcritically]

Thank you!!

[@dionisis1917]

I think that it's a bit easy for IMO.

[@deveshichaudhary2565]

Yes i also saw that

[@letsthinkcritically]

Probably that’s why it wasn’t chosen to be in the real contest?

[@Emir-life2]

Are you serious? I'm în 9 class, and I'm preparing for IMO, but this is so hard 😭!!

[@SimpMaker]

You wrote 3 when you multiplied by 9 at 3:41

[@mithutamang3888]

While, b=32 is also a solution? Reply now.

[@sohumsharma2892]

Wow, nice.

[@rk-ds4vl]

Very good

[@beautyofmath6821]

Great

[@danicharif7224]

Nice

[@anjaneyasharma322]

Guessing 1 2 3

[@nevomirzaihamadani2648]

Man please, i like what u are doing but explain it in a clearer way, speak more distinctly, and maybe present it in a better way if u can

[@Emre67511]

Fr I feel like he just writes it down without explaining at all

[@salarb.4604]

Improve your handwriting!😐