Almost an IMO Problem | IMO Shortlist 2019 N2
Вставка
- Опубліковано 22 гру 2020
- #MathOlympiad #IMO #NumberTheory
Here is the solution to IMO Shortlist 2019 N2!
------------------------------------------------
Follow my Instagram page for more Maths :)
Link: / lets.think.critically
------------------------------------------------
I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.
Subscribe: ua-cam.com/users/letsthinkcr...
Email me (address in video) your suggestions! lets.think.critically.27@gmail.com
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.
Can you explain why all three variables cant be higher than cbrt(3)? I dont quite get the implication
@@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).
@@typo691 using the AM GM inequality
Well done. Well presented. You are in Michael Penn territory. The road to 10k subs looks easily attainable.
there will never be another michael penn. He is far too powerful.
Thanks for these vids u make! U really do inspire)
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
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.
very interesting problem!
The most entertaining math video I've ever watched
Nice to learn it.
Amazing solution
Hope your channel grows soon
Your problems are great
Here's a homework:
a^b + b^c + c^a = (abc)^abc
do it urself
@@AyushKumar-ot3zr you think I'd give homework without knowing the answer myself? 🤣
How to solve it ? Give me a hint
This equation has only 3 solutions that [a; b; c]= [2; 1; 1], [1; 2; 1], [1; 1; 2]
@@KaranSingh-qw7cn use equality that a=>b=>c
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)
Yes, that’s exactly what I meant.
Thanks, I liked your way of solution, it's very nice.
a=1,b=2,c=3
Guessing 1 2 3
Good
Wow, nice.
Very good
Great
Its 1, 2 and 3.
2:29
how do you conclude a2
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
Nice
How did you determine that (abc)^2
Look again at the question statement and the previous line ;)
@@DoReMeDesignAh. I got it. Thanks for the reply!
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?
we don’t know whether a,b,c are consecutive so this formula does not actually works here
Nice solution! How long did it take you to solve that?
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 ?
You're a genius
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?
--- 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.
a(a^2-b)
@@cantcommute understand.
Where does it even say a < b^2 ? Or where does it come into play ?
Have you tried plugging a=-1, b=1, c=0,1?
0 is not an inetiger brother
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?
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!
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
@@fix5072 thank you, i get it now. :::)
@@fix5072 and one more question: if c²b
@@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
I don’t get how we get that a^2
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
@@reeeeeplease1178 thanks. What does he mean when he says that a^2 “is divisor” of …. and why can he say it?
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?
@@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)
@@reeeeeplease1178 riiiiiiight :). Thank you so much….
why (c/b)^3 Is Equal To 1 ?
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
Wow nice problem and soln too!
Thank you!!
3:40 why is b >= 2 when c = 2?
Because of the initial inequality a >= b >= c.
Please make a platform which we can write our favourite problems
Feel free to let me know here! I will make videos on problems that you suggest.
I really like your video. Thank you
Thank you!!
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 :)
Sure thing!
@@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 🙏
3:27 i dont get it, why can RHS exceed 2 ?
i think because C > B hence it (c/b) will be less than 1
@@icebeargt5142*C
man pls can you recomend me any book of maths which starts from basic to this level pls
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.
@@letsthinkcritically thanks !!!!
You can try pathfinder for prmo by prashant jain
@@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.
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
Hi
Why is it at 5:07,
a² - b | a³ + 1 - a(a² - b) ?
Thank you
Did you see my question?
Because we established a² - b | a³ + 1 and naturally a² - b | a(a² - b)
@@jm2270
Oh I see. Thank you
This UA-camr doesn’t reply at all
@@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 ?
@@Emre67511 x | y means x divides y.
how can we say that a^2 | a^3+b^3+c^3?
because from (abc)^2
Oh thanks lol@@Jan-vz5ge
Im sorry but why 1+ab>a(b+1)-b 6:26
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
Why does this guy sound like Uncle Roger
Why does it become 3a^3 in the beginning and not a^6?
Never mind I’m stupid
@@ryanclothier6853 ayoo broo 🥺
😅
I didn’t get anything after 2.00 min, can someone pls explain me how he got the other inequality at 3.35mins
He multiplied the previous result by 9.
I think that it's a bit easy for IMO.
Yes i also saw that
Probably that’s why it wasn’t chosen to be in the real contest?
You wrote 3 when you multiplied by 9 at 3:41
5:45 mistake
If (a)=1 then a^3(a-2) is = -1 which not 1
That’s my point, I was trying to say that this would lead to a contradiction.
@@letsthinkcritically thank u for your reply and yes now i get it 💓💓
Using AM>=GM we get (abc)^2 is < = 3 (Applying Am>=Gm on LHS). Hence all bounds found without any lengthy steps.
Wrong. AM-GM gives 3abc
@@alanjoelj1591 But if (abc)^2 is >= 3abc then isn't it contradicts original solution?
@@HemantPandey123 Where is the contradiction?
@@alanjoelj1591 Solution assumes (abc)^2
(only) solutions to the equation are (1,2,3) and permutations. (abc)^2 is clearly 36 which is > 3.
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
Fr I feel like he just writes it down without explaining at all
While, b=32 is also a solution? Reply now.
Improve your handwriting!😐