International Math Olympiad | 1998 Question 4
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- Опубліковано 29 чер 2024
- We look at a solution to a nice number theory problem from the 1998 International Mathematical Olympiad.
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5:00
15:37
For anyone needing to hear this today : you matter, you are strong, you can do it. Have a great Sunday.
ffs we can't beat this guy
Good place to start at 0:00
I just failed the entrance exam to my university. Nothing can help me now.
TNT I'm sorry to read that you failed your entrance exam. Don’t be too hard on yourself, don’t lose hope.
I would suggest you take some time to relax. Then, maybe think about the weak areas on your exam and try to find academical help.
It may seem a dark moment but this is the first day of a comeback.
Today is a great place to start
I really find myself improving with IMO problems after watching you for a few months. Originally I always found I’d understand the solution but would never know how to approach it myself without major hints. That’s slowly changing
That is great to hear. I have to admit, I have improved with these types of problem by producing these videos!
We can save steps when y=1 and y=2
We can use (xy^2+y+7) | (7x-y^2)
Which you find before
For y=2
(4x+9)|(7x-4)
(4x+9)|(7(4x+9)-4(7x-4))
(4x+9)|79
79 is prime
4x+9=1 or 4x+9=79
x= -2 or 17.5 both are not Natural number
So no solutions
Hi,
For fun:
1 "and so on and so forth",
6 "go ahead and ...", including 4 "let's go ahead and ...",
3 "great",
2 "that's a good place to stop", including the "and that's a good place to stop" at the end.
and some of "which tells us..."
@@stewartcopeland4950 I love those "which tells us", very poetic :) , an some times "which leaves us with..."
Oh guys... that "tells us"... for a non native speaker the sound of that "s" between "tell" and "us" is like a symphony... even because 9 out of 10 times we EAASL fail to pronounce it 😅
@@R0M4ur0 Relatable af
@@shangaiguarisnaque9277 that's what i meant
Thank you for the video!! learning a lot from the channel.
I love that this problem contains both an infinite family of solutions and some small ones. It is not often that you see such problems
from 5:30 , we could set Y-7x=(xY+y+7)m where X= x^2 , Y= y^2 and m is an integer , putting x=Y/7, we get a cubic on RHS which is fairly easy to solve
My teachers used this problem to train me and my friends for the Centroamerican Math Competition
It's a pretty good problem
Good one. Lots of solutions.
Wow. Love ur videos, really, u make everything look so easy, thank u sooo much!)
P.S. i have no solution when y=2
For y=2 there are no solution {x=-2;x=35/2}
Nice work, I came to the same conclusion.
Could you cover harder imo problems, like number 2 or 3
How do I verify these are the only solutions?
A no-hint solution: Write m(xy^2 + y + 7) = x^2 y + x + y. Define k s.t. m y = x + k. (to get this idea, observe that mx y^2 ~ x^2 y approximately, and so k must be "small"). By inserting and rearranging we get both (i) kxy = y - 7m - k and (ii) x = (y^2 - k (y+7) )/ ( 7 + k y^2 ). Obviously (ii) have no integer solutions for k > 0; for k < 0 it must hold that k y^2 < 7 (since x is positive) and we easily get the y=1 solutions by checking k=-1,...-6 and can rule out y=2, k=-1 by insertion. For k=0 we get x = y^2 / 7 and from (i) y = 7m. QED.
maybe i'm confused, but for case 1, it says that y^2-7x < xy^2 + y + 7, but we also rewrite y^2-7x as d*(xy^2 + y + 7), which that value is larger than xy^2 + y + 7 for d>0.
👍well done ✅
Sir bring questions from latest olympiads also.
Sir bring more and more imo problems.
Wow... A "nice" sol
4:40 if any of you wanna hear the notification
5:00 i got scared ngl
I have a question: We peek a couple of of numbers and derive a condition that the solution must satify. It seem to me that it's a necesesay but maybe not sufficient contition. For example if the selection implies that the number must be multiple of K , ca we infer that all the multiples of K are solutions?
No, but by plugging the infinite family back into the original condition we can see that all pairs from the infinite family work out.
As Tracy H said, we have to plug in the solutions we find into to the original equation to check that they work, for exactly the reason you mentioned. We may have extraneous solutions.
Good question answered well by Tracy h
No solutions here also, cool vids man.
Wanted to watch a video from putnam exam, but decided that this one could be more interesting :) also who wants to solve maths problems ,perhaps ,we can unite and do it(just a suggestion :)
Me :)
Ofc me
Me
@@h4z4rd28 your messages do not show up on my UA-cam can you comment from another device?
@@electrovector7212 now?
For y=2 there are no solutions in the natural numbers:
plugging in y=2 => (4x+9)|(a(4x+9)+b(2x^2+x+2))
Picking a=x and b=-2 => (4x+9)|(7x-4)
=> (4x+9)|(a(4x+9)+b(7x-4))
Picking a =7 and b=-4 => (4x+9)|79
So either 4x+9=1 or 4x+9=79, none of these have solutions in the natural numbers.
3：53 The formula is satisfied for all integer a, b, but you only choose some nice ones. How do you prove that this method would find all solutions?
Because the divisibility property he gave is true regardless of what a and b you choose.
If x is a solution to our original problem, then it is a solution for every particular choice of a and b. Therefore solving for any choice of a and b will get us all the solutions we need... maybe extraneous ones also. We have to plug back into the original to check that our solutions are correct. But there's no way we will miss a solution.
At the end you are just doing long division to divide x^2+x+1 by x+8.
It is easier than the method Mr Michael talked.
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120,
what is the value of s(3m)?
Can someone please help me with this?
It came in India. PRMO 2019 25th August Paper
Thanks this was a nice video! A quick question: shouldn’t we have plugged the solution (7m2, 7m) from the first case in the given formula to check its correctness? since we derived it from a chosen restriction of a and b and it is not necessarily valid for all values of a and b
I am not sure about the procedure in general, it seems to me that it finds necesary contitions not suficient.
Yes, that's right. It's the same with all the solutions. We may have extraneous solutions so we have to plug into the original to check.
You can also have y equals 0 and after the simplifications get the infinite set {7k, 0} for every natural k.
As for the 2 case, 4x+9 |2x^2 + x +2,
4x +9 | a (4x+9) +b(2x^2+x+2)
a=x, b=-2 ======>>
4x+9 |4x^2+9x -4x^2 -2x -4=7x -4
Here you choose a=7, b=-4 :
4x +9 | 7 (4x +9)- 4 (7x- 4)
= 63+16 = 79, which is prime, thus you have: 4x+9 equals 1 or 79.
Case 1: 4x +9=1 =====>>
x =-2, not acceptable, because -2 is not Natural.
Case 2: 4x + 9= 79 =====>
x is not an integer
So no solutions here
i think 0 isnt a natural number
@@jwy4264 actually it is. If you want to select all integers bigger than or equal to 1, you have the set N* for that
This is not true for the whole world
It is a convenience and not a fact@@user-ht1vg5we2p
nice
nice nice :3
What does it mean more solutions? There are 2 solutions on the blackboard x=11,y=1 and x=49,y=1. And what about x=y=7?
That fits under case 1 where x=7m^2 and y=7m
@@IanXMiller Thank you, I'm just looking for the end result. I wanted to think about it myself .
6:45 why is d >= 0 when according to definition of divisibility d should belong in the integers
Edit: my bad he cleared it up near 7:55
Since LHS is positive => RHS have to be positive too so d must be positive because xy²+y+7 is always positive
@@grecunarcis09 oh yeah, thanks!
x=70/4=35/2
How can (X+8) divide(X^2+X+1) with x e Z ,when X^2+X+1 doesnt have integer solutions?
that is some mombo-jumbo
If the one polynomial divided the other, that would just tell you that _all_ possible values of x were solutions. But we don't need that; we're looking for the particular values of x that make the two values divisible.
Indeed, if you divide x^2+x+1 by x+8, you get x-7 with a remainder of 57. So that means (x^2+x+1)/(x+8) = x-7+(57/(x+8)), which is an integer when 57 is divisible by x+8. It's precisely the fact that we _do_ have a nonzero remainder that implies the specific solutions.
👍👌
You missed another general solution of the form (7k, 0).
0 is not a natural number.
@@TheCrimsonMoose1 Depends on your definition. In Francophone textbooks it's considered a natural number.
@@Ideophagous In almost all textbooks I've read if you include 0 the set is called the "whole numbers" and if not it's called "natural numbers" I think that is why he did not include these solutions.
He says just 11 seconds in what definition of natural numbers he's using.
Don't we lose anything if we just pick "nice" coefficients for the linear combination?
We don't lose anything by picking nice coefficients. It's a rather counter-intuitive thought why, but it goes like this:
We assumed x and y 'worked' and picked nice coefficients. If we then derive 'solutions' from this picking, we know that the actual solutions are a subset of the found 'solutions'. The actual solutions need to work for all coefficients, thus the found 'solutions' do contain the actual solutions but maybe more. After plugging in your 'solutions' in the original problem, you'll see that maybe some are not valid and only work for the nice coefficients, while others are valid and thus work for all coefficients. These latter values are the actual solution.
In this case Michael did forget to check whether the results of case 1 and case 2 are actually valid solutions and not some results that just pop up by the nice picking of coefficients
If a linear combination is valid for any pair of values (a,b), then it is also valid for a pair of values judiciously chosen in order to simplify the mathematical expression, without loss of generality
If we check the solution (7m², 7m) we get (7³m⁴+7m+7)|(7³m^5+7m²+7m). While I'm not a number theoretic, my gut feeling is that this implies m=1 is the only valid solution. Therefore his infinite family of solution comes down to just (7, 7) and maybe one or two more
@@mathijs1752 Notice that you have p(m) l p(m).m, which is true for any integer m
@@JoseFernandes-js7ep that's true, thanks!
x=7, y=7 is a soln.
Yes, it's part of the infinite family of solutions of the form (x,y)=(7m^2,7m).
If I take it as [x(xy+1)+y]/[(y(xy+1)+7]= k, I see that y=7k;x=ky, so we have first solution (x,y)=(7k^2,7k)
If I take x as fixed number, there is linear function variable y/ quadratic function variable y, it's not interesting for division.
If I take y as fixed number, there is quadratic function variable x/ linear function variable x. Only for y=1 is division OK.For y=1 we have (x^2+x+1)=(x+8)(x-7)+57. So (x^2+x+1)/(x+8)=[(x+8)(x-7)/((x+8) ]+57/(x+8). We have 57/(x+8). 1.)case is when x+8=57, 2.) case is when we take 57/19=3, se x=11
A problem hunts me since more than 20 years. It's a problem from "Cono Sur" Olimpiads from 1996. Given the sequence on real numbers a_n+1 = a_n + 1 / a_n. For any a_0 real positive show that a_1996 > 63. Problem nro 2 in this link in Spanish www.oma.org.ar/enunciados/con7.htm
I think there must be an error in the original problem. It’s pretty easy to show that the sequence consists of the ratios of consecutive values from a Fibonacci sequence that starts {1, a_0, a_0+1,...}. But, this ratio converges to phi, which is much less than 63. :-(
@@leickrobinson5186 I'm pretty sure the problem is Ok. I have a (near) solution. You can say that there for each a_n it is true that if a_n = k, a_n < ⌈k⌉ and and there are less than k a_n between k and k+1 ... so... you sum n times each n from 63 to 1 and you reach... 2016, which is pretty close, but not close enough
You can prove by induction that a_n > sqrt(2n) for n>0, and then check that sqrt(2*1996)>63.
@@Notthatkindofdr Amazing... thank you very much ...
@Adam Romanov Single round. The outstanding participants of each national competition of South America were participating.
well u forgot about (x,y) = (0,7m)
No, the problem specifies positive x and y.
leaf blower
Slt les solutions sont (7k; 0); (11; 1); (49; 1); (7q^2; 7|q|)
I have strong doubt for step at 3:11
It's a basic result in elementary number theory
@@Ideophagous It is like starting with the assumption that needs to be proved
Not really x|a and x|b => x|u*a+v*b for x, a, b, u and v integers is a basic result and can easily be proven
@@suniltshegaonkar7809 Except he's not doing a proof. He gets to suppose the given conditions on x,y and deduce what those conditions imply. The converse of those implications _may_ not hold, but as long as every step in the chain of reasoning is biconditional, you don't even have to check.
You somewhat cheated because in case 1 you ignored the possibility x=0 (with solution x=y=0: definitely, 0 = 0*7 is a multiple of 7) which then you recover with m = 0 \in N.
In American mathematics, by convention, the set of natural numbers does not contain 0, but that's good to point out!
He says ten seconds into the video that that's the definition of natural numbers he's using.
Any one from 🇮🇳🇮🇳🤔🤔