International Math Olympiad | 1998 Question 4

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We look at a solution to a nice number theory problem from the 1998 International Mathematical Olympiad.
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4 июл 2024

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Комментарии : 129

@goodplacetostop2973 3 года назад

5:00 15:37 For anyone needing to hear this today : you matter, you are strong, you can do it. Have a great Sunday.

@adityamohan7366 3 года назад

ffs we can't beat this guy

@goodplacetostart9099 3 года назад

Good place to start at 0:00

@minh9545 3 года назад

I just failed the entrance exam to my university. Nothing can help me now.

@goodplacetostop2973 3 года назад

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.

@ianloree2784 3 года назад

Today is a great place to start

@tomatrix7525 3 года назад

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

@MichaelPennMath 3 года назад

That is great to hear. I have to admit, I have improved with these types of problem by producing these videos!

@skwbusaidi 3 года назад

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

@CM63_France 3 года назад

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.

@stewartcopeland4950 3 года назад

and some of "which tells us..."

@CM63_France 3 года назад

@@stewartcopeland4950 I love those "which tells us", very poetic :) , an some times "which leaves us with..."

@R0M4ur0 3 года назад

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 😅

@shangaiguarisnaque9277 3 года назад

@@R0M4ur0 Relatable af

@R0M4ur0 3 года назад

@@shangaiguarisnaque9277 that's what i meant

@wasselkun5015 3 года назад

Thank you for the video!! learning a lot from the channel.

@user-nm5ge9ht3c Год назад

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

@alejandrocampos7459 3 года назад

My teachers used this problem to train me and my friends for the Centroamerican Math Competition

@zafarb4219 3 года назад

It's a pretty good problem

@ravinderbhandari1974 3 года назад

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

@concretemathematics414 4 месяца назад

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.

@mcwulf25 3 года назад

Good one. Lots of solutions.

@newkid9807 3 года назад

Could you cover harder imo problems, like number 2 or 3

@deeznuts4333 3 года назад

For y=2 there are no solution {x=-2;x=35/2}

@anonymoususer9837 3 года назад

Nice work, I came to the same conclusion.

@tueherlau548 3 года назад

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.

@mukaddastaj5223 3 года назад

Wow. Love ur videos, really, u make everything look so easy, thank u sooo much!) P.S. i have no solution when y=2

@iridium8562 3 года назад

How do I verify these are the only solutions?

@rishavmondal8045 3 года назад

Sir bring more and more imo problems.

@TechToppers 3 года назад

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

@rishavmondal8045 3 года назад

Sir bring questions from latest olympiads also.

@DungNguyen-ti4hg 3 года назад

Wow... A "nice" sol

@gamedepths4792 3 года назад

5:00 i got scared ngl

@thomasstokes9412 3 года назад

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.

@123bluestorm1 3 года назад

4:40 if any of you wanna hear the notification

@suranjanlk 3 года назад

👍well done ✅

@jaimeduncan6167 3 года назад

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?

@tracyh5751 3 года назад

No, but by plugging the infinite family back into the original condition we can see that all pairs from the infinite family work out.

@otakurocklee 3 года назад

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.

@shreyathebest100 3 года назад

Good question answered well by Tracy h

@milanchrastina8014 3 года назад

No solutions here also, cool vids man.

@user-kl8dh7nt2e 3 года назад

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?

@jadegrace1312 3 года назад

Because the divisibility property he gave is true regardless of what a and b you choose.

@otakurocklee 3 года назад

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.

@narekbojikian5885 3 года назад

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

@jaimeduncan6167 3 года назад

I am not sure about the procedure in general, it seems to me that it finds necesary contitions not suficient.

@otakurocklee 3 года назад

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.

@Andrew-mx7vt 3 года назад

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 :)

@h4z4rd28 3 года назад

Me :)

@electrovector7212 3 года назад

Ofc me

@mehdifachel2057 3 года назад

@electrovector7212 3 года назад

@@h4z4rd28 your messages do not show up on my RU-vid can you comment from another device?

@h4z4rd28 3 года назад

@@electrovector7212 now?

@basedblueboy8770 3 года назад

nice

@laszloliptak611 3 года назад

At the end you are just doing long division to divide x^2+x+1 by x+8.

@rickshaw1641 3 года назад

It is easier than the method Mr Michael talked.

@tgx3529 3 года назад

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?

@IanXMiller 3 года назад

That fits under case 1 where x=7m^2 and y=7m

@tgx3529 3 года назад

@@IanXMiller Thank you, I'm just looking for the end result. I wanted to think about it myself .

@user-ht1vg5we2p 3 года назад

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

@jwy4264 Год назад

i think 0 isnt a natural number

@user-ht1vg5we2p Год назад

@@jwy4264 actually it is. If you want to select all integers bigger than or equal to 1, you have the set N* for that

@yuseifudo6075 6 месяцев назад

This is not true for the whole world It is a convenience and not a fact@@user-ht1vg5we2p

@puranjayjoshi5549 3 года назад

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

@grecunarcis09 3 года назад

Since LHS is positive => RHS have to be positive too so d must be positive because xy²+y+7 is always positive

@puranjayjoshi5549 3 года назад

@@grecunarcis09 oh yeah, thanks!

@MatriceMX 3 года назад

How can (X+8) divide(X^2+X+1) with x e Z ,when X^2+X+1 doesnt have integer solutions?

@MatriceMX 3 года назад

that is some mombo-jumbo

@TJStellmach 3 года назад

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.

@maxamedaxmedn6380 3 года назад

👍👌

@sidimohamedbenelmalih7133 3 года назад

x=70/4=35/2

@mehdifachel2057 3 года назад

Slt les solutions sont (7k; 0); (11; 1); (49; 1); (7q^2; 7|q|)

@ZeonLP 3 года назад

nice nice :3

@tgx3529 3 года назад

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

@Ideophagous 3 года назад

You missed another general solution of the form (7k, 0).

@TheCrimsonMoose1 3 года назад

0 is not a natural number.

@Ideophagous 3 года назад

@@TheCrimsonMoose1 Depends on your definition. In Francophone textbooks it's considered a natural number.

@TheCrimsonMoose1 3 года назад

@@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.

@TJStellmach 3 года назад

He says just 11 seconds in what definition of natural numbers he's using.

@maxtirdatov 3 года назад

Don't we lose anything if we just pick "nice" coefficients for the linear combination?

@mathijs1752 3 года назад

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

@stewartcopeland4950 3 года назад

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

@mathijs1752 3 года назад

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

@JoseFernandes-js7ep 3 года назад

@@mathijs1752 Notice that you have p(m) l p(m).m, which is true for any integer m

@mathijs1752 3 года назад

@@JoseFernandes-js7ep that's true, thanks!

@IgnacioIacobacci 3 года назад

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

@leickrobinson5186 3 года назад

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. :-(

@IgnacioIacobacci 3 года назад

@@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

@Notthatkindofdr 3 года назад

You can prove by induction that a_n > sqrt(2n) for n>0, and then check that sqrt(2*1996)>63.

@IgnacioIacobacci 3 года назад

@@Notthatkindofdr Amazing... thank you very much ...

@IgnacioIacobacci 3 года назад

@Adam Romanov Single round. The outstanding participants of each national competition of South America were participating.

@suniltshegaonkar7809 3 года назад

I have strong doubt for step at 3:11

@Ideophagous 3 года назад

It's a basic result in elementary number theory

@suniltshegaonkar7809 3 года назад

@@Ideophagous It is like starting with the assumption that needs to be proved

@Ideophagous 3 года назад

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

@TJStellmach 3 года назад

@@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.

@DilipKumar-ns2kl 3 года назад

x=7, y=7 is a soln.

@TJStellmach 3 года назад

Yes, it's part of the infinite family of solutions of the form (x,y)=(7m^2,7m).

@natepolidoro4565 3 года назад

leaf blower

@beyremmerdassi3148 3 года назад

well u forgot about (x,y) = (0,7m)

@TJStellmach 3 года назад

No, the problem specifies positive x and y.

@M-F-H 3 года назад

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.