This solution was extremely well explained. I fully agree with what you said about explaining each step. I hope you continue to post more IMO/high level competition problems on this channel!
I couldn't resist trying this problem when you called it "a nice exercise in number theory", but it took me a few days to crack. My solution was pretty much like yours, but I used synthetic division to divide out the polynomials, which is neater I think.
Hmm quite remarkable that you had the perseverance to stick with the problem for days - it's quite unusual to be able to do so, unless you have a background in olympiad mathematics of course.
Your right about that guy explaining hard competition math exercices like every of us know a bunch of hard math stuff. Very good job explaining every single step u make :)
Excellent! I really love IMO problems, they are just so fun to do even if the math needed is elementary concepts. You can actually start making a playlist of like Contest Problems.
@@V-for-Vendetta01 ohhhh my bad🙈... Yup it was from an entrance exam question (JEE mains) and I qualified that exam to get a seat in nationally funded institute. So yeah I got through that test😁😁
another method a^b + a + b = k*(ab^2 + b + 7) for some positive integer k. => ba^2 + (1 - kb^2)a + (b - kb - 7k) = 0 ---(1) let α, β are roots of a in equation (1) , α + β = -(1 - kb^2)/b = kb - 1/b since a,b,k are integers, 1/b must be integer. => b = 1 => a + 8 divides a^2 + a + 1 => (a^2 + a + 1) / (a + 8) = a - 7 + 57/(a + 8) => a + 8 divides 57(=3*19) => a = { 11, 49 } => (a, b) = { (11, 1), (49, 1) }
9:19 Papa Flammy throwing shade at other channels left and right But seriously, I really like that you always go through all steps and explain them, and only skip stuff that you covered already in other videos (like complicated integrals and stuff). That way it feels really like an accomplishment at the end of the video, and not as if we cheated by skipping the hard parts ^^
It’s not that you’re not smart enough. It’s just that standard mathematical curricula teach people to solve “exercises” rather than problems. It is mostly dull repetition of concepts, without ever caring for the creativity that is associated with solving real math problems. Competition math is all about solving problems, not exercises. And therefore most people who have not had a problem-solver’s mathematical education are thrown off by it.
Joe Horrell Yes, but those are generally details. I agree there is mathematical content that is not covered in school, but it is relatively simple to learn that theory on one’s own.
Joe Horrell There are tons of resources online. As I mentioned, I don’t think it’s learning the content that is difficult. It is more important to learn real mathematical problem solving. That is applicable at every level of math and is the essence of what math is all about.
The reality is that most problems from thoose contest are uninteresting, and to solve them, you need creativity but MOSTLY to know a lot of other problems. I don't think it's a very good introduction to mathematics, and you'll never do something usefull with mathelatical olympiads, apart from having (maybe) fun
This is so damn hard, I’m trying to get into the IMO but first I have to be with the best ~30 Dutch students to be eligible, then we’ll have a great course and the 6 students who get the highest scores go to the IMO, rn I am with the best ~120 Dutch students and we’ll have the next test in September, but I think this might help getting there. The explanation was great, I understood the steps even though I’ve never seen a “|” to use when something devides. You might be the reason I participate in the IMO one day! Thank you keep it up!
At 9:10 you should have specified that all (a, b) = (7r^2, 7r), for r € N, are solutions only because b divides a in those cases (since you multiplied the numerator by b in the beginning).
0:58 - 1:03 But 0 is a natural number too. In fact, (a = 0, b = 0) is a pair such that (ab^2 + b + 7) | (ba^2 + a + b), because 7 | 0. There is merit to looking at the cases (0, b) with b > 0 and (a, 0) with a > 0 as well. If b = 0 and a > 0, then 7 | a, in which case (7n, 0), where n is any natural number, solves this. If a = 0 and b > 0, then (b + 7) | b, which is only possible if b = 0. Therefore, the only solutions such that ab = 0 are (0, 0), and in general, (7n, 0) where n is any natural number, although the former is just a special case of the latter with n = 0. So (7n, 0) is a solution family. The other solutions family have a > 0, b > 0, which is what I suppose this video is meant to address. I just think there is no reason to talk about solutions in the natural numbers but exclude 0 from the set for no apparent reason. Peano axioms!
@@l1mbo69 1 hour and 30 mins to be precise. Besides, they aren't your ordinary schoolstars, the people giving the IMO are already trained at such stuff and have the past experience of clearing many national olympiad levels to write the IMO, soo maybe it wasn't too tough for them, considering that P1s & P4s are relatively eazy.
Thank you, you had hide some work but go direct to the answer. Once I just expandind calculation I found few unpossible causes where they need to be consider if we do not direct all the way to the answer such as b=-7 or b=-8 substitution causes non positive possible value and etc. 😀
Hmm, but if we multiply the RHS of …|… by b then wouldn't we possibly introduce new solutions here? Like if b had any common factors with the LHS, then it may 'spawn' some divisibilities where was none before. I.e. if ab²+b+7 had some common factors with b (it can't divide b tho) - e.g. if b=7 (-> ab²+b+7=7(7a+2) ) then ab²+b+7|b(…) isn't equivalent to ab²+b+7|(…), but to (7a+2)|(…)
At 11:50 you've just shown that b^2-7a is always < than ab^2+b+7 without using the fact that b^2-7a>0. So your numerator is always strictly less than your denominator. Isn't that enough to say that the case where ab^2+b+7 | b^2-7a is incorrect, thus the only case is the first one, b^2-7a=0 with sols (7r^2, |7r|) ? Or am I missing something?
It was a bit unclear. The implication that a|b means b>=a is not really correct, it actually implies |b|>=a, which is why you have to be careful. Only if b is positive you can say b>=a.
In case 1 why did you say r is a positive or negative integer and then take its absolute value? Wouldn't it be easier to say that r is a natural number and remove that absolute value?
Well he said r is positive or negative because in either case you get positive solutions. You can also define it as you said. It's the same. Edit: extending to real values, you can interpret (7r^2, abs(7r) ) as a parametric curve, and you might know there are infinitely many parametrizations for the same curve, so you can choose different expressions.
Why can you multiply on the right side by b in the first part of the solution without possibly adding extraneous solutions? If x|y then x|ky for x, k, y in N, but the reverse doesn't hold.
@@angelmendez-rivera351I don't think I quite understand what you mean. It's possible that the LHS doesn't divide b but nonetheless needs the factor, though. To use a tangible example, 15 doesn't divide 20 or 3, but divides their product.
Nicholas Miklaucic That is not a valid counter example, because you are postulating an example in which 15 divides neither 20 nor 3, while in this exercise, ab^2 + b + 7 does divide ba^2 + b + a.
In other words, while it is true that x|ky does not imply x|y. However, the problem itself tells you that x|y, so the only possible way to create extraneous solutions is if x|k. What I'm saying is that in this particular problem, x|k is false.
So I was watching an episode of spongebob and sandy had the following on her clipboard f(x) = +-sqrt((x+x^2)^3/pi)) as x -> infinity. Papa can you find a function that has such a limit and make a video out of it!
Einmal tiiiief durchatmen - dann ist doch wieder alles gut. Da haben sich ein paar Fehlerchen eingeschlichen, aber schön jemanden mit solch Enthusiasmus zu sehen.
I think it's a meme based on a famous mathematician who supposedly stated 57 as a prime in a conference. Edit: this mathematician was Alexander Grothendieck. You can Google 'Grothendieck prime'
IMO? I didn´t even pass the first round of my school math olympiads( in 6th grade)( answering half of the questions, and half of the ones i answered were rong, only 8 questions).
@@PapaFlammy69 Using the method of natural deduction in a mathematical proof is actually quite laborious. I think every mathematician would get strongly annoyed while trying it out, like proving that the square root of any prime number is irrational using natural deduction. (Without jumping steps, obviously)
![An Ingenious Solution Strategy! [ Putnam 2018 A1 ]](http://i.ytimg.com/vi/1DrqYshkcXw/mqdefault.jpg)
2:00 not equivalent. a|b implies b>a, but the reverse implication is false
yes, my mistake ^^
@@PapaFlammy69 also, it was a long way to say a|b exists k in N such that b = k * a ? :D
it made me screech lol
Yes I saw this too, but I got the point he was making.
This solution was extremely well explained. I fully agree with what you said about explaining each step. I hope you continue to post more IMO/high level competition problems on this channel!
3:27 >mfw c|a+b dab , approximated by the human eye.
:vvv
Bad kerning.
I couldn't resist trying this problem when you called it "a nice exercise in number theory", but it took me a few days to crack. My solution was pretty much like yours, but I used synthetic division to divide out the polynomials, which is neater I think.
Hmm quite remarkable that you had the perseverance to stick with the problem for days - it's quite unusual to be able to do so, unless you have a background in olympiad mathematics of course.
Your right about that guy explaining hard competition math exercices like every of us know a bunch of hard math stuff. Very good job explaining every single step u make :)
:))
I love the IMO videos, they're my favourite
Papa : -1 is not Zero if you didn't know already
Anime girl : WOW!!
nani?!
@@PapaFlammy69 lmao😂
Excellent! I really love IMO problems, they are just so fun to do even if the math needed is elementary concepts. You can actually start making a playlist of like Contest Problems.
Will do!!
Flammable Maths don’t, it’s a common scam
@@PapaFlammy69 why didn't you heart my comment ma bro?
bruh
@@PapaFlammy69 lol
Very nicely explained. Even I had this type of question in one of my university entrance exam. But couldn't solve 🙈
Thank you Sumukh! =)
Did you end up getting in?
@@emilyscloset2648 not back then, but now after seeing the solution method, I think I could have got that :)
@@sumukhhegde6677 I think he asked you if you got through the entrance exam.
@@V-for-Vendetta01 ohhhh my bad🙈... Yup it was from an entrance exam question (JEE mains) and I qualified that exam to get a seat in nationally funded institute. So yeah I got through that test😁😁
Man, thank you for the Birthday upload man. Another great and informative maths video man!
Happy fkin birthday RC!! :)
@@PapaFlammy69 Why thank you so my man!
@@PapaFlammy69 :D
another method
a^b + a + b = k*(ab^2 + b + 7) for some positive integer k.
=> ba^2 + (1 - kb^2)a + (b - kb - 7k) = 0 ---(1)
let α, β are roots of a in equation (1) , α + β = -(1 - kb^2)/b = kb - 1/b
since a,b,k are integers, 1/b must be integer. => b = 1
=> a + 8 divides a^2 + a + 1
=> (a^2 + a + 1) / (a + 8) = a - 7 + 57/(a + 8) => a + 8 divides 57(=3*19)
=> a = { 11, 49 } => (a, b) = { (11, 1), (49, 1) }
9:19 Papa Flammy throwing shade at other channels left and right
But seriously, I really like that you always go through all steps and explain them, and only skip stuff that you covered already in other videos (like complicated integrals and stuff). That way it feels really like an accomplishment at the end of the video, and not as if we cheated by skipping the hard parts ^^
Thank you Hal! :)
Who's he throwing shade at, tho? I genuinely couldn't figure it out.
I think the meme is supposed to say "pure mathematicians are just analytic philosophers who are good at math".
F in the chat for the analytic philosophers
;_;
Then give me a Z
15:20 Just divide the polynomials to get residue of 57/(a+8) quickly getting a=11 and 49.
Contest Math turned me away from math, cos I wasn't smart enough. Please do more of these problems, Papa. btw Loved this one 😘.
Will do! :)
It’s not that you’re not smart enough. It’s just that standard mathematical curricula teach people to solve “exercises” rather than problems. It is mostly dull repetition of concepts, without ever caring for the creativity that is associated with solving real math problems.
Competition math is all about solving problems, not exercises. And therefore most people who have not had a problem-solver’s mathematical education are thrown off by it.
Joe Horrell Yes, but those are generally details. I agree there is mathematical content that is not covered in school, but it is relatively simple to learn that theory on one’s own.
Joe Horrell There are tons of resources online. As I mentioned, I don’t think it’s learning the content that is difficult. It is more important to learn real mathematical problem solving. That is applicable at every level of math and is the essence of what math is all about.
The reality is that most problems from thoose contest are uninteresting, and to solve them, you need creativity but MOSTLY to know a lot of other problems. I don't think it's a very good introduction to mathematics, and you'll never do something usefull with mathelatical olympiads, apart from having (maybe) fun
14:00 it should be b²-7 < 0 if it is equal to 0 the former inequality doesn't hold
2:03 so 2 divides 3
Eccoci
my mistake ^^ Just an implication
Bruh
It does in universes where peano's axioms result in natural numbers that start 1, 2, 4, 3, 5 etc 😁.
@@davidbrisbane7206 clearly the case
Please make more IMO videos 👏
will do! =)
2:00 equivalent ??!!!
my mistake, just an implication ^^
@@PapaFlammy69 plzz try this problem : consider f(x)=3x^2+1 . prove that the product from i=1 to i=n of f(i) has at most n distinct prime divisors
This is so damn hard,
I’m trying to get into the IMO but first I have to be with the best ~30 Dutch students to be eligible, then we’ll have a great course and the 6 students who get the highest scores go to the IMO, rn I am with the best ~120 Dutch students and we’ll have the next test in September, but I think this might help getting there.
The explanation was great, I understood the steps even though I’ve never seen a “|” to use when something devides.
You might be the reason I participate in the IMO one day! Thank you keep it up!
good luck broski
How did it go?
At 9:10 you should have specified that all (a, b) = (7r^2, 7r), for r € N, are solutions only because b divides a in those cases (since you multiplied the numerator by b in the beginning).
That's just unnecessary, though, since |r| € N anyway. And ab^2 + b + 7 never divides b unless b = 0.
0:58 - 1:03 But 0 is a natural number too. In fact, (a = 0, b = 0) is a pair such that (ab^2 + b + 7) | (ba^2 + a + b), because 7 | 0. There is merit to looking at the cases (0, b) with b > 0 and (a, 0) with a > 0 as well. If b = 0 and a > 0, then 7 | a, in which case (7n, 0), where n is any natural number, solves this. If a = 0 and b > 0, then (b + 7) | b, which is only possible if b = 0. Therefore, the only solutions such that ab = 0 are (0, 0), and in general, (7n, 0) where n is any natural number, although the former is just a special case of the latter with n = 0. So (7n, 0) is a solution family. The other solutions family have a > 0, b > 0, which is what I suppose this video is meant to address. I just think there is no reason to talk about solutions in the natural numbers but exclude 0 from the set for no apparent reason. Peano axioms!
I liked your statement at 9:32 to 9:36.
:^D
imagine trying to do this as a teen under time constraints:
mindf"ck 💯
People giving IMO have 1hr+ to do one question
@@l1mbo69 1 hour and 30 mins to be precise.
Besides, they aren't your ordinary schoolstars, the people giving the IMO are already trained at such stuff and have the past experience of clearing many national olympiad levels to write the IMO, soo maybe it wasn't too tough for them, considering that P1s & P4s are relatively eazy.
Thank you, you had hide some work but go direct to the answer. Once I just expandind calculation I found few unpossible causes where they need to be consider if we do not direct all the way to the answer such as b=-7 or b=-8 substitution causes non positive possible value and etc. 😀
a divides b therefore b>a .
b>a, therefore a divides b
It´s called the reverse statement theorem.
18:36 messed up and and or again
sry hehe ^^'
damn Notation is messing with me for one reason: and in german means "und" starting with a "u", which looks extremely similar to the or symbol :x
I almost had a stroke reading the phrase "and and or"
@@PapaFlammy69 😐now we are all gonna do the same mistake
IMO, a very nice video
mfw papa says 57 is prime ...
then I remember what channel I am watching
:^D
>tfw the outtake comes before the main video
messed the publishing up hehe XD
Good QESTION
Sir u r great teacher
1:55
Whoa, that's a suspicious equivalence. Or I'm stupid, idk.
In the case where b²-7a>0, the guy had a greater than symbol between two quantities, when they should both be less than symbols.
@TNys I have a math reference for you guys in the video! Comment the theorem!!
Help this Papa Flammy Fan get started with this channel
Very cool
:)
"This case sucks balls."
~ Flammable Maths, 2020
:D
Excelente
That at 2:05 isn't an equivalence btw.
2|3 is false but 3 is greather than 2.
oh, you're right!
a condition is missing I think for it to be an equivalence
I Want to know how can i improve myself to solve this problems ..?
Nice
eciN
Papa did you take part in math olympiads etc. when in highschool/middleschool? And if yes, how did you do?
nope^^
a
9:30 What does that little 2 mean :)
Hmm, but if we multiply the RHS of …|… by b then wouldn't we possibly introduce new solutions here? Like if b had any common factors with the LHS, then it may 'spawn' some divisibilities where was none before.
I.e. if ab²+b+7 had some common factors with b (it can't divide b tho) - e.g. if b=7 (-> ab²+b+7=7(7a+2) ) then ab²+b+7|b(…) isn't equivalent to ab²+b+7|(…), but to (7a+2)|(…)
what's with the triple line equalities?
didn't you at around 10.30 mean that a is 49r^2 and not 7r^2?
At 11:50 you've just shown that b^2-7a is always < than ab^2+b+7 without using the fact that b^2-7a>0. So your numerator is always strictly less than your denominator. Isn't that enough to say that the case where ab^2+b+7 | b^2-7a is incorrect, thus the only case is the first one, b^2-7a=0 with sols (7r^2, |7r|) ? Or am I missing something?
It was a bit unclear. The implication that a|b means b>=a is not really correct, it actually implies |b|>=a, which is why you have to be careful. Only if b is positive you can say b>=a.
How did they create such questions😅😅😅
dunno lol
can we start by using inequality? and try to simplify it from there? i have no clue but ya
Wait, didn't you spend a lot of time doing what is essentially polynomial long division in a really convoluted way?
In case 1 why did you say r is a positive or negative integer and then take its absolute value? Wouldn't it be easier to say that r is a natural number and remove that absolute value?
Well he said r is positive or negative because in either case you get positive solutions. You can also define it as you said. It's the same.
Edit: extending to real values, you can interpret (7r^2, abs(7r) ) as a parametric curve, and you might know there are infinitely many parametrizations for the same curve, so you can choose different expressions.
I try to learn about this Exam .
Why can you multiply on the right side by b in the first part of the solution without possibly adding extraneous solutions? If x|y then x|ky for x, k, y in N, but the reverse doesn't hold.
Because b < ab^2 + b + 7, so ab^2 + b + 7 | b can never be true.
@@angelmendez-rivera351I don't think I quite understand what you mean. It's possible that the LHS doesn't divide b but nonetheless needs the factor, though. To use a tangible example, 15 doesn't divide 20 or 3, but divides their product.
Nicholas Miklaucic That is not a valid counter example, because you are postulating an example in which 15 divides neither 20 nor 3, while in this exercise, ab^2 + b + 7 does divide ba^2 + b + a.
In other words, while it is true that x|ky does not imply x|y. However, the problem itself tells you that x|y, so the only possible way to create extraneous solutions is if x|k. What I'm saying is that in this particular problem, x|k is false.
@@angelmendez-rivera351 Gotcha! Thanks for the clarification I get it now.
So I was watching an episode of spongebob and sandy had the following on her clipboard
f(x) = +-sqrt((x+x^2)^3/pi)) as x -> infinity. Papa can you find a function that has such a limit and make a video out of it!
lol I can try
Are you asking for a function that is asymptotically equivalent to sqrt((x + x^2)^3/π) as x -> ♾? Or are you asking for something else?
@@angelmendez-rivera351 Asymptotic equivalence could be one for sure. It makes sense that there could be multiple f(x) that have such a limit.
Bro is pro
*N O I C E*
Einmal tiiiief durchatmen - dann ist doch wieder alles gut.
Da haben sich ein paar Fehlerchen eingeschlichen, aber schön jemanden mit solch Enthusiasmus zu sehen.
16:36 57 is the best PRIME??? HOW?!
I think it's a meme based on a famous mathematician who supposedly stated 57 as a prime in a conference.
Edit: this mathematician was Alexander Grothendieck. You can Google 'Grothendieck prime'
v notation means and/or and u cant say b is 1 and 2 because this is MATHS not quantum mechanics, DUH!
i wish there were same proofs but in lean
Can you do a video about this problem
Find all natural numbers n,k and p prime such that the sum
2^2 + 3^2 +... + n^2 = p^k
Ill try that
Sumukh Hegde i’m new to NT so it will really help me!
@@DragonKidPlaysMC sure but see p should be more than 1 I think because 1^anything = 1... in that case we will have infinite solution
so if p>=2 and k>=2,
n=24, p=70 and k=2 is a solution
Sumukh Hegde can you please illustrate your thought process upon arriving at the solution?
At 14:04 it has to be less than 0
IMO? I didn´t even pass the first round of my school math olympiads( in 6th grade)( answering half of the questions, and half of the ones i answered were rong, only 8 questions).
meaning?
Wait am I watching this on the wrong channel? I'm confused 🤔
haha :D
nice channel.
b(ab + 1) + 7 | a(ab + 1) + b
bk(ab + 1) + 7k = a(ab + 1) + b
b - 7k = (ab + 1)(bk - a)
b = a = 7
Only solution i could find and i'm done
I keep thinking this is in your opinion because of the internet, but is it really?
hmmmmm
4:12 b squared missing ... Uhm, this sucks. LOL
Pure mathematicians are just analytic philosophers who are bad at formal logic
lel
@@PapaFlammy69 Using the method of natural deduction in a mathematical proof is actually quite laborious. I think every mathematician would get strongly annoyed while trying it out, like proving that the square root of any prime number is irrational using natural deduction. (Without jumping steps, obviously)
Absolutely magical great video lovely question,beautiful explanation ,gr8 effort. algeBRUH
:)
0:06 say that shit again and I'll forcibly give you a 3-hour-long lecture on Gödel's incompleteness theorem and its philosophical implications 😤😡
Also 7:36 the second case subsumes the first since everything divides zero, or zero is a multiple of every number
Nice class sir .I am Indian.My country is India. You are best math teacher. Thank you sir.please replay
Where is the proof that the number is not divisible by any non-multiples of a?
you look a bit tired dude, them undereye bags are getting more and more purple everyday
;_;
@@PapaFlammy69 spending most of time gazing at screen during this lockdown thing, eh?
Bruh 5|25=5 and 25/5=5
T
i have a math channel check it out. i´m doing a false proof I found in an algebra book at my school library that 1=2