Solving an IMO Problem in 6 Minutes!! | International Mathematical Olympiad 1979 Problem 1
Вставка
- Опубліковано 2 лют 2021
- #IMO #NumberTheory #MathOlympiad
Here is the solution to IMO 1979 Problem 1!!
------------------------------------------------
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
“Notice that 1979 is prime” of course I noticed
Yaa of course, we all noticed it 😂😂😂
Well, on the IMO they have a lot of time and checking if 1979 is prime takes about 5 minutes.
They were all ready to see 1979 in a problem of 1979. You always learn things about the number of the year.
A rule of thumb:
Always study the properties of the year number. Trust me you will need it a lot!😅
@@littlefermat Yes, for this specific year 2021 we have it is factorized as 43*47, a very useful thing. If you dont know that you may think 2021 is prime
The solution of the problem is so elegant and simple. This is why I've always enjoyed math, yet the spark of finding such a solution often eluded me.
To say nothing of the limited time! I can't do almost any problem if I have a limited time because of my anxiety
I mean you have 1.5 hours
The reason b is not divisible by 1979 is because all the denominators in the sum are not divisible by 1979, in general all numbers less than a prime p are not divisible by p. It may be obvious but necessary for a rigorous proof.
IMO problems get progressively more difficult as you progress from question 1 to question 6. Question 1 is a warm up one and as such this is a good warm up one.
It's just because it's ancient
Nitpicking note: problems 1 and 4 are supposed to be the warmups, 2 and 5 medium, and 3 and 6 TOUGH. The test is split into 1, 2, 3 and 4, 5, 6. Section 2 is generally harder than 1, but 4 usually isn't that hard.
Yes true
Problem creators are brilliant, too!
😮🙏@@bairnqere7u
That's incredible !! I was a participant at IMO 1979, and made 0 point to that question (I thought it was the most difficult of that event). Now I just had a look at your question and solved it mentally without any pencil/paper in less than 1 minute !!
you should have made a better story
Fake
how come lol. it's like a typical Chinese high school math problem - not an easy one but the last problem or one before the last problem.
You took too much. These kind of simple questions must be solved while sleeping.
Yep You learn this type of problem in Sequence & series in high school typically in Asia, you can factor it with (n+2)-(n+1) and creating difference like it. .( just giving example) @@kekehehedede
Superb, Awesome! ❤❤.. Let's all think critically 🔥
What a beautifully elegant solution to a lovely problem!!! 🥰😍
wow pairing the terms up is genius
Really great video,nice content
Nice resolution, elegant and simple
I would like to share some finding(s) with all when I tried to "create" some similar questions. Last term of the question- 1/1,319, 1,319 is/must a Prime! (1,319+1) x 2/3 = 660. 660 is the denominator of the 1st term of modification of the question- i.e. 1/660 + 1/661 +.....+1/1,319 Ha! Ha! Then re-arrange to (1/660 + 1/1,319)+(1/661 + 1/1,318)+.....(1/989 + 1/990).
The numerators in each & every bracket would become 1,979! The proof would be easy. If the denominator of the last term of the question isn't a Prime, or if it is a Prime, after adding 1 to it & multiplying by 1/2, the no. obtained not equal to the denominator of the 1st term of the rearrangement of the original question....p is/may not divisible by the pre-set#.((last denominator in the qn. +1) x 3/2, - 1, Not 2/3!)
Very nice solution. I don’t believe one can get a solution which consists of fewer number of steps.
You are purely genius 👏👏
imo, the proof should be left as an exercise to the grader!
if the result is a reduced fraction a/b where a and b are integers, then we can also write it as 1979a/(1979b). We can then substitute p=1979a and q=1979b: p/q. Because a is an integer, 1979a and therefore q are divisible by 1979.
The task is to solve for all p and q that satisfy given equation, your solution works only for trivial cases ;)
I put the final sum in my calculator and the result is a fraction with enormous numerator and denominator. Without the 1979 factor It approximates to 3.5e-4.
What makes it a worthy IMO problem is that (a) the mathematical prerequisites to understand and solve the problem are quite modest: summation, cancellation and basic prime number facts and (b) that it is very hard to actually find the solution on your own under the time constraints of a written exam.
For the non-mathematically gifted there is another way to show the proposition: Caclulate the sum p/q with a computer to the lowest terms and check the remainders mod 1979. They are 0 for the numerator (as expected) and 1044 for the denomitator. Now we just have to notice that every presentation p/q for the sum can be written as p=k*p_min, q=k*q_min for some integer k where p_min/q_min is the presentation to the lowest term. As p_min is divisible by 1979 also p=k*p_min is divisible by 1979 and the proposition follows. For practical matters, p_min has 577 digits in decimal representation and q_min has 578. Python using the fractions module and a few lines of code help us out.
Sometimes seeing beautiful solution like this remind me why i love math so much
Cool solution i saw this trick on 2014 olympiad but i forget the name of country
Graceful solution
I do realise that this is the first problem, but have the problems gotten harder over the years or are the first questions still this easy?
"so yay we are done"
- every math olympic when they finish a problem
Magical.
It is rigorous enough cause natural series which start and finish by numbers are both not even or both not odd have even number of terms and this grouping can be made.
I love the content
Nice solution although I don't think it is obvious that the denominator 'b' in the final line is not divisible by 1979, this would need to be argued about as well
What do you mean? If you have 1/a + 1/b you can split the denominator into common factors c and coprime factors p and q i.e. 1/a+1/b=1/cp + 1/cq =1/c * (1/p + 1/q)=(p+q)/cpq. Since p and q are coprime and c is the greatest common factor of a and b, no divisor of cpq can divide both p and q at the same time, so this is actually already the maximally reduced form. Since a and b are a product of two numbers less than 1979, they can't contain 1979 and so can't c, p and q. The rest would follow inductively.
@@digxx you just spelled out what he believes should be spelled out instead of just stated.
@@tongchen1226 ah, ok.
Yes, he seems to assume, that a/b is an integer
@@mthimm1no he doesn't
Very interesting is how this problem was generated. It turns out that the prime p must be of the form 6k+5 and also 3l+2. Also the reason for 1979 and 1319 in the problem is that 1319 is of the form 4k+3 and 6k+5 happens to be prime.
Interesting if anything can be said about any of the other cases (number of terms in the sum is 4a+1 or 6k+5 does not happen to be prime)
What board do you use?
The 2nd power to the digit equals let the division in the equation equal the answer.
The creator of this problem is very clever
It is Riemman
Could you cover some geometry problems?
this is brilliant
Love it
0-1/12 has a cool pattern function
Ramanujan sum
4:45 why do you write 1979/k.(1979-k) instead of 1/k.(1979-k) ?
I am missing something in that step
Not sure. For the number to be dividable by 1979, a/b has to be integer. Not sure if that summation at the end results into an integer.
a/b doesn't have to be an integer (in fact, it isn't). You're only checking if the numerator p is divisible by 1979, not the whole number.
Nice problem!
Nice job
At first, I thought the first case to consider is to find 1979, which is 1 and 1979 and somehow prove 0(mod 1) and 0(mod 1979)
Thanks guy
Here's how I did it:
If you group the terms 1,2 3,4 etc. you get these terms plus the last 1/1319 term:
1-1/2=1/2
1/3-1/4=1/12
1/5-1/6=1/30
Making a function to define nth term:
f(n)=2n(2n-1)=4n^2-2n
So our sum is just:
[Sum n=1 to 659 f(n)] + 1/1319
= 4(1^2 + 2^2... 659^2)-2(1+2+...659)+1/1319
Using the sum of squares and normal sum formula:
= 4(329^2) - 2(659)(660)/2+1/1319
= 4*329*329 - 659*660+1/1319
= [4*329*329*1319 - 659*660 + 1] / 1319
So p = 4*329*329*1319 - 659*660 + 1
Then you do mod 1979 on it and viola!
The first trick is not bad, but you have to sum 1/f(n), not f(n)... And there is no inverse square sum formula. Your is solution is also way too big, how did you not notice that? It should be smaller than 1 because it is 1-(1/2-1/3)-(1/4-15)-...
The idea of adding and subtracting the even terms was great. How did you come up with it ? Any structured way of thinking or just a fluke / irrational-realization that worked ?
I like it
I don't see it written anywhere that p over q has to be the reduced fraction. Simply note that a sum of fractions is rational, call it a over b, and let p=1979a and q=1979b.
This is a beautiful problem:
At the solution,
Can you say that actually, a = 1.
… and then it’s a little suspect what youre doing with this.
Since one is summing over integers strictly not on the prime ideal,
The denominator isn’t carrying factors of this prime.
So youre looking at a rational: something in lowest form
1979/…
… by just observing its definition:
No spooky logic implied,
Doesn’t have any factors of this prime.
… which makes me think its not too good of a problem after all these years.. I know thats not what theyre asking.
It’s interesting to see if you can find this symetry elsewhere in this summation to find out if one can write this as 1979*a where a isn’t actually 1:
This would beget spooky logic.
If 1979 isn’t actually a prime: then none of this holds … - but one can’t be unsure about this pre-argumentation
The only thing one can call beautiful is:
The sameness between
(1) Taking away a subsequence, additively
(2) giving that same sequence, additively, whilst taking away
.. the ring element ‘2’ multiplied by the same thing one is giving.
This is the kind of trick one is using here.
You sound like from HK
he is actually
Whats HK?
@@gamedepths4792 Hong Kong
confirmed by hker, but i study in international sch so i have proper eng
I am from hk and his accent sounds so familiar to me, especially that tone
Find all a,b,c,n natural number such that:
a^3+b^3+c^3=n*a^2*b^2*c^2
(1,1,1,3) and (1,2,3,1)
@@raffaelevalente7811 The first one should be (1,1,1,3).
Nice but how did you get to that path?
can somebody explain why only some a,b in natural number but not all. Another question is why we should mention a.b=1
can anyone explain how he got the numerator to be 1979 in the summation?
Maybe we could write the first terms as the sum of 1+3+... over 1979!! ??
The handwriting looks familiar...
The problem statement didn’t say gcd(p,q)=1, so we can take p/q = (1979p’)/(1979q’)
Where p’/q’ is equal to the given series. (lol)
Why are your 9s mirrored ?
In this question one thing is observable that for the first S(n) number decided by s(n-1) gives result 1.like
Elegant
whatever the value of the series is, multiply the the top and bottom by 1979. that means that the top will be divisible by 1979. done
At 6:11, how do you know that 1979 does not divide b?
Take a look 4:33 try to undersand it by yourself😉
from fractions import Fraction
total = Fraction(0)
for den in range(1, 1320):
total += (-1)**(den + 1) * Fraction(1, den)
print(total.numerator % 1979 == 0) #True
wait does finding the derivative, simplifying then finding the integral work? a bit like the infinite version of this problem which converges to ln(2)
Wow thanks from India , my school teacher of 8th grade gave me this homework, I didn't know it was from imo 😅
How come every JEE doing Indian claims to be so smart but their country is so polluted? Look at Ganga river, they can’t even solve that. Smart for what, tech sup?
@@benyseus6325 rip logic .. dumbhead
@@benyseus6325 what does that have to do with math? and how did you come to the conclusion that Indians claim themselves smart?
@@alanyadullarcemiyeti lol. Just look at the comments, they are always claiming to be smart. But they are hypocritical because their country is in turmoil right now. They are not smart because that can do JEE
@@benyseus6325 If all Indians have an ability to perform better on tests like the JEE, then they *are* smarter. However, there is no evidence to suggest that Indians have such an advantage. So, these claims are wrong (if anybody's really made them). However, the *individuals* who are working their arses off to clear JEE have definitely developed an expertise in those subjects that others don't have. This makes them smarter (arguably, because some people tend to perceive 'smart' as 'gifted') than others. The *individuals* who are actually able to do great in JEE (advanced. not mains. mains are easy) are some of the smartest we (human beings) have.
As far as pollution and such things are concerned, smartness cannot be determined by the condition of the homeland. Most people have nothing they can do about it (not enough influence or money or smarts to overcome political corruption and stuff like that).
Of the ones who can actually do something about it, most (most, not all) are selfish, and would rather have a carefree life in some developed nation.
That's selfishness, heartlessness, maybe even hypocrisy in a lot of cases. The patriotism promoted in India is limited to cheesy stuff like standing up when the national anthem plays in the theatre. Very few people in the society are actually doing something for the country.
But all of this definitely does nothing to show that Indians are not smart.
why 4:12 write 989 on the top? I did not learn about it , hope you can teach me , thanks
On the line above the summation he grouped terms so that the first term in each group goes from 1/660 to 1/989 thus the summation on the next line is from 660 to 989
Nice proof
Neat
Smart asf
Big fan bro
Very pro bhai
Make me look very chutiya
Hard work bro
Very liking your nice work 👍
hahahahhhah 🤣🤣🤣🤣🤣🤣🤣 wtf bro
Xd
Actually it is not difficult after seeing this but this is tricky one
To get full credit for this problem did you have to include a proof that 1979 is prime? Or does it suffice to strategically memorize the prime factorization of the year you're taking the IMO? :-)
It would be assumed that you came with that knowledge without need to prove it, but it's also easy to check since 1979 isn't too large. You only need to check all primes less than the square root of 1979. I don't know the square root of 1979, but I know 45^2 is 2025 so that's close enough. If none of the primes from 1 to 44 are divisors of 1979 then 1979 is prime.
Why doesn't prime number 1979 divide b?
If I got your question it's because the denominator is made of only smaller factors than 1979
wait this is in my book
Turn on postifications
Cool Solution but as a imo student from where should u know that 1979 is prime?
Was thinking the same thing, how would you figure it out?
You could check if any primes devide 1979 below sqrt(1979). It's not that fast but it will get the job done
@@koenmazereeuw4672 yes that should work i mean you got more than an hour for one problem
At least as far as I can tell, in most mathematical olympiads (at least nowadays) the participants should know the prime factorization of the year, e.g. 2021=43•47. So if the year is a prime number, they should know that, too. I even think that in many contests you can then simply state that as given. At least here in Germany it is like that.
@@TM-ht8jv Yeah that's true. Thanks for the info btw I have to do the first round tomorrow :)
Good morning
Lesson learned: Before going to IMO, check if the current year is a prime number.
The particular problem arrangement is true for any prime number in place of 1319....
No
We will not have the sum equal to 1979 then
I am confused about why 1979 not divides b.
Because prime factorization of "b". Content numbers All smaller than 1979
@@karanagrawal8499 but can't the sum of them become bigger than 1979.
I got it.Not need to explain.
Could you help me with problem, i send you a message on email, but you didn t response me?
Amazing problem but i have a question how we could now a large prime number like 1979 😥
You "only" need to check it isn't divisible by any prime up to sqrt(1979), i.e. up to 43; fourteen divisibility checks is boring, but doable.
Yes thank u but i wonder if there are any easier way !!!!!
@@tonyhaddad1394 Easier than just 14 divisions?? Don't think so.
@@landsgevaer your e right but when i sayed easier i mean in a bigger number situation
There are polynomial time primality tests. I think that is the best we can do asymptotically.
en.m.wikipedia.org/wiki/AKS_primality_test
But I don't think anyone would ever do this by hand in a math olympiad. ;-)
I just learned that we can think "critically" in 6 minutes. In school, critical thinking was often associated with "long thinks".
So the answer is 1?
I didn't get it how you prove that p must be divisible by 1979 in the end ?
The sum of fractions in the end would be (some great number)/(680*681*682....989*(1979-680)*(1979-681)*...*(1979-989) (because of how addition of fractions work) and 1979 is not divisible by this because it is prime (sorry for my english)
@@martinkopcany6341 but why if 1979 is not divisble by b then p must be divisible by 1979
@@abirliouk8155 p/q=(1979*a)/b so p=((1979*a)/b)*q so q is divisible by 1978
@@martinkopcany6341
but p/1979 =(a/b) *q and a/b is not an integer
@@abirliouk8155 a/b is not an integer but (1979a/b)*q is an integer because p is an integer. And if (1979*a*q)/b is an integer and b is not divisible by 1979 then (1979*a*q)/b is divisible by 1979.
You sounds 5 years younger than me, but 500 times smarter.
easy problem considering imo.
How did tou go from 1978 is not divisible by b to p being divisible by 1978? Is there some sort of a trick? Otherwise great vid
So basically you ended up with p/q = 1979a/b. Since you know that 1979 is not divisible by b, and a and b are coprimes (hence a is not divisible by b) 1979a/b is in its most simplified form AND is equal to p/q so you get that p=1979a ---> p/1979 =a :D
@@albertoferis8250
What happened to the b and q. Could you please explain though?
@@RogerSmith-ee4zi the response is in the first reply but I will try to reexplain (make it clearer) : the fact that b does not divide 1979 and that also a and b are coprimes that makes 1979a and b coprimes therefore 1979a/b is in its most reductible form but so does p/q and there is only one unreductible form so p/q=1979a/b => p=1979a and b=q and therefore the result p/1979
P/q doesnt have to be unreductible mate if we can divide p by 1979 we can divide 2p or 3p am i correct pls reply
@@onurbey5934 yes what you did say is correct but that was not what we were trying to prove in fact p/q = 1979a/b will give you that p = 1979. aq/b but that last part q/b might not be a natural number (if that what you were saying)
I came here not even knowing times table
I got lost on the sumation part (note im finishing calc 1 so xd)
Are you uncle roger in disguised ?
this was easy :D
You are piro
Suggestions: you should be more detailed for the answer since you skipped many crucial steps.
I think the people who watch these kind of videos don't need those steps
What crucial steps did he skip? On the whole this solution seemed to cover just about anything. I wasn't left with any questions.
Why 1979 not divisible by b??
because 1979 is a prime number
why 1979 not divisible by b?
cause as it say 1979 is a prime number and b is a factor of number that is less than 1979
The denominator of a sum of fractions is the least common multiple of the denominators of the summands (or a factor of the least common multiple).
Because 1979 is prime, this means that in order for the denominator of the sum to be divisible by 1979, the denominator of at least one of the summands must also be divisible by 1979. However, the summands here only have denominators between 660 and 1319, all of which are less than 1979 and therefore are not divisible by 1979.
Therefore, the denominator of the sum is not divisible by 1979.
🎉
Does 1979 being a prime matters in this question?
Yes, if it was not a prime it could be divisible by some factor in the denominator, thus the fully reduced fraction's enumerator wouldn't be divisible by 1979 (like, imagine if it was 1980, the denominator has both 660 in its factor so you would only have 3*something in the enumerator(and even this 3 would go away with 663), which could be not divisible by 1980
@@chamsderreche5750 @5:48 what does (a.b) = 1 mean
@@Nothingeverything192 sorry for the VERY late reply, I guess he meant gcd(a;b)=1(greatest common divisor)
Hk?
The key is that 1979 is a prime number. this is very hard to know.
p/q = 0 (Zeros) 456789 = 6
pourquoi 1979 ne divise pas b ?
You can think of it this way, while adding all the fractions, the denominator is basically all the individual denominators multiplied, and cuz there’s no 1979 term in individual denominators, and 1979 is prime, so it can not be constructed by multiplying two different numbers, hence b is not divisible by 1979.
1🎈7🎈
I solved It in 2 min and 46 seconds
(Insert current year here)