Solving an IMO Problem in 6 Minutes!! | International Mathematical Olympiad 1979 Problem 1 

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

But seriously though, before any math competition, we always memorize the prime factorization of the current year, the next year, and the previous year

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

Yeah very true

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

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

hahahahhhah 🤣🤣🤣🤣🤣🤣🤣 wtf bro

Xd

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.

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

Any 4 digit number is not hard to check if its prime if you already know that primes till 100 Here, since square root of 1979 is somewhere between 44 and 45 (since 44^2 = 1936 and 45^2 = 2025) We can check if any of the primes less than our equal to 44 divides 1979 In this cases the primes are given by 2,3,5,7,11,13,17,19,23,29,31,37,43 (13 in total) Since 2,3,5 don't divide 1979 for obvious reasons, we just have 10 primes to check which won't take more than 5 minutes to do

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

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)-...

Take a look 4:33 try to undersand it by yourself😉

If I got your question it's because the denominator is made of only smaller factors than 1979

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.

2024 is an even number i.e. it is divisible by 2. As 2 is a prime number, 2024 is also a prime number 🤓🤡

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. ;-)

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.

It is Riemman

No actually this is Euclid itself.

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)

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.

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 :)

Ramanujan sum

You are piro

The task is to solve for all p and q that satisfy given equation, your solution works only for trivial cases ;)

the problem says "Let p,q be natural numbers" which means that for any natural numbers p,q so that the fraction p/q is equal to the sum then p should be divisible by 1979. What we are trying to argue is that for any pair of p,q so that p/q is the sum, then p should always be divisible by 1979. So actually the problem is equivalent to asking whether p/q if written in simplest form, is p still divisible by 1979? Because if that's the case then any other fraction p/q that is equal to the sum will always have p divisible by 1979. While it is true that you can always multiply by 1979 and get another fraction where it trivially holds, the question is asking you to prove that it is always the case for any p,q pair not just those trivial cases. To illustrate, 1/2 for instance can be written as 1979/3958 as well as 3/6 or 2/4. But in the case of 3/6, 3 is not divisible by 1979. Which is a counterexample. The problem equates into proving that there is not a single counterexample in the case when p/q is supposed to equal that sum.

@@jimallysonnevado3973 Yes I see what you mean. There are technically infinitely many different solutions for the fraction because kp/kq = p/q and when p and q are prime between them, we have to proove that p = 0 mod 1979. But even when I knew that, I didn't managed to solve the problem :,).

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.

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

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.

(1,1,1,3) and (1,2,3,1)

@@raffaelevalente7811 The first one should be (1,1,1,3).