Japanese Math Olympiad

Japanese Math Olympiad | 2020

Рет қаралды 32,519

Күн бұрын

Пікірлер: 72

@littlefermat 3 жыл бұрын

In other words, when you are given some ugly quantity is a perfect square, Always try proving that this quantity is between two consecutive perfect squares. And your problem will be done!

@TechToppers 3 жыл бұрын

Yup! A typical bounding trip which I got to know 3 months back.

@mukaddastaj5223 3 жыл бұрын

Lol, i dunno why, but we call it policeman's lemma😂

@AlephThree 3 жыл бұрын

Nice. I like the style of leaving some of the easier bits as HW and avoiding a 20 minute video.

@kevinmartin7760 3 жыл бұрын

Around 9:00 when he offers a^2+2a+1 as being greater or equal to a^2+a+2 this is only true of a>0. By this point though we could have observed that the formula m=2^(a+1)-3 gives a negative m when a is zero and so excluded this case.

@mcwulf25 3 жыл бұрын

Well explained. I like it when the answer isn't just 0 or 1.

@GhostyOcean 3 жыл бұрын

5:48 I don't understand why the middle term must be a perfect square. I follow the proof up to that point, but this step confuses me. Edit: OH! It's part of the requirements on the left. I couldn't see because Michael was standing in front of it.

@TechToppers 3 жыл бұрын

We want it to be. It's not necessary. According to the problem statement.

@Luxaray2000 3 жыл бұрын

My favorite proof of the "exponential outpaces polynomial" fact is the following: Let P(x) = sum a_i x_i be a polynomial and E(x) be some exponential function e^(kx). Then lim as x tends to infinity of E(x) / P(x) is infinity by L'Hospital. This is because the quotient is infinity / infinity, so applying L'Hospital we get an exponential over a polynomial of degree 1 minus what we started with. Continuing this process will leave a constant polynomial in the bottom and infinity in the top.

@GhostyOcean 3 жыл бұрын

Important caveats: (1) this is directionally dependant. So if x is going to positive infinity, k>0 or else e^(kx)

@mirkorosner1044 Жыл бұрын

Yes, but this does not give a quantitative bound.

@GreenMeansGOF 3 жыл бұрын

What happens if we allow m,n to be negative? EDIT: The fact that we always want a perfect square forces n to be positive. If we allow m to be negative, then it can be argued similar to how it was done in the video that m=-1 and n=1 is the only other possible solution over the integers.

@vinc17fr 3 жыл бұрын

No need for calculus to prove that 2^a > (a+1)^2 for a ≥ 6. This can be done by induction: true for a = 6, and then 2^a doubles at each iteration while the factor for the square is ((a+2)/(a+1))^2 < 2.

@davidbrisbane7206 3 жыл бұрын

That's what I thought too.

@JMTchongMbami 3 жыл бұрын

8:58 a²+a+2

@ShinySwalot 3 жыл бұрын

I was just warming up to studying Japanese, this sort of counts right?

@ΓιώργοςΚοτσάλης-σ1η 3 жыл бұрын

Inequality proved by calculus could also be proved by induction more easily

@robonthecob5092 Жыл бұрын

4:10 why is this possible? Why can we just put that bound on m

@GeoPeron Ай бұрын

This is confusing, so I recommend switching the orders in which he made the assumptions. First note how the polynomial that bounds m is less that 2^(a+2), then substitute that in for m in the bottom to create an inequality and notice how it factors into (2^a + 2)²

@goodplacetostop2973 3 жыл бұрын

13:57 それはいいところで止まっている

@jplikesmaths 3 жыл бұрын

That’s so good 😂

@reshmikuntichandra4535 3 жыл бұрын

Hey Michael, It'd be great if you tried some problems from the 2021 INMO(Indian National Mathematical Olympiad). Regards, Adarsha Chandra(Fan from India)

@morbidmanatee5550 3 жыл бұрын

Brilliant! Love this channel

@ekramulhaque8028 3 жыл бұрын

One more way to solve; (n^2+1) is divisible by 2m So n^2+1 is even; consequently n^2 is odd. Therefore n is odd. n-1 is even. Let (n-1)/2=k ( eq.1). Where k is some natural no. n-1=2k. 2^(n-1)+m+4= 2^2k+m+4= (2k)^2+m+2^2. Now; 2^(n-1)+m+4 is a perfect square. So (2k)^2+m+2^2 is a perfect square. So 'm' must be equal to +2*2*2^k or -2*2*2^k. as it will give (2^k +/- 2)^2 since m is natural. We have m=2*2*2^k 2^k=m/4. 2= kth root of (m/4). Now k cannot be grater than 2 other m/4 will not be a natural no. and cannot be equal to 2. so k=1 or 2. putting (n-1)/2 =k= 1 or 2 (by eq.1) Therefore n=3 or n=5

@kensmusic1134 2 жыл бұрын

but this is false. 25+1/2^k+3 which is not an integer

@Reliquancy 3 жыл бұрын

I thought this was going to be one of the ones where all the possible values are like 2 and 3, but that 61 and 11 works is kind of surprising.

@danielleza908 Жыл бұрын

They probably wrote the question intentionally to suit to those numbers.

@anshuldeshmukhansh7390 3 жыл бұрын

I got one answer (m,n) = (1,3) by just observation But second answer is quite surprising 😅😉😁👍 Thank you...

@goodplacetostop2973 3 жыл бұрын

HOMEWORK : Let n be an even number which is divisible by a prime bigger than √n. Show that n and n^3 cannot be expressed in the form 1 + (2x + 1)(2x + 3), i.e., as one more than the product of two consecutive odd numbers, but that n^2 and n^4 can be so expressed. SOURCE : Suggested for the 25th Spanish Olympiad but that didn’t make the cut.

@noahtaul 3 жыл бұрын

The conclusion is equivalent to the number being an even perfect square, so obviously n^2 and n^4 are, but n (and therefore n^3) can’t be because it can only have one copy of the large prime in its factorization.

@keyanaskar1981 3 жыл бұрын

Video was uploaded 1 week ago, then how is your comment 3 weeks old?

@goodplacetostop2973 3 жыл бұрын

@@keyanaskar1981 Time travel

@particleonazock2246 3 жыл бұрын

Nice brother, but wth.

@andy-kg5fb 2 жыл бұрын

1+(2x+1)(2x+3) is (2x+2)². Let p be the prime larger then √n that divides n. Then we have p²>n; Therefore p² does not divide n. Hence n is not a perfect square. Similarly you can prove that n³ is not a perfect square. N²,n⁴ are perfect squares therefore we can express them as (2x+2)² with x=(n/2)-1 and (n²/2)-1 respectively. Both of which must be natural.

@fatimahmath4819 3 жыл бұрын

1:13 why did you already delete the m value??

@Ardient_ 3 жыл бұрын

Here is an extremely hard math problem you can try: Find all (a,b,c) which are natural numbers which satisfy both of these equations: 1/a=(1/b)+(1/c)+(1/abc) and (a^2)+(b^2)=c^2 dosen't seem like much but trust me it is a monster to prove.....

@shivansh668 3 жыл бұрын

Pythagorean triplets

@Subrankur 7 ай бұрын

The intervention prism in math is equal to zero. Not equal one of n And one value.

@fatimahmath4819 3 жыл бұрын

2:15 why did you say that the denominator is equal or less than the numerator

@harivatsaparameshwaran4174 3 жыл бұрын

Otherwise you can get fractions / decimals and according to the question you can get only natural numbers.

@fatimahmath4819 3 жыл бұрын

@@harivatsaparameshwaran4174 thank you very much

@matthewdodd1262 3 жыл бұрын

Its not hard to see how m

@Jack_Callcott_AU 3 жыл бұрын

Nice little sojourn!

@xCorvus7x 3 жыл бұрын

Well, we don't need to check the perfect square condition at all, since m was constructed to be a perfect square.

@hongphitrinh6015 2 жыл бұрын

Why 2^(2a)+m+4

@amiasam3354 3 жыл бұрын

I am a time traveller

@timetraveller2818 3 жыл бұрын

@@aashsyed1277 liar! he is just smart

@amiasam3354 3 жыл бұрын

@@timetraveller2818 🤣😂🤪

@timetraveller2818 3 жыл бұрын

@@amiasam3354 xd

@amiasam3354 3 жыл бұрын

@@timetraveller2818 🤪

@helo3827 3 жыл бұрын

can someone please explain to me how to prove 2a^2+2a+1

@SSGranor 3 жыл бұрын

This is a little inelegant; but, here goes. Assume there is some value of a such that 2a^2+2a+1=2. So, what this shows is that if the inequality holds for any a>=2, it also holds for all higher values of a. And, it just remains to check 0, 1, and 2, which is simple enough and returns 1

@yukihyde1 3 жыл бұрын

a mathematical induction!

@egillandersson1780 3 жыл бұрын

Great !

@stewartcopeland4950 3 жыл бұрын

the couple m = 5, n = 57 seems to work but you will see below why it isn't , through informed responses

@erikpedersen2245 3 жыл бұрын

sqrt(2^(57-1) + 5 + 4) = 268,435,456.0000000167638063... so m = 5 n = 57 doesn't work, 2^{n-1} + m + 4 isn't perfect square

@stewartcopeland4950 3 жыл бұрын

@@erikpedersen2245 Indeed, my digital calculation software rounds this number to an integer...

@richardsandmeyer4431 3 жыл бұрын

You seem to be incorrect here. (5,57) satisfies the first condition, namely (57^2+1)/(2*5) = 325 which is an integer. However, the second condition yields: 2^(57-1)+5+4 = 2^56+9 which is not a perfect square. It lies strictly between two consecutive perfect squares, namely (2^28)^2 and (2^28+1)^2, and therefore cannot be a perfect square itself. Did sqrt(2^56+9) perhaps look like an integer on a computer with insufficient precision?

@athysw.e.9562 3 жыл бұрын

He proved mathematically there's no other solution, Python won't help you...

@serdarbozdag3749 3 жыл бұрын

Harder one next time please!

@mukaddastaj5223 3 жыл бұрын

Выручили!) Мне как раз сегодня дали эту задачку, я порешала, но ничего не вышло( а это видео быдо на очереди к просмотру уже месяц, и вот как пригодилось🤩🤩