A Fun One from the British Maths Olympiad

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2 жыл бұрын

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Today a fun one from the 2009 Bri'ish MO. Given the equation sqrt(a)+sqrt(b)=sqrt(2009), find all solutions (a,b) out of the positive integers. Enjoy! =)
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Пікірлер: 268

@bowkenpachi7759 2 жыл бұрын

2:37 “That didn’t work out, it can fuck off” made me a subscriber

@xainy3500 2 жыл бұрын

I loved it too🤣🤣

@samuelatienzo4627 2 жыл бұрын

@@xainy3500 I subscribed too because of this...

@ivanperkovic8668 2 жыл бұрын

same 🤣🤣🤣 had to rewind back to check if he really said that

@christiansrensen5958 2 жыл бұрын

It's how we Germans get through a day efficiently without any fuss and wasted time.

@peterosborne1302 2 жыл бұрын

amen

@ironici 2 жыл бұрын

I love how the Olympiad includes the current year in many of its problems

@holomurphy22 Жыл бұрын

you better learn the factorisation of the present year

@moth5799 8 ай бұрын

@@holomurphy22 I did that for the SMC this year and it made the first question so easy, since it was entirely about 2023 = 7*17²

@clintonmead 2 жыл бұрын

37 years old and I never have heard this divisibility trick for 7. Not sure I learn something new every day anymore but I have today!

@PapaFlammy69 2 жыл бұрын

nice! :D

@wherestheshroomsyo 2 жыл бұрын

here's another nice trick. 1001=7*11*13 when checking divisibility by 7, adding or subtracting any multiple of 7 is fair game. 2009-2*1001=7 7-1*7=0 0 remainder thus it is divisible. same trick works for divisibility by 13 since it too is a factor of 1001 here's two more great numbers. 111111 and 10^6 - 1 these are also multiples of 7 and 13 for very large numbers separate the digits into groups of 6, then add together each group, then check if the sum is divisible by 7 this trick is equivalent to subtracting multiples of 10^6 - 1 This is the definitive way of check 7 and 13 divisibility. it's the same idea as summing all the digits but now it's groups of digits.

@NUSORCA 2 жыл бұрын

You can instead start learning languages and writing systems and I guarantee you theres something new every day 끈氣가 必要하다

@pucek365 2 жыл бұрын

But it is quicker to just divide by 7

@GreenMeansGOF 2 жыл бұрын

Yeah. It’s relatively new knowledge. Search for Chika’s rule for divisibility by 7 to learn more about it.

@noreng9333 2 жыл бұрын

At 8:40, you can easily conclude that a = 41 * n^2 due to the fact that the square root of (41 * a) must be a natural number. At which point you really only need to plug in numbers for n and check if they form a solution.

@musicalchicken1024 2 жыл бұрын

yea that's what I was thinking. This guy is taking the more mathematical approach whereas we're taking the more constructive algorithm type approach

@bestcreations4703 2 жыл бұрын

@@musicalchicken1024 I think it could be for one of two reasons, one is that he is trying to teach and so explaining it in depth mathematically is helpful which it was for me or two, he is writing it more as a proof and so the more mathematical approach is favoured. I’m not sure though.

@musicalchicken1024 2 жыл бұрын

@@bestcreations4703 It really just depends on what context you are in. If you were actually taking this olympiad, i am sure that you would save time and just take the constructive approach that the original guy suggested. But if you were a teacher or someone trying to prove your answer, you would surely not care as much about time and explain your mathematics in depth

@Sunsaparilla 2 жыл бұрын

FM: "41 is not zero" Me: "why?" FM: "it's the successor of 40" Me: "oh"

@lucador977 2 жыл бұрын

We can also decompose 2009 in factors of prime numbers and factorise the sums left by root of a

@leif1075 2 жыл бұрын

I thought of factoring but ehst donyou mean by sums left by a?? What about sums left by b..what do you mean by sums left by something??

@lucador977 2 жыл бұрын

@@leif1075 i just traduced what I had in head at the moment , a representation of an equation with (√a +√b) on the left and 2009 on the right . I usually speak French this is why I did that error , thank you . Now I will be more careful

@hammadsirhindi1320 2 жыл бұрын

Solve for x and y x^2 +y^2 = a x^3+y^3 = b

@lucador977 2 жыл бұрын

@@hammadsirhindi1320 Y can be considered like that : y=(b - x^3)/(a - x^2) Thanks 👍 that was really interesting If y and x are Real numbers and not only digits, I don't think we can solve, having precises and finite solutions, but we can notice that the function Y is symmetric by the curve Y=X because we can just replace X by Y and this is always true . Now I hope I helped you like you hoped 😂

@hammadsirhindi1320 2 жыл бұрын

@@lucador977 actually this problem has 6 solutions in general, maybe real or complex according to different values of 'a' and 'b'

@Henrix1998 2 жыл бұрын

5:10 the content inside the square root could be form of N/2 as well and still be element of natural numbers as a whole. 8:45 N/7 as well.

@commanderoreo-6388 2 жыл бұрын

@@tusharmuraliold I didn't realize the root of any natural number has to be irrational or a natural number until I read your comment. Great thinking!

@HagenvonEitzen 2 жыл бұрын

@@tusharmuraliold Yes, the claim was correct, though the argument was wrong

@MyOneFiftiethOfADollar 2 жыл бұрын

Never had seen the divisibility by 7 test either. One can prove it with modular arithmetic using the fact -20==1 mod 7. What a fun channel you have spawned! Social aspect of your channel is above the stratosphere.

@PapaFlammy69 2 жыл бұрын

hehe, thx :)

@holomurphy22 Жыл бұрын

2:37 but it does work out. You get ab is a square, so that a=n^2*x and b=m^2*x, with x a squarefree integer. So (n+m)sqrt(x)=7sqrt(41) and thus x=41, n+m=7, yielding the solutions.

@sbunny8 Жыл бұрын

I don't think that works. Given ab is a square number, that's not enough to conclude that a and b are multiples of square numbers. Counterexample: a=2, b=18, ab=36.

@holomurphy22 Жыл бұрын

@@sbunny8 Hi :) In your example a = 2*1 and b = 2*9 Notice that 1 is a square! A proof of my statement can be as follow : let p be a prime number that divides the product ab. Let's write vp(ab) the exponent of p in the prime decomposition of ab. Since ab is a square, 2 divides vp(ab) (to see this, you may write ab=k^2 for some integer k, and use the prime decomposition of k. Keep in mind that this decomposition is unique). Now, since vp(ab) = vp(a) + vp(b), we get that vp(a) and vp(b) have the same parity (both odd or both even). Let's define x as the product of every prime p such that vp(a) is odd. One can show that x verifies the property stated. Edit: notice that every number is a multiple of a square (multiple of 1). The property I use is a stronger statement. Both a and b are the same multiple of some squares

@dneary 2 жыл бұрын

Another nice way to do this is to take into geometry space: square both sides, then isolate and square 2sqrt(ab) to get a parabola with oblique axes a^2-2ab+b^2-2019a -2018b +2019^2 = 0 - then transform the axes to get a y=x^2 parabola

@dneary 2 жыл бұрын

Although this doesn't help so much with rational or integer solutions...

@Anonymous-wj6bu 2 жыл бұрын

This problem boosted my confidence. searched up the prime factorisation of 2009 and got sqrt(2009)= 7sqrt(41) from there I knew sqrt(a) and sqrt(b) would have to equal (something)sqrt(41) so I just did sqrt(a)= sqrt(41), 2sqrt(41), 3sqrt(41), 4sqrt(41), 5sqrt(41), 6sqrt(41), and 7sqrt(41) and same thing for b except in reverse order From there it’s very easy to obtain the values of a and b

@fahrenheit2101 2 жыл бұрын

Not quite. You found some values, but have no evidence to suggest that you found them all, which is the actual challenge of this problem. As obvious as it may seem to you that this is the only way to pick the as and bs, you still have to prove it to get full credit. Each problem in this competition is marked out of 10, and 'incomplete or poorly presented solutions will score few marks'. Your job as a mathematician is not to convince yourself of the truth, but to convince anybody, even the biggest sceptic, that you are correct.

@Anonymous-wj6bu 2 жыл бұрын

@@fahrenheit2101 you’re right. I’ll try to see if there’s any way to prove that my solutions are the only valid ones.

@fahrenheit2101 2 жыл бұрын

@@Anonymous-wj6bu I mean, the video does do just that, but fair enough - it's worth working through on your own.

@Anonymous-wj6bu 2 жыл бұрын

@@fahrenheit2101 I didn’t really watch the whole video so I could leave it unspoiled

@colinbalfour1834 2 жыл бұрын

@@fahrenheit2101 I mean with some decent intuition you’d know it’s the only solutions. If you wanted to prove it more formally you just go from sqrt(a) + sqrt(b) = 7sqrt(41) to ksqrt(41) + (7-k)sqrt(41) = 7sqrt(41) and then a = 41k^2, b = 41(7-k)^2. And then since b > 0 k < 7, k >= 0. Yes they didn’t prove it in math terms because they probably don’t know how but the thinking is right there it’s just not formalized

@grandpoisson50 2 жыл бұрын

5:00 "If 2 times s is an integer, then we also must have that (since 2 is an integer) that s is an integer." 😳 (Natural numbers are closed under multiplication, not division)

@grandpoisson50 2 жыл бұрын

same at 8:33

@grandpoisson50 2 жыл бұрын

but yeah, we are just missing the argument that the square root of a number can never be a half integer (or a non-integer multiple of 1/7). Basicly we can just mention that sqrt(integer) is either irrational or integer

@lambda2693 2 жыл бұрын

Replying twice to yourself . Try to fit it all in one comment

@prestonang8216 2 жыл бұрын

@@lambda2693 no but why does it matter if its all in one comment

@lambda2693 2 жыл бұрын

@@prestonang8216 just saying its better to be concise and not just on youtube but this is a major problem that people even smart people cant organize their thought process which dampens their efficiency

@Fireworker2K 2 жыл бұрын

really liked the video, there were a lot of moments in the video where your line of thinking really surprised me as I could see myself doing the exact same thing. Oh and thanks for the divisibility trick for 7. I'll keep this problem in the back of my head if I ever get into nerd talk with other maths fans

@Foxxey 2 жыл бұрын

2:37 my fav moment

@suka_sukaGaming 2 жыл бұрын

I really enjoy your explanation, thanks!

@PapaFlammy69 2 жыл бұрын

@amandacoelho0609 2 жыл бұрын

I started giggling at "3 hundred ....and.... 69, nice!!" Need to get my head out of the gutter😂😂

@nicolascampailla 2 жыл бұрын

Well I came up with a different solution, I factorized 2009 as 7•√41, so √a+√b would have to be all of the different combinations of 0•√41+7•√41, 1•√41+6•√41, etc and to get the values of a and b we would square each of these so all of the number pairs would be (0,2009) (41, 1476) (164, 1025) (369, 656) and the same thing reversed

@tahjsimon9058 Жыл бұрын

That was exactly how I did it 😃

@hannesthorsell3566 2 жыл бұрын

As for the very first approach you mentioned, squaring both sides, that's how I solved the problem initially. It's essentially the same solution as yours, except it starts at the "other end". I'll post the start of the solution here: Square both sides to get (1) a + 2\sqrt{ab} + b = 2009. Thus \sqrt{ab} is an integer. Consider the prime factorization of a and b. Since \sqrt{ab} is an integer, the prime factors of a with odd exponent must also occur in the factorization of b with odd exponent, and vice versa. Let the common prime factors with odd exponent be p_1, p_2, ..., p_n and define c = p_1p_2...p_n. Thus we can write a = a_0^2c and b = b_0^2c. Plug this back into (1) to get c(a_0 + 2a_0b_0 + b_0^2) = 2009, or (a_0+b_0)^2 = 2009/c. Since the prime factorization of 2009 = 7^2*41, we must have c = 41, for the left-hand side is a square. After this, we proceed just like in the video, since we've gotten a and b in the form 41*k^2.

@hbpr714 2 жыл бұрын

im in class 8 and thanks this is the first time i actually understood what you did

@mr.inhuman7932 2 жыл бұрын

That video was amazing!

@donmoore7785 8 ай бұрын

Outstanding!

@Tactix_se 2 жыл бұрын

“It can fuck off” 😂😂

@elisabethranne2377 Жыл бұрын

Very cool problem!! Enjoy it very much !! Since i did my math final exam in February i am more interest to do some math in my freetime and your channel help me to do this!! ❤️❤️☺️☺️

@whaddoiknow6519 2 жыл бұрын

Fantastic! A simple algorithm giving a divisibility rule for 7. I have never seen that before -- not even in Hardy and Wright.

@wherestheshroomsyo 2 жыл бұрын

for very large numbers, another great trick is to separate the digits into groups of 6, then sum all the groups, then check if the sum is divisible by 7 this process works because 10^6 - 1 is a multiple of 7. the same trick works for all prime factors of 10 ^ 6 -1 as well. namely 3, 7, 11, 13, 37 since 3*3*3*7*11*13*37 = 10^6 - 1

@itismethatguy 2 жыл бұрын

@@wherestheshroomsyo bro I don’t get why it doesn’t work with 14 minus 8 is 6 so is it only till 3 digits and higher cuz that doesn’t make sense to me

@probropalzlive6961 Жыл бұрын

@@itismethatguy you need to remove the last digit being doubled. 14 implies 1-8 (14 with 4 removed, 4 doubled) which implies -7 which is 7×-1

@holomurphy22 Жыл бұрын

Hardy and Wright... dude this is basic modular arithmetic. I dont think it is worth publication

@yoloswaggins2161 Жыл бұрын

Another way to do divisibility by 7 for decimal numbers is to pair two-and-two digits and treat it as base 100, 100 is 2 mod 7 so you reduce to binary number system. 2009 -> 20 09 -> 6*2 + 9 -> -1*2 + 2 = 0 mod 7 It's pretty tractable to do in your head. Another thing you can notice here is that the numbers get these neat bases like 1, 2, 4, 8, 16, 32, ... = 1, 2, 4, 1, 2, 4, ... mod 7 so that you can you treat every 6 digit number as an independent unit. Indeed this means that for +6 digit numbers you can permute every 6-digit group without changing the value mod 7, just like you can for every 1-digit group in mod 3.

@martinshoosterman 2 жыл бұрын

There's a small shortcut that you could have theoretically done in a problem like this. I'm not sure if it would ever save time but it's interesting. You technically don't need the prime factorization of 2009, you only need to know if 2009 is square free or not, and if it is not square free then you need to know what it's non square factors are. (Which means now you do need the prime factorization). If it was square free though, then this can be checked much faster than checking if its prime. If you first check divisibility by 2, 3 and 5. (Which is automatic) you then only need to check divibility by 4 (which you've already done), by 9 (which you've already done) by 25 (which you've already done) and by 49. For any number less than 2500, if it's not divisible by 2, 3 and 5 and not divisible by 49 then it's square free.

@hammadsirhindi1320 2 жыл бұрын

Solve for x and y x^2 +y^2 = a x^3+y^3 = b

@any_think 2 жыл бұрын

you can deduce by the fact that sqrt( 2009 ) = 7 sqrt(41) that a, b must both be of the form n sqrt(41) for some integer n. This is because a,b must be members of the group Q sqrt(41).

@mr.kansaltant 2 жыл бұрын

Right, this is how I solved for it further. Let a > b and since they are both natural numbers then sqrt(a) - sqrt(b) = n×√41 sqrt(a) + sqrt(b) = 7×√41 ==> sqrt(a) =√41 × (7+n)/2 ==> sqrt(b) =√41 × (7-n)/2 Possible values of n are; 1,3,5,7 And voila you have your 4 pairs

@puikihung5882 2 жыл бұрын

not group, should be field

@sbunny8 Жыл бұрын

I found something interesting along the way. From a + b + 2sqrt(ab) = 2009, we can divide both sides by 2 and get (a+b)/2 + sqrt(ab) = 2009/2. On the left side, the first term is the arithmetic mean of a and b, and the second term is the geometric mean of a and b. I don't see how that gets us anywhere, but it did seem interesting to me. The fact that the two means add up to 1004.5 suggests that each pair (a,b) would have one number less than 500 and the other greater than 500. Hence, a brute force attack would require searching approximately 500 possibilities. That's totally doable with an Excel spreadsheet, but not practical for a Math Olympiad.

@mehdinadjafikhah7314 2 жыл бұрын

By square both side, we find that: a+b+2*sqrt(ab)=2009. sqrt(ab) must be perfect. Thus, we can assume that a=c^2d and b=de^2. Therefore, d(c+e)^2=2009. But 2009=41*7^2. Thus, d=41 and c+e=7, or d=1 and c+e=2009.

@holomurphy22 Жыл бұрын

true

@chazzbunn7811 Жыл бұрын

Here's a lazy solution. Two roots on the LHS and one root on the RHS means a and b must be multiples of a number in common, let's call it c. sqrt(a) + sqrt(b) = sqrt(2009) n*sqrt(c) + m*sqrt(c) = sqrt(2009) => a = cn^2 and b = cm^2 n*sqrt(c) + m*sqrt(c) = 7*sqrt(41) (n + m)*sqrt(c) = 7*sqrt(41) This implies n+m=7 and c=41. The solution set are the ordered pairs (41n^2, 41m^2) such that n & m are nonnegative integers and n + m = 7.

@mathsprofabderrahim 2 жыл бұрын

Thanks

@anujjaiswal1101 2 жыл бұрын

When I solve any equation, that time even is equal to odd number. In this condition what is the answer of this problem.

@themibo899 2 жыл бұрын

Dude you need to do an mit integration bee challenge which is timed that would be fun

@longsteinpufferbatch4949 Жыл бұрын

Nice question

@guest2649 5 ай бұрын

what i would’ve done in this question was take the second power of the both sides, put a and b next to 2009 to seperate 2*sqrt(ab), then take the second power once again. It’s a rigorous process with some big numbers, but I’ve done some problems using that method!

@dushyanthabandarapalipana5492 Жыл бұрын

I am so greatful to you!

@gemmaburger65 2 жыл бұрын

Ok so before watching I’m gonna try and solve it. Quick prime factorisation gives 7 x 7 x 41 = 2009 Therefore sqrt(2009) = 7 x sqrt(41) 3sqrt(41) + 4sqrt(41) = sqrt(2009) sqrt(369) + sqrt(656) = sqrt(2009) a and b = 369 and 656

@jakobj55 2 жыл бұрын

2:35 **sad square root equation noises intensify**

@suharsonoofficial8400 Жыл бұрын

Very good my friend

@user-nh8vd4nv5t 2 жыл бұрын

I substituted sqrt(ab) with x^2 and solved x^2 + 2bx + b^2 - 2009 = 0, In order for x to be in N got b = 41^(2n-1)*m^2 with a = 41(7 - 41^(n-1)*m)^2 where n=1,2,3,.., m belongs to N

@mastershooter64 2 жыл бұрын

(integheral of e^x dx from 0 to 1.098)st

@veer56170 2 жыл бұрын

In India, mobile numbers are ten digit numbers starting with 9,8,7or 6 .Assume all such possible numbers are issued. Find the probability of four mobile numbers each having sum of it's digits 39. Please solve this one if you can

@pingger1de1tl 2 жыл бұрын

Mistake at 5:15 sqrt(2009*a) may not be N_0, because there is the factor '2' in front, which also allows sqrt(2009*a) to halfs (ends in x.5)

@PapaFlammy69 2 жыл бұрын

a is an integer, so it trivially follows

@angelrodafanboy5757 2 жыл бұрын

Mega Video!

@Amine-gz7gq 2 жыл бұрын

can you give more explanations on why b = 41 * r^2 Thanks

@imRJD14 2 жыл бұрын

Same process for a = 41*n^2 Put root a = root2001 -root b

@revelations2044 2 жыл бұрын

because it's identical to a in the initial equation. So everything that applies to a should apply to b.

@asheep7797 2 жыл бұрын

Doing some factorization, i found out 2009 = 7^2 * 41 So... If a + b = 7 and a and b are non-negative sqrt(41a^2)+sqrt(41b^2)

@helifynoe9930 2 жыл бұрын

I could not understand a thing of what was going on here, all due to me being a high school dropout. But years later I did in my spare time analyze motion to determine what it was, and in doing so I independently discovered the special relativity phenomena, and independently derive the SR mathematical equations at the same time, including deriving the Lorentz transformation equations. No one else seems to have derived the equations in the same manner. You simply have to create a geometric representation of the motion that takes place within the 4D environment known as Space-Time. It is constructed with motions vectors and length scalars combined. You then use it to derive all the equations. It makes things so simple, that the job is completed in mere minutes.

@holomurphy22 Жыл бұрын

you did suppose that there is a speed limit?

@diskont9605 2 жыл бұрын

That was beautiful indeed

@nayjer2576 2 жыл бұрын

Habe mir wegen dir brilliant geholt und bin sehr zufrieden

@RexxSchneider 2 жыл бұрын

Since 2009 = 49 x 41, we can write √a + √b = 7√41 Since √41 is irrational, that must imply that √a and √b are multiples of √41. So let √a = c√41 and √b = d√41, where c, d ∈ ℕ0. Now we have c√41 + d√41 = 7√41. So c + d = 7. That gives us the following possible solutions: c=0, d=7; gives a = 0 x 41 = 0, and b = 49 x 41 =2009 c=1, d=6; gives a = 1 x 41 = 41, and b = 36 x 41 = 1476 c=2, d=5; gives a = 4 x 41 = 164, and b = 25 x 41 = 1025 c=3, d=4; gives a = 9 x 41 = 369, and b = 16 x 41 = 656 Swapping a and b gives the other 4 solutions.

@albertogarcia4177 Жыл бұрын

At 5:27 is used the fact p.sqrt(c)€Z wlth c€Z, p a prime number then sqrt(c)€Z (p=2, c=2009a). It is not obvious (at least for me). Here a proof: Let p.sqrt(c)=m€Z then p²c=m², since p is prime and c is an integer, p|m so for some n€Z we have m=pn => m²=p²n² => p²c=p²n² => c=n² => sqrt(c)=n Is integer

@SAKEISUDMathee 2 жыл бұрын

I have a quesion: you define n and r as elements of the natural numbers except zero. However on 12:33 you include 0 as a solution for n and r. Did i miss something or should zero not be there? Or does the notation ℕ index 0 include zero? I was always thought that ℕ index 0 meant the "Naturals without zero". I am a little confused. Thanks

@viktorandersson4945 2 жыл бұрын

N subscript 0 generally means that 0 is included. This is because mathematicians can't agree on whether 0 is natural or not

@npip99 2 жыл бұрын

The subscript zero is how you explicitly say "Naturals with zero"

@GirishManjunathMusic 2 жыл бұрын

Naturals doesn't include 0. N0 is the way you say Naturals + zero.

@viktorandersson4945 2 жыл бұрын

@@GirishManjunathMusic well the naturals can include zero, and a lot of people use that convention as well. The problem is that there is no universally agreed upon definitely of the natural numbers

@GirishManjunathMusic 2 жыл бұрын

@@viktorandersson4945 I learnt that the Naturals were the positive integers starting at 1. The way you'd start counting naturally if you were counting, naturally. lol. So that N = Z ∩ [1,∞). And then Nsub0 includes 0 in that.

@genocider5868 2 жыл бұрын

I'm really confused about what he did to factor out 2009 at 6:20 can someone please explain this method? Why does it work?

@genocider5868 2 жыл бұрын

Okay, after watching it for the third time I realized it's just checking if it's divisible by 7

@wyerscor5599 2 жыл бұрын

Yeah, it’s kind of irrelevant because he just could have checked it by the same method he deduced the divisor later. Just think about a near number which you now is divisible by 7 (here 2100 for example) and just check if the difference is divisible by 7. Then you have the check and solution with one thought process ;)

@HershO. 2 жыл бұрын

14:32 nice

@erfanmohagheghian707 2 жыл бұрын

No matter if your final answer is correct, your reasoning is not. 2*sqrt(2009*a) has to be a non-negative integer does not imply that sqrt(2009*a) must be a non-negative integer! obviously it can be a half-integer!

@fakharyarkhan5848 2 жыл бұрын

But then that would mean that a half integer would square to 2009*a which is an integer but squaring a half integer can't give you an integer.

@erfanmohagheghian707 2 жыл бұрын

@@fakharyarkhan5848 Thanks.

@mihairoatapatrata69 2 жыл бұрын

PAPÀAAAAAAAAAA FLAMII plz do integrals again plz plz plz

@HopUpOutDaBed 2 жыл бұрын

I solved this in like 5 minutes. Key note is that sqrt(2009) reduces to 7 * rt(41) = rt(41) + rt(41) + rt(41) + rt(41) + rt(41) + rt(41) + rt(41). After that it's just grouping the root 41's. For example, rt(41) + 6rt(41) = rt(41) + rt(36*41) => a=41, b=1476. Easy problem

@holomurphy22 Жыл бұрын

you need to prove it's the only solutions

@HopUpOutDaBed Жыл бұрын

@@holomurphy22 how does this not prove it's the only solutions? By exhausting all combinations of the 7 sqrt(41)'s it's clear that these are the only integer solutions

@holomurphy22 Жыл бұрын

@@HopUpOutDaBed What I mean is "how do you prove that a=41*c, where c is some integer"? Because this is what you implicitly use

@vivaviiv Жыл бұрын

2\sqrt{2009a} is an integer does not imply \sqrt{2009a} is an integer too💀

@ukmathentertainmentclass8950 2 жыл бұрын

😭😭❤️ UK math entertainment class 🙏❤️🙏❤️

@ebonyphantom6977 2 жыл бұрын

Can anybody help me out, how did he go from a=k^2/41 to a= 41/k^2

@holomurphy22 Жыл бұрын

41 divide k*k. But 41 is a prime. By Euclide's lemma, 41 divides k or k

@mr.inhuman7932 2 жыл бұрын

I really concider getting brilliant

@MyOneFiftiethOfADollar 2 жыл бұрын

Like the way you parameterized in terms of n,r. n+r=7 is “pretty damn easy” but wonder if BMO competitors would have time to even get to that point and then deal with the slightly onerous arithmetic?.

@nikolaninkov3341 2 жыл бұрын

5:00 2009 + a is always greater than or equal to 2sqrt(2009a), so you don’t have that issue

@whitecomet3036 2 жыл бұрын

90 % of the math teacher that i know in school or online are cool people, My man just said: *"this equation is useless it can fvck off"* am losing it lmao

@alvaro6358 2 жыл бұрын

14:34 👌

@jasoncheng7964 Жыл бұрын

Wow the divisibility trick is amazing nowadays people just use a calculator

@krishgarg2806 2 жыл бұрын

nice

@Happy_Abe 2 жыл бұрын

Why can’t n and r be negative integers? That would still give us positive values for a and b Edit: we can actually, but then we get the equation |n|+|r|=7 which has the same solution as this video

@heewahhin7470 2 жыл бұрын

sqrt(41a)=k, which means k is the element of natural numbers U 0, k=41n, which means n also has to be the element of natural numbers U 0

@Happy_Abe 2 жыл бұрын

@@heewahhin7470 oh true, good point!

@salterchaotica3795 2 жыл бұрын

“I feel like I’m missing something here and each step has me re-checking a bunch of work. 2*sqrt(2009*a) as a term has to be a natural number (including 0, denoted as N_0). The claim that sqrt(2009*a) must also be part of N_0 doesn’t follow for me. If sqrt(2009*a) can be written as c/2, c E N_0, that means the whole term would return as an element of N_0. Possibly meaning we’ve missed an entire set of answers. Basically, if the square root term spits out a number with a decimal value of exactly 1/2, it would still be a valid answer. Haven’t found if there are any useful solutions here but yeah. Confuzzled.

@fahrenheit2101 2 жыл бұрын

Nice, as much as I expected that to be the answer, I didn't quite know how I'd prove it - this solution is pretty neat.

@edwardhudson815 Жыл бұрын

9:45 - if 12 divides 36 = 6^2, that doesn't mean that 12 divides 6 ... right? edit i got it, it only works for primes and 41 is prime

@ricardosefa4186 2 жыл бұрын

I thought when you square both sides it cancel the square root leaving a+b=2009

@Earthzooka 2 жыл бұрын

I assumed you already watched the video, but just in case and for future comments: When you square both side, the LHS becomes (sqrt(a) + sqrt(b))^2, which you need to expand as a quadratic equation. it becomes sqrt(a)^2 +2*sqrt(a)*sqrt(b) + sqrt(b)^2 = a + 2*sqrt(ab) + b

@trillionman2105 2 жыл бұрын

Not gonna lie, I applied boundaries as there was no tomorrow

@gamerxani9575 2 жыл бұрын

at 8:40, since 7*root(41*a) must be an integer, cant we conclude that a is 41

@RexxSchneider 2 жыл бұрын

We could conclude that a = 41n^2, where n is an integer, but we can't just assume that n=1 without more restrictions.

@hexagon8899 2 жыл бұрын

a=0 b=2009

@fal4784 2 жыл бұрын

a=b= 2009/4

@mhm6421 Жыл бұрын

Isn't this guy the mathematician with jojo and shirts

@evap304 Жыл бұрын

funny how u say five like fav

@evap304 Жыл бұрын

and u keep using overall like people keep using literally

@purple_sky 2 жыл бұрын

5:02 2√(2009a) is natural doesn't imply √(2009a) is an integer - what if 2√(2009a) is odd?

@holomurphy22 Жыл бұрын

it implies sqrt(2009a) is rationnal, and this is true iff 2009a is a square

@dimoztarra 2 жыл бұрын

But 2*sqrt(2009*a) can also be an integer if sqrt(2009*a) ends in .5 I may be wrong idk

@LunizIsGlacey 2 жыл бұрын

Yes, but 2009a is an integer, so sqrt(2009a) cannot be a non-integer rational.

@dimoztarra 2 жыл бұрын

@@LunizIsGlacey true, nice answer

@LunizIsGlacey 2 жыл бұрын

@@dimoztarra Thanks :)

@edwardhudson815 Жыл бұрын

Doesnt anyone else find this really unintuitive that the addition of the square roots of 2 integers can equal the square root of another integer

@werneryc 2 жыл бұрын

At 5:33 you say: if 2 sqrt(2009.a) is an integer then sqrt(2009.a) must be one, that is untrue. if sqrt(2009.a) = 0.5 then doubling this will result in an integer yet the square root isn't

@frag.mp4953 2 жыл бұрын

I'm watching this like I understand what he is saying

@sshilovsky 2 жыл бұрын

At 4:50 the transition is incorrect. The result is still true, but it's not enough for the integers to be closed under multiplication to claim it. For example, you could have sqrt(2009*a) be a fraction over 2, and still have an integer after multiplication by 2. It needs a little deeper reasoning here.

@PapaFlammy69 2 жыл бұрын

a is an integer by definition, so my reasoning is completely true. All statements were prposed in the context of given restrictions.

@sshilovsky 2 жыл бұрын

@@PapaFlammy69 That's what I'm saying, it needs additional comments. That a square root of an integer can't be a rational (non-integer) number is used here, and it can't be summed up just with closedness of integers under multiplication, because by applying square root and (implicitly) division by 2, we're walking away from the realm of integers here from two sides.

@holomurphy22 Жыл бұрын

I strongly agree with Sergei. Please Flammable you should look at it twice... A more understandable argument is that sqrt(2009a) is rationnal, and thus is an integer since 2009a is an integer

@pflasterstrips7254 2 жыл бұрын

41 being prime isn't that easy to know so people would waste a bit of time there too.

@holomurphy22 Жыл бұрын

it is in fact easy to know if a given number less than 121 is a prime. You only need to check divisibility by 2, 3, 5 and 7 (if n is not a prime, there exist a divisor less than the square root of n, or equal to it). The first three are obvious, so taht you only need to check for 7, which is quite easy to do

@Mark-cz2qe 2 жыл бұрын

2009=7x41^1/2 then easy

@hurktang 2 жыл бұрын

sqrt(2009)=44.8, which mean that 1 of the 2 integer number will be smaller than 45. you can then brute force the 46 possibilities. And be done by the time this guy attempt to explain his solution.

@holomurphy22 Жыл бұрын

you would take more time than him without a computer

@hurktang Жыл бұрын

@@holomurphy22 no, i'd be faster. You only need to calculate up to 45*45. You just start at 1 and go +n+n+1 45 times on a paper. substract the 45 lines to 2009 and than see if the resulting answer is in the 45 lines.

@Tagos2005 2 жыл бұрын

a= 2009 b=0

@Ready4Music 2 жыл бұрын

Haha, the 69 nice part. :D

@Mathemagical55 2 жыл бұрын

Not allowing 0 into N really makes me sad

@andreydrobjev2760 2 жыл бұрын

Ez done it without a pen

@bglfine7902 Жыл бұрын

🤫

@classsix6491 2 жыл бұрын

0 is not in the natural group

@PracticeAptitude 2 жыл бұрын

Cumbersome

@CTJ2619 2 жыл бұрын

you went through the long winded algorithm for di visibility for 7 when i knew instantly that it was divisible by 7 take the number in the thousands place and subtract from the rest of the number eg 2009 ==> 9-2 =7 and 7 is divisible by 7 so 2009/7 = 297 QED

@ultimatedude5686 2 жыл бұрын

That only works because 1001 is divisible by 7. 108 isn’t divisible by 7, even though 8-1=7.

@CTJ2619 2 жыл бұрын

@@ultimatedude5686 yes it works for 4 digit numbers or larger