Two problems from the world's happiest country.

[@almafater]

And that’s a happy place to Finnish

[@nguyentruongphu1383]

i see what you did there

[@almafater]

@@nguyentruongphu1383 there’s Norway you see it!

[@maroonshaded]

@@almafater haha some Sweden innofensive jokes

[@GaborRevesz_kittenhuffer]

i scanned the navy and found no be better puns.

[@roberttelarket4934]

almafater: Very very very clever!

[@AntoshaPushkin]

The first problem isn't rigorous enough. You need to check a few more cases (WLOG a≤b≤c) a=2, b=3,4,5 a=3, b=3,4 a≥4 take minimal c in those cases and show that max is achieved for a=2, b=3, c=7

[@karlstrecker6002]

I think Mike‘s proof works: wlog a

[@AntoshaPushkin]

@@karlstrecker6002 min a doesn't necessarily lead to optimal a, b, c

[@AntoshaPushkin]

@@karlstrecker6002 I have a counterexample for your logic. Assume it works the way you describe it. Then it also works for 4 numbers, 5 numbers, etc. Let's try to use your logic for 5 numbers. We would get min a = 2, bin b = 3, bin c = 7, then min d = 43 and min e = 1001 1/a + 1/b + 1/c +1/d + 1/e ≈ 0.99954... However there is a better solution (a,b,c,d,e) = (2,3,7,61,135) such that 1/a + 1/b + 1/c +1/d + 1/e ≈ 0.99991... So your proof is invalid, and picking minimal a then b, ... is not a valid strategy, as almost always with Diophantine equations

[@karlstrecker6002]

@@AntoshaPushkinRight. I am convinced.

[@HoSza1]

There is more: if you choose 2, 3, 7, 43 and 1807 then the sum of reciprocals is even more close: 0,999999694 (rounded).

[@QuaDue]

concerning 5076, well we have already factors 9 and 4 from the filters so we know thta 5076 = 36*141 and that is not a perfect square

[@TitanOfClash]

For the first part, why does the maximum of ths first two numbers mean that the third has to be on top of that? Why couldn't it be a completely different combination of numbers?

[@Happy_Abe]

I have the same question. It should require more proof. There’s I reason the maximum combination can’t have a, b that are bigger than 2,3 with a c that is smaller than 7

[@adogonasidecar1262]

@@Happy_Abe Note that the formula is symmetrical in a, b, and c. Also 3/7 is less than 41/42. Therefore at least one of the digits has to be smaller than 7. One way is brute force with numbers from 2 to 7, computing all the combinations. I am pretty sure there is a smarter way.

[@hyperpsych6483]

I agree it's really not clear. Just work through it case by case. Prove there must be one 1/2 term: 1/3+1/3+1/3 = 1. 1/3 + 1/3 + 1/4 doesn't reach our maximum. No other options. Prove there must be one 1/3 term: 1/2 + 1/4 + 1/4 = 1. 1/2 + 1/4 + 1/5 doesn't reach our maximum. No other opions. In both cases, we try to make the maximum possible sum without using one of the terms and show it doesn't work out.

[@adogonasidecar1262]

@@hyperpsych6483 yep, need to go through cases

[@Happy_Abe]

@@hyperpsych6483 yes that’s sufficient but video’s proof doesn’t have this and I think it’s lacking because of it

[@warmpianist]

For the first problem, do we need to show that there are no more tuples of (a,b,c) where a >= 3 that makes the sum more than 41/42? This seems like a greedy algorithm where you can't be sure that smaller a will surely give a largest sum.

[@hyperboloidofonesheet1036]

a, b, and c are interchangeable, so you can always impose a ≥ b ≥ c without loss of generality (and then permute the tuples at the end). With that, you know that a < 4 because otherwise the sum could not exceed 3/4 (=1/4+1/4+1/4), which is certainly less than 41/42. This means you only need to check cases where a=3. Starting with b=3, c must be greater than 3 or the sum will be too large. With c=4 the sum is 1/3+1/3+1/4 = 11/12 which is less than 41/42. Any larger value of c results in a smaller sum. If b = 4 then c ≥ 4, which results in a sum of not more than 10/12 (=1/3+1/4+1/4). Any larger value of b will result in a smaller sum.

[@monzurrahman8307]

@@hyperboloidofonesheet1036 this is the missing part from the proof. Unless I'm missing something, there doesn't seem to be an obvious guarantee that 1/a + 1/b + 1/c achieves its max where 1/a + 1/b does, given the constraint in the problem.

[@kappasphere]

​@@hyperboloidofonesheet1036I think you meant to write a

[@Bodyknock]

@@monzurrahman8307 I agree, this was an important oversight in Michael's solution. It's quite possible at the stage he left it on the board that, for example, there was some a and b that weren't 2 and 3 respectively where 4/6 < 1/a + 1/b < 5/6 in which case you could possibly have that adding 1/6 to that total was between 41/42 and 1.

[@lgooch]

You do, Michael’s solution would not yield full points.

[@哀恩欸芙屁]

Amazing solutions for the 2nd problem!! But, if we were sitting in the contest: 1. How exactly did we get to the step of multiplying "2001n + 2" by "2001n - 2" at 5:38 to create a difference of squares at the LHS of the equation? 2. And how could we prime factorize 8008006 completely and correctly (just like at 8:43) within the time limit?? 🤔🤔🤔

[@williamperez-hernandez3968]

For the filter of perfect squares, as with 00, should have been 25, since a number ending in 5 has a square that always ends in 25.

[@pierreabbat6157]

Also, if a square is divisible by 3, then it is divisible by 9. That would not have killed 5076, which is divisible by 9. However, dividing by 9, then by 4, yields 141, which is not a square.

[@goodplacetostop2973]

17:02 Happy, happy, happy 🇫🇮

[@seanbastian4614]

For the first problem, I feel part of the proof needs to include if a pair of a,b, or c are equal and see why 41/42 is still a max in that case. In the problem, it didn't explicitly say that a,b, and c can't be equal to one or another.

[@topquark22]

Good point. We want a, b, c to be as small as possible. The (only?) next case is say a=b=3. Then 1 - 1/a - 1/b = 1/3, the distance we have to make up. But c=3 already doesn't work. And if c=4 then 1/a + 1/b + 1/c = 11/12 which is not as good as 41/42

[@GaborRevesz_kittenhuffer]

apparently this works in general! "sylvester's sequence" a(n) given by a(1) = 2 and a(n+1) = 1 + a(1)·...·a(n) has the property that if b(n) are positive integers with S:= 1/b(1) + ... + 1/b(n) < 1, then S is maximized exactly when {b(1), ..., b(n)} = {a(1), ..., a(n)}. en.m.wikipedia.org/wiki/Sylvester%27s_sequence ...pretty cool.

[@aromeran]

Still, choosing 1 is not allowed, repeating 2 either and 1/3 + 1/3 + 1/4 = 11/12 < 41/42, i firmly believe the solution is still optimal given any a b and c integers but ill leave it as a proof for the reader as to debunk or proof

[@johnchessant3012]

The second problem was already covered on this channel on Nov. 17, 2022.

[@Magnus-727]

I know my squares by heart. 5076 is between 5041 (71^2) and 5184 (72^2). I also recognized the 4,004,001. Notice 21^2 = 441, 201^2 = 40,401, 2001^2 = 4,004,001 etc. Just add one more zero to each 0 block each time.

[@keedt]

numerical experimentation shows that the smallest modulo filter that works is that of modulo 13, which says that perfect squares belong to the eq. classes {0, 1, 3, 4, 9, 10, 12} mod 13. and 5076 = 6 (mod 13). This is clearly more work than noticing that as 70^2=4900; 71^2=70^2+141=5041 < 5076 < 72^2= 71^2+143=5184, so 5076 is bounded by two consecutive perfect squares.

[@Bodyknock]

7:07 Shouldn't C = 2001, not 2001² ? In other words multiply 2001n + 2 by 2001 to get 2001²n² + 2*2001 = (n² + 2)C. (Michael wrote 2*2001² here) I don't get how he managed to get the right answer after this step since it really seems like the way he did it would have used 2*2001 instead of 2*2001² at this step. 😵

[@blue_sand6854]

This is my question, too (or, it is nor n^2).

[@dneary]

For 5076, you only need to know that 70^2 = 4900, and (10x+4)^2 = 10y+6 - so the smallest possible number that could square to 5076 is 74 - but 74^2 = 5476.

[@pdpgrgn]

For the 2nd problem, just like we used that fact that a perfect square ending in 0 has to end in 00, we could have used the fact that a perfect square ending in 5 has to end in 25.

[@davidblauyoutube]

5076 = 4*1269 = 4*9*141 and 141 is not a square because it is divisible by 3 but not 9.

[@LeviATallaksen]

Calculating products of every combination of {2, 19, 83, 2539} and subtracting 2 from each of them is fine, but taking square root at the end of each calculation is too much work??? I agree that reducing how many combinations you have to check is a good idea, but it feels like you're working through each combination anyway. I might do some of the modular arithmetic before doing the multiplication, but it's not "obviously better" (since if we can't eliminate a lot of combinations, we'll end up doing more multiplications that way; the extra ones are very simple though): 2 === 2 (mod 4) 19 === 3 (mod 4) 83 === 3 (mod 4) 2539 === 3 (mod 4) So we can eliminate every combination of two factors congruent to 3 mod 4, since 3^2-2 = 7 is not 0 or 1 mod 4. 2 === 2 (mod 3) 19 === 0 (mod 3) 83 === 2 (mod 3) 2539 === 0 (mod 3) So we can eliminate 2*83, since 2^2-2 is not 0 or 1 mod 3. The mod 10 check feels like it's not worth doing this way, since we're getting less overlap. And we can't check for 00.

[@rohkofantti8673]

Interesting mathematics task from my country from the USA. Thank you again for an excellent video lecture, Dr. Michael Penn! 😊

[@QuaDue]

Is it just me or is the first proof incomplete, e.g. one needs to prove that 1/2 + 1/4 + 1/6 =1 or am I missing something here?

[@viktorsmets29]

Yeah the first proof is not complete, but you should actually check for 1/2+1/4+1/5 and also for 1/3+1/3+1/4

[@scragar]

That's obvious though, because 1/4

[@viktorsmets29]

The point is to be rigourous, so also explaining the simple steps

[@Zack-xz1ph]

1 - 5/6 = 1/6, so the third fraction must be less than 1/6

[@nixlarfs1002]

For the first problem, you can assume wlog that a

[@l.h.308]

42 = 2 * 3 * 7 and we need to have small denominators. Check these and there is the answer - in 10 seconds. (6 permutations of a, b and c)

[@awesomechannel7713]

Solution of problem 1 doesn't seem rigorous at all. Why we are maximizing the 1/c given maximized 1/a+1/b? Why cannot 1/a+1/b+1/c be greater if 1/a+1/b+1/c was less? I.e 1/3+1/3+1/3 = 1 is larger. Yes, it doesn't satisfy the conditions but nothing in the solution even implicitly ruled that out.

[@pierreabbat6157]

The Sylvester sequence is relevant. The Tweety Bird sequence is not! Wikipedia has a red link to Vardi's constant, which is the number that, when raised to power-of-2 powers, yields numbers close to the Sylvester sequence. Who's Vardi?

[@johns.8246]

At 7:07, why is C equal to 2001^2 instead of 2001n ? I'm lost.

[@barutjeh]

It doesn't follow from the first equation. He's just using the fact that 2001² n² + 2001² ×2 is what you get if you work out 2001²(n²+2) (and is therefore a multiple of (n²+2) as well).

[@peterhall6656]

This is typical of his delivery. He will an inordinate amount of time on eye glazing banal arithmetic and then he will make a jump and not explain with the same level of detail somethng that is much less trivial. It is his signature move! @barutjeh explains the move.

[@restcure]

I've had my "no, it CAN'T 'easily be seen that'!'" moments in many math classes I've taken.

[@adogonasidecar1262]

@@peterhall6656 Absolutely, I have noticed that many times, where I get bored in the slow delivery of basic calculations, and then have to pause or rewind the video for one significant leap that comes out of nowhere and is not sufficiently explained or justified. Drives me nuts :) ALso, the sloppy mathematical writing... Argh Still, great and varied problems, so it's entirely worth it.

[@59de44955ebd]

Concerning how to find out quickly if 5076 is a perfect square, I don't know about a filter, but I think it's obvious that 4900 is 70^2 and that 5076 is 4900 + 176. So if 5076 is a square, then (by binomial theorem) there must be some k that solves 176 = k * (2 * 70 + k). And it's obvious that there is no solution, since k=1 is not a solution and for k=2 the right side is already bigger than 176.

[@martinnimczick839]

In the first task... Why a=2 and b=3 ist the only case to check? Why is a=3, b=3 and any other an appropriate c less than 41/42? Or a=2 and b=4 and so on?

[@Zack-xz1ph]

I believe the numbers have to be different

[@wolfmanjacksaid]

5076 is between 71^2 and 72^2, so it can't be a perfect square. ✓

[@danjwheatley]

how do you (quickly!) find the prime factors of that huge number in the 2nd problem?

[@khoozu7802]

Small mistake: 166-2=164=/=163 However, 164 is not perfect square mod 3

[@xwtek3505]

There is another filter. The principle used is: if n | k and k > 0, then k >= n, and n | k - n Note that 2001n+2 > 0, so 2001n+2 - n^2+2 = n(2001-n)>=0 If n=0 or n=2001, we found a solution, so ignoring those two solution n(2001-n)>0 If n is not a multiple of 3, n^2+2 is a multiple of 3, but n(2001-n) is not a multiple of 3. So n has to be a multiple of 3. Additionally GCD(n^2+2, n)= 2 if n is even, or 1 if n odd. So, we split cases for n: If n = 6k, k > 0: 36k^2+2|6k(2001-6k) 18k^2+1|667-2k Note that 667-2k >0. So, we get: 667-2k >= 18k^2+1 But 18k^2+2k-666=9k^2+k-333 does not have an integer solution as the value of the polynomial at k=6 is smaller than zero, but the value of the polynomial at k=7 is larger than 0. This also means k 18k^2+1 18k^2+1|667-2k-18k^2+1 18k^2+1|666-2k-18k^2 18k^2+1|333-k-9k^2 333-k-9k^2>=18k^2+1 27k^2+k-332

[@呂永志-x7o]

8：33 Factoring is not so easy for us.

[@Zack-xz1ph]

do a, b and c have to be different numbers? on the first problem

[@fgp693]

They will always be happy with the two problems. They say, no problem, because they are Finnish.

[@hazalouldi7130]

anyway there is an error in the7 th minute around 2001 squared

[@egillandersson1780]

Even if this is the correct result, the first demonstration is not rigorous enough, I think. In fact, why not achieve a max for 1/a + 1/b + 1/c WITHOUT achieve the max for 1/a + 1/b ? It may be impossible but not prooven here, is it ?

[@Blabla0124]

I don't find the first proof convincing ...

[@cycklist]

The map on the thumbnail is wrong. The shaded part is the EU, but the UK left the EU some time ago.

[@eldanchy3782]

And Norway is missing in the map

[@birdbeakbeardneck3617]

From the happiest country to the happiest community*

[@pojuantsalo3475]

The World's happiest country has only two problems?

[@pederolsen3084]

That's why it's so happy

[@Zaxx70]

Your voice was barely audible on this vid :(

[@vke6077]

Suomi mainittu :D

[@JoopWilkens]

Your "second equation" involving c, in the second problem, seems to come out of thin air. On a problem solving contest, this would get you a 0 on 10 :) Very poor explanation, way below your level.

[@minwithoutintroduction]

7:36

[@ikarienator]

Sir your volume is too low.

[@tamasdhgebrq5968]

You can also start from the fact that 42=2*3*7, and 2*3=6, 2*7=14 and 3*7=21. However, 6+14+21 = 41.

[@advaykumar9726]

S U P E R C E L L

[@GaborRevesz_kittenhuffer]

i somehow find the explanation of problem 1 not fully satisfying. also, it apparently generalizes in the most straightforward way! "sylvester's sequence" a(n), given by a(1) = 2 and a(n+1) = 1 + a(1)·...·a(n), has the property that if b(n) are positive integers with S:= 1/b(1) + ... + 1/b(n) < 1, then S is maximized exactly when {b(1), ..., b(n)} = {a(1), ..., a(n)}. en.m.wikipedia.org/wiki/Sylvester%27s_sequence ...pretty cool.