From the Benelux Math Olympiad.

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[@corsiKa]

"Honey, let's go camping." "But my KZbin series? These kids need math!" "I got you a big ass chalkboard and some rope." "Bags are already packed, let's go." I think that's how this happened.

[@greysonaron5556]

i guess I'm quite randomly asking but do anybody know of a good site to watch new tv shows online ?

[@stephenr7548]

Math in the woods? The best way to do!

[@goodplacetostop2973]

0:01Drone footage PogChamp 12:02 Good Place To Stop

[@goodplacetostart9099]

You just copied my time stamp

[@goodplacetostop2973]

@@goodplacetostart9099 Not my fault if the footage starts at the beginning of the video 🤷

[@goodplacetostart9099]

@@goodplacetostop2973 yeah anyways it would be a total copy if you had mentioned good place to start though

[@littlefermat]

Alternative for 37|a_3k+2: Note that since we have 11...1 we should think of 37 because 37|111 Note: you don't have to divide, just take mod Ex: 3811 is divisible by 37 Proof: 3811 = 38*100 + 11 = 100 +11 = 111=0 mod 37 Now 3811...111 (3k+2) 1s is divisible by 37 Proof: 37|111 and 37|3811 so you can write the number as sum of multiples of 37

[@randysavage1011]

If a math problem is solved in a forest and no one is around to hear it...

[@megauser8512]

lol

[@caluire24]

From "it's a nice place to start" to "it's a good place to stop"... Excellent

[@domc3743]

Before watching, perhaps setup a geometric series to create a general expression for the number and hence show that it cannot or can be prime edit: this works out really nicely, we obtain the expression: (1-10^(k+1) * (343)/ -9 ) the rest is left as an exercise to the reader

[@MathElite]

Awesome intro! I've never seen that before a math video before :D Quality of the video is insane too

[@yoav613]

Michael,all of your hard work in making this videos should be appreciated and it is!

[@praharmitra]

I feel like the important step here is to identify some sort of pattern. For instance, looking at a_5 and a_6, it is not at all clear that they are divisible by 37 and 233! How did you determine that? Do you just somehow check all the prime factors till you find one? Does that method actually work under exam conditions? Are there any tricks one can use to find them somewhat quickly? I wish you would go through that stage of the solution process in a bit more detail. Once you find a pattern and have a conjecture, proving that conjecture is much easier.

[@Mephisto707]

A well prepared Olympiad participant will know by heart all primes up to 50 at least, preferably up to 100. Then, he only has to check factors up to the square root of the number he wants to factorize.

[@praharmitra]

@@Mephisto707 Agreed. But one of the factors here was 233. How are you supposed to figure that one out?

[@praharmitra]

​@@Mephisto707 BTW, do you mean primes up to 50/100 or the first 50/100 primes (because the former is super easy) ?

[@Mephisto707]

@@praharmitra I meant primes less than 100. Regarding the 233 factor, it would indeed be pretty hard for someone to find it during a contest, but in this question specifically it wasn't really needed to solve the problem. The way the professor proved the 3m case never cared what the actual factors of the number were.

[@praharmitra]

@@Mephisto707 I know! I am not worried about the proof. But the insight with 233 led to the conjecture that 3811...11 was NEVER prime and was always factorizable. Once you have that conjecture, the rest of the proof was of course easy. The main issue is arriving at the conjecture in the first place. Further, the problem was worded as "Can 3811...11 ever be prime?", not "Show that 3811..11 is never prime." If I was taking the test, the wording of the problem might well lead me to believe that there ARE some cases for which 3811...11 is prime, which would lead me down a different path. Generally, this is fine because I would of course soon realise this is wrong, but in exam conditions, this wastes time. It is for this reason that I asked my original question (reworded) - "How would one approach and solve this problem in an exam?"

[@iabervon]

"Why are you doing math problems in the woods?" "Because it's too hard to write legibly on a rock face."

[@killermakd2015]

1:12 37*103 = 3811

[@AlephThree]

Got quite close on this one. Started by observing that a third of the cases would be divisible by three due to the digit addition trick. I then got stuck for about half an hour before considering the the residues mod 3, and proving the 37 multiple for the residue of 2 case. Just didn’t quite finish it off. Thank you @Michael Penn for these videos. If I’d tried this six months ago, would have got nowhere with this.

[@wasitahmid749]

2:00 A2=3811=37*103.not 37*107..li'l mistake sir

[@CauchyIntegralFormula]

Oh, this is a really nice problem. We can write f(n) := 381...11 (with n 1's) as (343*10^n - 1)/9. If n = 0, f(0) = 38 = 2*19, which is not prime. Otherwise, we break the problem into three cases, depending on n's residue class modulo 3. If n = 3k, then the numerator of f(n) is a difference of two cubes, and neither factor in the standard factorization divides 9, so f(n) is composite. If n = 3k + 1, then the numerator is congruent to 0 mod 27, so f(n) is divisible by 3 but greater than 3 and so composite. If n = 3k + 2, then the numerator is congruent to 0 mod 37, so f(n) is divisible by 37 but greater than 37 and so composite. In any case, f(n) is composite, for any n.

[@leif1075]

You meant 381 times 10 not 343 times 10 right. Otbereise i have nonidea where you got 343 from...

[@CauchyIntegralFormula]

@@leif1075 No, I didn't mean that. If you calculate (343*10^n - 1)/9 for any n, you'll see it has the format I described

[@leif1075]

@@CauchyIntegralFormulasorry but why on earth would you think of 343 what's the connection to 381?? Does t tbst come out of nowhere

[@CauchyIntegralFormula]

@@leif1075 I wanted to classify the numbers we're talking about in the problem, which are 38 and then n 1's for some non-negative integer n. If you multiply such a number by 9, you get 342 and then n 9's, which is one less than 343 and then n 0's. I know how to classify "343 and then n 0's"; it's 343*10^n. So our numbers are (343*10^n - 1)/9

[@leif1075]

@@CauchyIntegralFormula OK thanks wpuld.you agree o don't see how anyone could.deduce that's what you are doing if you didn't tell them? I thought it had to do with 343 being 7 cubed or something .a d why 9 kind of irrelevant or unrelated since we have all 1s after 38 and not all 1s..would you agree ot would be ewually.logical.and smart to mjltiply.by something thst give you all 1s?

[@elbuenfercho4286]

You make all these problems look very easy and yet when I try to solve one of them on my own, they're freaking hard !!!

[@leif1075]

Doesn't that piss you off and discourage you? How do you not give up?

[@Osirion16]

@@leif1075 Because one day you'll solve one and feel good about it, then your brain wants more

[@littlefermat]

How do we even come with the number 233...3 in the third case? Well, you don't have to. Assume that 38111...1 (3k 1s) = p (prime number) 38111...1(3k 1s) = 38*10^3k + 111...1(3k times) = 38*10^3k + (10^3k-1)/9 = p Multiply by 9 both sides to get: 343*10^3k - 1 =p (Note 343 = 7^3 !) So by factoring: (7*10^k-1)((7*10^k)^2 + 7*10^k + 1) = 9p And a contradiction because: p =(7*10^k-1)/3((7*10^k)^2 + 7*10^k + 1)/3 And so we are done!! (funny fact : (7*10^k-1)/3 = 233...3 (k times)) :-)

[@leif1075]

Yea where does he het 2333 for the case 3m shouldn't it just be 233 since that's what divided the third example where'd he get those extra threes from?

[@Tiqerboy]

One thing I have figured out with mathematics: 1) If you can follow the solution, means you 'could have gotten that'. 2) If you can't follow the solution, no way you would have gotten that. I see how the first two cases were not prime, but he lost me on the 3rd case. I need to watch it again a few times, and the sum of a geometric series is definitely NOT something I remembered from my undergrad mathematics.

[@zygoloid]

At the end, we can notice that (7*10^m-1) = 6, 69, 699, 6999, ... so a_3m is divisible by 2, 23, 233, 2333 as predicted (because those factors are always 2 mod 3).

[@vinc17fr]

At 10:05, there is much faster: the first factor is larger than 9, and so is the second factor (since it is larger than the first factor). So, after the division by 9 (whatever the contribution of 9 in the factors), you'll still get two factors larger than 1, so that the number is not prime.

[@goodplacetostop2973]

HOMEWORK : Let m, n be positive integers and let x ∈ [0, 1]. Prove that (1 - x^n)^m + (1 - (1 - x)^m)^n ≥ 1. SOURCE : The 10th Annual Vojtěch Jarník International Mathematical Competition

[@moonlightcocktail]

if we observe mn trials of an event that happens with probability x, then the outcome will satisfy one or both conditions: when subdividing into m groups of n trials each, the event does not occur exactly n times in each group when subdividing into n groups of m trials each, the event occurs at least once in each group. Since at least one of the events must happen (and it is possible for both to happen), the sum of the two probabilities must be greater than or equal to one. ... I am very unsure if this proof is valid. I think it is not, but that a correct proof could be written on these lines.

[@moonlightcocktail]

Edit: if the first condition is not satisfied, then there is at least one group of n trials where the event occurs exactly n times out of the m groups of trials. But then if we partition the mn trials into n groups of m trials, each group has at least one occurrence of the event. I think?????? this could be considered equivalent to filling the cells of table with m rows and n columns with a specific probability X. Then it is true that either no row is completely filled or no column is completely empty, if the proof I have made is valid.

[@goodplacetostop2973]

SOLUTION We will prove that (1 - x^n)^m ≥ 1 - (1 - (1 - x)^m)^n. Take an m . n chessboard. The probability, that one particular square is black, is x ∈ [0,1], the probability of being white is x − 1. Assume this for all squares. Then : * (1 - x)^m is the probability that the whole row is white, * 1 - (1 - x)^m is the probability that there is at least one black in the row, * (1 - (1 - x)^m)^n is the probability that in each row there is at least one black, * 1 - (1 - (1 - x)^m)^n is the probability that at least one row does not contain a black. Denote by A the last event, in which some row does not a contain black (it is all white). We continue: * 1 - x^n is the probability that the column contains at least one white, * (1 - x^n)^m is the probability that each column contains at least one white. Denote by B the event in which each column contains at least one white. It is clear that A ⊂ B, because if one row is white then each column contains some white. Therefore P(B)≥P(A), written in the other form: (1 - x^n)^m ≥ 1 - (1 - (1 - x)^m)^n

[@parthdeshpande2966]

"And that's a good place sto-" I don't think that THAT'S a good place to stop.

[@johnchessant3012]

So glad I took the time to try this myself! I was so proud to find 9N = (70...0)^3 - 1 in the 0 (mod 3) case. P.S.: At that step you can just divide both sides by 9 because both 70...0 - 1 and (70...0)^2 + 70...0 + 1 are divisible by 3.

[@leif1075]

What is N in 9 times N and why dk you have 700 if this number is 38111 where does 700.and 3 exponent come from?

[@tomatrix7525]

Wow, those drone shots ate beautiful.

[@JorgeGomez-li9td]

I love this videos, I always learn something new, thanks from Colombia.

[@Tehom1]

Good video. The first and second cases were perhaps more easily shown by induction. You already have the base cases on the board, and the A_n+1 case is just A_n * 1000 + (3 * 37) = 3*k*1000 + 3*37 for some k; similarly with 37.

[@tomasstride9590]

I think this was a very good problem and in some sense a very doable problem. I wander if others agree with me that although the problem was something that could be cracked open this is not something that can be easily done in a few minutes. If I was given the problem cold I would probably firstly try to think of any general results I could use to help. After considering this I probably try looking a examples just to see if there was any structure that I could use. However, with no calculator to use there is quite a bit of number crunching to do . The point I am making is that doing the problem in just a few minutes would seem a tall order.

[@dkravitz78]

7*10^m-1 = 699...999 which is divisible by 3 but not 9. Therefore 3a[3m] = 233..333 * n N must be divisible by 3, leaving a[3m] = 233..33 *(n/3) Since the first factor is not a[3m] the second factor is not 1, thus not a prime.

[@andreferreira1758]

I loved everything including the intro music. I learned so much, thank you Professor!

[@rajatadak7826]

a_2 =3811 = 37x103

[@charlottedarroch]

I thought it would be easier to solve the third case by just doing the multiplication, but it turned out to be a complete mess. We know that a_(3m) = 38*10^(3m)+(10^(3m)-1)/9, so if we let b_n = 2*10^m+3(10^m-1)/9 and c_n = 16*10^(2m)+5*10^m+7+(3*10^(m+1)+60)(10^(m-1)-1)/9, then we see that the b_n looks like 233...3, the c_n looks like 1633...3566...67 and formally multiplying we obtain the result that b_n c_n = a_(3n), thereby proving that a_(3n) has proper divisors b_n and c_n and therefore is not prime.

[@dppowers7880]

Michael- Love your videos. The amount of content you are cranking out is impressive. At 10:15 i think you can more cleanly note the two factors on the right have digit sums 6 and 21. So 9 divides neither factor. So a_3m divides neither one either, therefore it is not prime. And to tie back to the 23, 233, 2333, ...., we can see those are (7x10^m - 1) / 3.

[@Cloiss_]

given that 9 divides neither factor, couldn't you still theoretically have A = 3a B = 3 you need to also show that 3 divides neither factor to conclude that 'a' divides neither factor, I think

[@mcwulf25]

Cases 1 and 2 are simply proved by long division. 3 divides 38 with remainder 2 so 381 and any number of 111 after that is a multiple of 3. 37 divides 38 remainder 1. So 38 and any number of 111 after that is a multiple of 37.

[@mcwulf25]

Sorry that should be 3811 then any number of 111 after that.

[@mihaipuiu6231]

Michael....you are a magician!

[@swingardium706]

0:38 the screams of the damned have begun making cameo appearances in KZbin videos, I see

[@fariesz6786]

funny observation: assume a_-n is when instead of concatenating n 1s to 38 you instead subtract a sequence of n 1s from 38. a_-1 would become 37, thus fall into case 2 and the divisibility rule still holds, and a_-2 would become 27 and the divisibility rule for case 1 still holds. in fact, the divisibility rule for case 1 will always hold, but that for case 2 will break down at a_-4 = -1127 which is not divisible by 37. a_0 = 38 is divible by 2 and a_-3 = -127 is divisible by -1 (duh). wonder if those are pure coincidences or if something underlying here actually makes sense of the extension. (would outline it nicer but too tired, sorry)

[@Eric-dt7bt]

How do you obtain the first cases in the first place? I mean, in a competition how should one quickly find a large prime factor of a ridiculously large number (in this case, 37 and 23)?

[@rosiefay7283]

Or even 233?

[@ezequielangelucci1263]

factoring: The first case is easy noticing that each time m grows in 1 then the sum of the digits changes in one. then 1/3 of the times that is a multiple of 3 the second case is a bit more difficult but when you try to factorize numbers you always start fot the lowest ones so it doesnt take so much time to get up to 37 the third case it is indeed the hard part

[@VibratorDefibrilator]

In the third case [ n = 3 * n ] yuo may also use the fact that these 1's are repunit, which is presented as (10^(3n) - 1 ) / 9... instead using geometric series. Both ways are fine, just saying.

[@alainrogez8485]

I just found the first case with the multiple of 3.

[@242math]

awesome problem, you did a great job solving it

[@goodplacetostart9099]

Good Place To Camp at 0:01

[@BigAsciiHappyStar]

And Job said "Hitherto shalt thou come but no further" bwahhahahahahahah :D :D :D :D :D

[@antormosabbir4750]

well, can't we use that p^2 is congruent to 1 mod (12)?

[@muhammedmrtkn]

that is pretty nice

[@replicaacliper]

wow

[@advaykumar9726]

Loved the video

[@malawigw]

How do people do these calculations without calculators in a limited time?

[@ramakrishnasen4386]

Well I had a completely different idea of using repunits, but I'm not quite sure if they do help somewhere

[@leif1075]

What are repunits?

[@easymathematik]

I have a problem for you guys which I have constructed. :) Let p(x) = x^2021 + 2021x^1512 - 1512 be a polynomial with 2021 real and complexe roots r_i, i=1, ..., 2021. Show: sum from i=1 to 2021 ( r_i ^(-3024) ) is rational and the numerator of this fraction in lowest terms is a perfect square. Which one?

[@holyshit922]

My C# program checked first 64 of these numbers and they are not primes

[@Mathcambo]

Good solution

[@happygimp0]

From the thumbnail, i thought about how to solve it, fond one which was a prime but had no idea how to prove which one are prime and which are not. Only to realize while watching the video that only 1 can be inserted and that there don't need to be at least 2 1's. I thought 38141 would be a candidate, i didn't understand the rules.

[@leif1075]

Hiw did you figure out thst 381111 js divisible by 3..but working it out by hand I guess?

[@Lokalgott]

If m = 1 : 7× 10^ - 1 = 69 69 divisble by 3 What i did not get right here?

[@Yash42189]

But how would you know that they arent prime without a calculator? Like how do u come up with 3811 = 37 * 107? Is it some magical calculation skills those kids have?

[@Nothing_serious]

I don't understand it but interesting.

[@Dudu_cmrd]

we can use induction in this problem?

[@ZainAlAazizi]

3811=37*103.

[@user-bf7zo1lh1z]

107x37 is not 3811 it should be 103x37

[@egillandersson1780]

🇧🇪Hooray for Benelux ! 🇧🇪

[@Ardient_]

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..... (Day 5 of asking) please can you see me

[@MichaelPennMath]

I see you... Do you have a hint?

[@Ardient_]

@@MichaelPennMath try subbing a and b in terms of m and n when m²+n²=c.....

[@srijanbhowmick9570]

Yes I can see you but don't think Michael can unless this comment gets at least 20 likes

[@alanyadullarcemiyeti]

@@srijanbhowmick9570 he saw

[@yashvardhan6521]

Just guessing, not tried But the substitution for the primitive Pythagorean triples may work

[@luisrosano3510]

Oh you say "follow by ones". I understand wrong the all exercise. Sorry man!!!! I think that dots were spaces for complete with any cifer you want, not just with a secuence of ones.

[@moros_gamer2874]

👏🏻👏🏻

[@kasun1998lk]

One of the first to comment