Today we are doing the number 2 777 777 88888 99 So I guess if you want to split that up, two hundred seventy seven trillion seven hundred and seventy seven Billion seven hundred and seventy eight million eight hundred eighty eight thousand eight hundred and ninety nine. This is a record-breaking number when it comes to the multiplication persistence of a number. so we'll do a smaller one just to get a handle on it. We'd like to give me a small. I don't know like three or four digit number. Would you like Brady? Oh, I always regret asking you this 5428, alright, so the first thing we do is we multiply all the digits together. So 5 and 2 is gonna give us 10, 4 and 8 to give us 32, so it's gonna be 320 and then we continue we multiply all the digits together now, there's a zero so zero So you made it two steps, before we hit a one digit number and if we have any one digit number You just you can't multiply the digits anymore and you stop so the question is I'm so disappointed Yeah, can I have another try did you want another go? Yeah, let's try this. It's not called persistence for nothing. What would you like? See where you went wrong brady is a 502 spells doom because that's gonna chuck a 0 on the end straight away So don't make that mistake again 327? 327. Nice, you're going for primes. look at you? Okay. Let's see how that goes So that's six sevens which is 42. Someone will correct me if I'm wrong Four times two is gonna give us eight and then we're stuck Two steps. At home if you want try and pick a number. See if you can beat Brady's current record of 2 So, how long do you reckon this one goes for? Your big number? Yeah big number What do you what do you think we're gonna get out of that one. have a guess, what do you reckon? You've already told me it's a record holder. That's right But because you're willing to do it for me now makes me think it's not crazy big. When have I shied away from a crazy-big calculations? So I'm gonna go for something surprisingly smaller. I'm gonna go for like 10 Getting my calculator out two times seven equals There's one seven two four nine nine Six. ... times six times two.... Oh, okay. So now we're down to Four times three times eight.... What's very exciting about this is I've never actually checked this This is the first time I've actually and now I'm feeling a little nervous like that's the level of preparation yet on numberphile how much prep I put into these videos pretty is like What numbers are you into currently? I like are kind of interested in this one... two, seven seven times eight 54 5 times 4 is 20 Is 0 1 2 3 4 5 6 7 8 9 10 11 Oh 11 and that's the correct answer. Thank goodness. It has a persistence of 11, which is the current world record for multiplication persistence of a number other numbers equaled the record But this is the shortest number with the biggest currently known Persistence and so that is the current champion. Hang on so you're with no limit on the number of digits? No-knock is about managers everyone and often in these videos I say I went away and I programmed a thing and then I calculate it and I found this I found that I have not Programmed this yet because as you may have those little underprepared today, I haven't coded it up But we could should we do it live do me do it? Do it cuz I always say I code something and I find them. Let's code something. Let's find it I'll get my laptop so funny story last week my Keyboard stopped working and the trackpad on my laptop, which is a bit awkward. Okay So first things first, we're gonna set something going which is going to do this process over and over and over again But we want it to stop once it gets to a thing which is only one okay, so this is gonna take some number and the first thing we're gonna say is if the the length of the number of digits to the string n so the great thing about n is, it's not only a number When we care about digits I'm going to turn it into a strings that goes from being like a number represented in binary or whatever base to just being The base ten digits in a string and if the length of that string equals one Then we're done right? We're at the very end of the thing, right? So that's at that point print whatever n is and then finish so return I don't know like "DONE" All right, that's just gonna tell it you know, your job here is done. Okay. Otherwise we need to multiply together all the digits. digits equals Let's do it as a list. i for i in The string version of the number, right? So they're turning into a string of digits and then it's taking out each one individually Oh, but we want them as as numbers So let's turn them back into numbers So bizarrely I'm turning it into digits as a string then taking out each one separately then turning them back into numbers, right? Which is kind of, because this is a base-10 thing one of the sad things about stuff like this is its base Specific so now we've got all the digits for j I'm just using j is the placeholder in Digits so now we want to multiply them all together. Okay, you know, let's have Current result Result equals one to start with and then each time result equals result Multiplied by that digit and you can do x equals just means make it equal to this times J whoops J. Okay, I think that's it and that's gonna give us a new result, but then we've got to repeat the process So this is where we can cheat And I haven't genuinely haven't tried coding this before so I don't know if this is going to work I'm going to try and get recursive and then put that new result into the same function. So, if it's one long, it'll stop otherwise It'll multiply all the digits together and then put it back into the beginning and it will Keep repeating through this process and you know what? Let's make it print the result each time So we get, we get to see them all as it as it loops through and then when it hits here It'll stop and say done. Okay? That can't go wrong Let's find out. I really should have checked this in advance. Okay copy. I'm just gonna fire up terminal Okay. So what I'm actually gonna do this is the laziest way to run something; literally paste it into terminal. Let's do the Persistence of 327 which is the second one, you said. 42 8 and then 8 forever, right? And so it's done twice and then going we're out Okay, I couldn't fix the code to not get the the last one twice. I just changed where the check is So actually what I could have done is have another check in here So I don't print the result and then print it again before stopping I could put the check in but I'm not gonna It fit for purpose. Alright for a first pass it's fine. And then the very first one Let's just check the first one just to make sure I've messed this up five four two eight It goes to 320 goes to 0. right? And now the ultimate test can it handle 2 six sevens six eights two nines There we go, right so and that's that's so much quicker! Right? so it's now spat out Exactly these all the way down and then it stops doesn't give us a number Ah, do we want like a, do we want a number of steps at the end? number of steps? Okay, okay Brady will cut this out and put it on the second channel 2 one two, three, four five six oops 1 2 3 4 5 6 1 2 There, total steps 11, okay, right So now we can put a number in and instantly we get everything. we could no longer print every single step along the way. It's kind of fun to see and we get the total number steps So Brady, let's put on what would you like? You know, let's Put put as we just mash it for a while and see what we get. No, cuz you can be strategic Ah, you're right to put a 5 Just put in a bunch put in Fifteen 9's. Fifteen 9's. one two three, four, five six seven eight nine 10, 11 12 13 14 15 Which is just nine to the 15. No, I put in another 10. Oh another ten nines. Yep Well, let's just see if that one works Two steps. Oh because we got the zero in the answer next time but what I can do in your in terminal just push up you get the Previous one stuck a few more 9s on the end two steps now what if we um, Put in eight instead two steps What about, what about the string you used like at the current record holder? Yeah, but put a three on the front I love it. Okay, so 3 2 1 2 3 4 5 6 1 2 3 4 5 6 9 9 ah brilliant I love like I love you thinking ready and 2 steps This is harder than it looks! So it's zeros are like a Zeros are the land mines Boom, hit a zero, you're out. This is minesweeper, but for number searches, okay So this is what I would do I would now play with this for a while and it seems to be whenever you put in some random string of Numbers and boom, right because it was a zero the next one it's gone We need to be more strategic What about all the digits of pi. All of them? I will Get cracking on that. three eight one four one five nine two six five I forgot, that's that's enough - 2 Wow, 11 seems a lot more impressive all of a sudden, no wonder it's the world record. so does this mean that you could set up a program that would just Put every number in one after another and just leave it, you know, leave it for an hour or two so what we've done now is we've built the basics of checking Next, we want to build something to do the search. so we could just get something to generate random numbers of a certain size Shove them in and send us an email if it gets a good one. The next step would be to be Strategic about what numbers we were putting in because already you realized Don't put it in a 5 you put in a 5 it's not gonna work so we could create a search Which doesn't put in fives? Doesn't put in any combination we know will definitely give us Zero and actually we can get even smarter than that because if we're looking for the smallest number that it works for It doesn't matter what order the digits are in So in fact all the current record holders the shortest number of digits for different persistence values So there's a record holder for 10 and for 9 and so on They're always, the digits are always in ascending order Because you want the smallest number with those digits. So in fact, you don't have to search for Shuffled around versions; you just need that set of digits in that order and then there's other things So for example this one here we'll use your first one you put in 5428, that's going to give you exactly the same result as First of all, you could have put in two four Five eight, which is smaller But actually two times four is going to give you eight, so you could actually have just put in 588 That's going to give you exactly the same sequence as that and it's smaller So if all you care about are the smallest possible values for that sequence of persistence afterwards (multiplication persistence) Then you never want to have a 2 and a 4 You never want to have two threes because I could be a 9 you never want a 5 at all In fact, you only ever end up with a few small numbers at the beginning Never more than one 2 never more than one 3 Never more than one 4 I think and then all sevens eights and nines for the rest of it so we could reduce our search space dramatically with a little bit of logic the current search has gone as far as 233 digits so if you do code it up you've got to start searching from 233 it's not gonna be smaller than that. We've already checked and it's currently The conjecture is you would never be 11 So if people want to have a go, I mean, I'm always one to give it a go See if you can write some code see if it does a clever search and you know It would be a major breakthrough. If someone could find a number with a multiplication of persistance of 12 Will you be the person to make that major breakthrough? Certainly the sorts of people who cracked tough nuts are those who think outside the box, creative thinkers, people who don't follow the flock and Brilliant wants to make you that type of person. Their courses quizzes puzzles, like the ones you see on the screen at the moment They're carefully crafted to mould people into smarter thinkers. Not memorizers, not people who just know all the equations and how to pass a test, but problem solvers. People with better wired brains. if you'd like to find out more about brilliant; what they're making and how they might help you, Go to brilliant.org/numberphile. There's free stuff on their site, but that /numberphile URL That'll get you 20% off one of their premium memberships and it'll also let them know you came from here.
Seeing Matt just bang out a program like that is cool as hell.
My first thought was: "what is stopping me from adding a digit "1" in front of the current record number?"
But then I realised that I misunderstood the challenge. We are not looking for the next number that also results in 11 steps. We are looking for smallest number that results in 12 steps.
He's not wrong. Here he is calculating pi by hand. Fair warning: the final result is rather disappointing.
I created a little F# program to search for the persistence 12 number. I found some 11s very quickly.
I haven't run into anything yet after that last 11.
I think you might be interested in my small blog post about it:
http://opensourc.es/blog/persistence it contains some code of getting the persistence and also how to create a list of interesting numbers as well as a visualization.
i Thought out something from this, you can check it out here
I hope to see Matt do more coding soon. It's so neat to see.
This is the longest chain I found starting with a 4 digit number:
9,797 -> 3,969 -> 1,458 -> 160 -> 0
4 steps
Are there 4 digit numbers that result in longer chains?
Edit: formatting
I've been using a program I wrote to check numbers for multiplicative persistence, starting at 200 digits.
The program just finished checking the 400-digit numbers, and it found nothing with a multiplicative persistence higher than 11 so far.
If anyone is interested, here's the code I'm using: github.com/Davipb/MultiplicativePersistence