Check out Brilliant (episode sponsor): brilliant.org/numberphile I love their problems of the week at: brilliant.org/NumberphilePOTW/ Truncatable Prime T-Shirt US: teespring.com/truncatable-prime-US And EU: teespring.com/truncatable-prime-EU
Numberphile Numberphile wearing at t shirt with a prime number on it is now a thought crime in the US. Although your chances of being detected are slim (how many members of the thought police even know what a prime is, and how would they explain their knowledge without implicating themselves?). The penalty is to be sent to a de-education camp (trailer park flooded with meth).
Let's start searching for videos by just entering in truncatable primes in other bases, and see what comes up. Even if you don't use base 10 by default, you can still find the video if you expand your search to other bases, assuming you at least use a similar number system.
Four two-digit deletable primes are "arbitrarily" deletable, i.e. removing either digit will return a prime number: 23, 37, 53 and 73. Any higher-digit arbitrarily deletable primes cannot, therefore, contain any numeral but 2, 3, 5 or 7, because to do so would introduce the possibility of reaching a two-digit number other than those four while deleting. But we can't go any further. Adding any of those numerals to any of the two-digit numbers introduces the possibility of reaching a different two-digit number. It can therefore be posited that the list of *all* arbitrarily deletable primes is finite, containing only 23, 37, 53 and 73. (And if empty string counts as a possible value, 2, 3, 5 and 7.)
Tantusar Just to further thid argument a bit: 2 and 5 must be the first digit because any number 2 digits or greater ending in 2 or 5 is non-prime. No digit can be repeated or you can end up divisible by 11. Since all the 2 digit numbers have a 3, you can't have another 3. You can't put a 7 on anything with a 7 and you can't put a 7 on anything starting with 2 or 5 or you'll get 27 or 57. You can't put a 2 or a 5 on anything with a 7 and you can't put a 2 or a 5 on anything with a 2 or a 5. That eliminates all the 3 digit numbers.
Yeah, I did some coding on Mathematica, those are the only ones in which you can keep deleting digits forever and only get primes. but I did find several which are primes for at least one deletion of a random digit, here are the results among the first 1,000,000 primes: 23, 37, 53, 73, 113, 131, 137, 173, 179, 197, 311, 317, 431, 617, 719, 1013, 1031, 1097, 1499, 1997, 2239, 2293, 3137, 4019, 4919, 6173, 7019, 7433, 9677, 10193, 10613, 11093, 19973, 23833, 26833, 30011, 37019, 40013, 47933, 73331, 74177, 90011, 91733, 93491, 94397, 111731, 166931, 333911, 355933, 477797, 477977, 633317, 633377, 665293, 700199, 719333, 746099, 779699, 901499, 901997, 944777, 962233, 991733, 1367777, 1440731, 1799999, 2668999, 3304331, 3716633, 4437011, 5600239, 6666437, 6913337, 7333331, 7364471, 7391117, 13334117 Remove any digit form these and they remain prime.
First I thought, "of course they EXIST -- 37 is one. The question is, do LARGE ones exist?" ("They" being "arbitrarily deletable" primes that remain "arbitrarily deletable" all the way down the chain to one digit.) Then I immediately realized they could have no repeated digits, since if they did, you could delete all the other digits and get down to something divisible by 11. They also can't have any digits but 2, 3, 5, and 7 (the one-digit primes), so you can't possibly get beyond 4 digits. And if they have a 2 or a 5, it has to be the first digit, since any number that ends with 2 or 5, other than 2 or 5 itself, is not prime. That leaves very few possibilities to check, and checking all the 3-digit possibilities quickly shows that you can't even get to 3 digits.
I believe all random digit remove primes are: 2, 3, 5, 7, 23, 37, 53, 73. Considering the sum of digits must never be divisible by 3 (else the number is divisible by 3) and there can be no repeated numbers (else divisible by 11 if removing anything else), there can't be many of the any order digit (random) deletable primes. You can't have any 0, 1, 4, 6, 8 or 9. (for 0, remove all after, result divisible by 5) The 5 and 2 can only be in the beginning. (else remove all after, result divisible by 2 or 5) Can't have 5 and 7, 2 and 7. (else remove all other, divisible by 3) We only get 2, 3, 5, 7, 23, 37, 53, 73, larger is impossible.
i got the same answer for base 10. the list ends at 73 because no 3-digit numbers can give 3 2-digit numbers that are random digit remove primes which also chain reaction means no 4-digit numbers exist for this either. i want to see if i can go further in higher bases but cant find a list of primes in base 12 :/
I just recently learned how big primes go. I always thought they didn't go very high, because it's hard to believe that a number on the billions doesn't have any single number that decided into it besides one and itself. If the prime number is on the billions, that means there's billions of numbers that could have divided into it and none of them do. That's pretty amazing to me.
Yep, the primes go on forever! The easiest way to see this is to just take the product of all the primes you’ve found so far, and add one. This must always be a new prime!
More like the new number is either prime, or it divides into a prime thats larger than the largest prime in your list, either way youve shown that theres always a larger prime than what you originally thought was the largest prime
Hey, Brady! I've always wondered why time isn't "metric", so I googled it and found "Decimal time". I totally fell in love with the concept. It would be nice to see a video in your format where you ask your friends what they think of it. Potential pros and cons, a bit of history and so on. (Please upvote, so he can see this)
@@XenophonSoulis You do lose a lot of functionality if you're very strict about sticking to powers of 10. It's nice having highly composite divisions instead, so you can talk about quarters and thirds of an hour for instance.
Let's just scrap hours and minutes and use kiloseconds instead. Sure, it'll mean a day on earth is 86.4 ks long, but there's nowhere else where a day comes out to a nice round number of seconds either.
@@Roxor128 there's no reason we'd have to keep using seconds though. Use a unit of time that's slightly shorter and have 100,000 of them in a day instead of 86,400.
Every once in a while I like to enjoy the joy you guys get out of numbers. I am not gifted with such abstract numeric talent, but find it fascinating to watch. Cheers and thanks.
373 is the largest prime such that each sub-string is prime. So you can truncate both side *simultaneously*, or *left*, or *right*, and still get a prime. You however can't erase the 7 and keep the 3s. The others are 2, 3, 5, 7 (obviously), 23, 37, 53 and 73
8:02 I actually got to work. I have a list of all the primes where you can remove any digit, and no matter which one, it will still be a prime, all the way to 1 digit. 2, 3, 5, 7, 23, 37, 53, 73. That's it.
A TH-camr uploads infinite videos on their channel. First video is a minute long, second one is two minutes long, third one is three minutes long. He puts all of those videos in a playlist. A viewer finished watching all of them 5 seconds before he made the playlist.
In the long System this number would be spelled out like that : 357 trilliard 686 trillion 312 billiard 646 billion 216 milliard 567 million 629 thousand 137
How does it feel unnatural? A billion is a million to the power of two, a trillion is a million to the power of three, and so on... This makes the long system feel much more natural than the short system (at least to me).
Counting "million, millionard" is unnatural. That's like counting "one, oneard, two, twoard, three, threeard". The value millionard represents isn't a subset of million or anything, so why would it look like there's a much stronger relation? Also, I know that qunitillion is million to the power of 5 in short system, that's easy. I have no idea what that would be in the long system, to the power of ten? How does that make sense?
8:13 There exists such numbers 23, 37, 53, 73. I believe these are all of them that don't include 0, which I excluded because if 0 is last to be deleted, that's not prime.
My first question was if you would have more or less digits with a larger base. I thought more because there would be more possibilities with each step, but then I thought maybe less. Happy you answered that for me.
I checked on the anywhere-deletable primes. The complete list of anywhere-deletable primes (in base 10) is 2, 3, 5, 7, 23, 37, 53, 73. So there are no 3-digit anywhere-deletable primes in base 10. Though you may end up with more interesting results in other bases.
It seems there is no number with at least three digits that can be truncated anywhere. It has to obey some rules: -Each digit has to be prime -2 or 5 can only appear in the first digit, because otherwise you would get a two digit number ending with 2 or 5 and that can't be prime. -All digits other than the first can only be 3 or 7. -No digits can appear twice, because otherwise you will get a number that is divisible by eleven after having deleted all other digits That gives us quite a short list: 237 273 537 573 None of those works, because you get either 27 or 57, which are not prime. ==> There is no number with more than two digits that you an truncate in any possible way and always get primes.
Cedros I think he was just listing the only numbers that fit the criteria, and he may have accidentally glossed over the fact that they aren't prime which is even more proof that it can't exist.
Yes, it felt easier for me, because I saw the 27 and 57. Checking if they are prime only turned out to be easy, as they are all divisible by 3. Otherwise the prime check would have been more complicated.
Largest one I could find that is always prime with random truncation: 717 (but it has the potential of spitting out 1). Here I've made the start keep going James
For numbers where it doesn't matter what digits are deleted you can not use any duplicate numbers, because the you can always get a number that can be divided by 11. You also can't use any non-prime digits because when you leave only one digit it still has to be prime, no matter what digit is left. So the maximum number of digits would be 4, since there are only 4 prime numbers with one digit (2, 3, 5 and 7). If you use 2 it must be the highest digit since if it isn't you can always make an even number, by removing numbers and leaving the 2 as the unit. However, if you use 5 it also has to be the highest digit, since any number ending in 5 (except 5) is not prime. So you can't use 5 and 2 in the same number, leaving the number with the maximum of 3 digits: a 3, a 7 and a 5 or a 2. That leaves us with four numbers with 3 digits: 237, 273, 537 and 573, but none of these are prime. Thus the only numbers that are prime and remain prime when any digits are removed have a maximum of 2 digits. (Btw they are 73, 37, 53 and 23)
Primes that always result in primes when you delete any digit are pretty few. First of all, all digits must be primes, as you can end up with each of them. Second, the last digit can't be 2 or 5 if you have more than one digit. Third, no prime can appear twice, as that would allow you to produce a multiple of 11. This is a comprehensive list of such primes: 2, 3, 5, 7, 23, 37, 53, 73. There are no three digit ones, as 237 and 537 are divisible by 3 and 237 can produce 27, which is not a prime. Any number with more digits would need to contain those, but as they already fail, no such numbers exist.
Without thinking too much, 73 for example is prime whichever digit you remove, and any such number is left (and right) trunkable, hence there are finitely many and at least one, there's a biggest.
The restraint that makes this less interesting is that you can’t repeat any digit and you can’t have a 1 anywhere, nor an even number else you can end up with 1, or a factor of 2 or 11. That limits the whole thing to subsets and combinations of 3,5 and 7.
11, 13, 17, 23, 31, 37, 53, 71, 73 113, 131, 137, 173, 311, 317 These are the "remove any and still prime" numbers I could find if you consider 1 being a prime 23, 37, 53, 73 Are the only real ones
You summed up why I watch your channels nicely at the end there. I've made a lot of different choices in my life since 2012 or so when I started watching you and Vsauce and et al. Maybe all the high maths and physics isn't being used in my daily life, but I think that I look at problems differently as a result of TH-cam.
It seems like for the left-truncatable primes, you should be able to add a 0 to the left, as long as you ended up with a non-zero integer at the very left as a stopping point.
kiffe22 The thing with the 'Deletable Primes' is that you can deliberately _choose_ which one you delete. If the number that's deleted is random, we can quite quickly find out all of them. Let's call them Random Deletable Primes (RDP) or whatever. Say a number is RDP. That means it is a Left Truncatable Prime, because deleting the numbers from left to right is a random possibility. For the same reason it is also a Right Truncatable Prime. Therefore an RDP is in both the Left and Right sets (RDP is a subset of intersection(LTP, RTP), for those that like notation 😜). Both these sets are finite. Hence, the set of RDPs is also finite.
He's not saying that any number you remove will leave a prime. He's saying that you can try deleting any number, and if one of them is prime, you can continue to chain. Another way to say this is that at least one choice of digit can be removed and create another prime.
To answer Brady's question on numbers where removing any digit keeps it prime: 1. there are finitely many. Given that you cannot repeat digits (else there's a scenario where you're left with a number divisible by 11), this limits us to 4-digit numbers (using the digits 2,3,5,7). 2. However, We can't use 2 in any but the leftmost position, since otherwise we can end with an even number. Same story with 5, though, meaning we're now stuck at 3-digit numbers. 3. Any combination of 3, 7, and one of the others is divisible by 3. That means no 3-digit numbers starting with 5 or 2. It also means no 3-digit numbers containing both 3 and 7. 4. Given the above, it is impossible to construct a 3-digit number which does not break one of those rules. If we start with 5 or 2, we either have to put the other somewhere not in the leading space, or use both 3 and 7. If we start with either 3 or 7, we have to put either 2 or 5 (or both) in non-leading positions. 5. So, in short, we're left with 2-digit numbers. The largest of which is 73. Sheldon Cooper will be happy to hear the news.
And if you're referring to the sharpening thing, that's definitely gonna cause more confusion for the people that doing get the reference. Besides that, that pencil was made after the logical name was already given.
Could including 0 lead to a potentially infinite number of these truncatable primes? For instance maybe 3000...trillions of 0s later...0007 is a prime so using the truncated method we would only need to check it and the number 7. Who knows how many of these types of numbers there are? You could then of course add any of the numbers found using the method in the video to the front of these other long numbers as well.
That did occur to me, but - at least from my point of view - it's a bit of a cheat. 7, 07 ,007 etc. are the same number, so you're effectively skipping a digit. A 30 digit number with 28 zeroes would only contain two different prime numbers.
To answer Brady's question at the end of the video (re: deletion in any order), the number would need to have digits consisting solely of single-digit primes (2, 3, 5, 7), since we can choose any digit to be the last one standing. The digits 2 and 5 could only appear as the first digit (otherwise deleting all the digits to their right would yield a composite). Furthermore, no digit can appear more than once (since an appropriate deletion would yield a number divisible by 11). Two-digit candidates for satisfying these conditions are 23, 27, 37, 53, 57, and 73, of which only 23, 37, 53, and 73 are prime. Since the order in which deletion occurs is arbitrary, we can get all the three-digit candidates by right-appending to these four two-digit primes, while adhering to the earlier conditions: 237 and 537. Both of these are divisible by 3, so the list is complete: 2, 3, 5, 7, 23, 37, 53, and 73.
45, 56, 57, 63, 456, 453, 567, 573, and 4563, also aren't prime. The point with that example is you don't have to remove digits from the ends, but you do still have to remove them in a specific order.
One of the rules that they forgot to mention is that the number can't contain a zero for it to be considered left-truncatable, and you obviously can't have a zero in a right-truncatable prime because at some point you'd end up with a number divisible by whatever base you're using.
It is a very funny and interesting video, but what I liked the most was Brady's last words on why there has to be a point to study these numbers (or any field of study, for that matter).
For numbers for which there is some sequence of digits you can delete and always get a prime, I propose "startling primes", because there is a sequence in which you can delete letters from "startling" and always have an English word.
“Random digit deletable” primes can only include the digits 2, 3, 5, and 7 and must: Include 2 at the beginning of the prime (if at all) Include 5 at the beginning of the prime (if at all) Not include two or more 3’s or 7’s 237, 273, 537, and 573 are all composite, so 73 is the greatest of this kind of prime. Full list: 2, 3, 5, 7, 23, 37, 53, 73 *QED*
Roger Wang If you allow 1 to be prime, there are 20 numbers which work: 1, 2, 3, 5, 7, 11, 13, 17, 23, 31, 37, 53, 71, 73, 113, 131, 137, 173, 311, 317
A prime that will remain a prime if you delete any of its digits must: 1) Have only prime digits: 2,3,5,7 2) Not have 2 or 5 after the first digit because if then we right-trunkate it we will get a number ending in 2 or 5,which is not prime. _If_ it is in the first place it does *not* matter, because even if we right-trunkate it we will get 2 or 5 themselves, which are primes. 3) *Not* have repeating digits, because if we delete all other digits the result will be divisible by 11, 111, etc. Numbers that only include 1 as a repeating digit would be allowed, but they would not fulfill the first two criteria. 4) Pass the prime test. The only numbers that fulfill the three first criteria are 2,3,5,7,23,27,37,53,57,73,237,273,537,573. Of these, 2,3,5,7,23,37,53,73 pass the first prime test and it so happens that all of them pass all required prime tests. So, the largest number that will always be a prime even if we delete all its digits is prime is 73.
Lets find the largest "fully" deletable prime, this is a prime where you can delete any digit in any order among the chain and always get a prime. First a few facts that make this problem easier: 1. All digits must be a prime, otherwise the digit by itself would be not a prime (and therefore it would be non-deletable) 2. There can be no replicated digits, if you replicate a digit then you can delete all the digits but the replicated one which would then be divisible by 11 and not a prime (except for 9 but 9 isn't a prime and so we won't use it via rule 1) 3. All subsets of digits within the deletable prime must also be deletable primes (this should be obvious why). 4. If the number has a 2 or a 5 in it, the 2 or 5 must be the left most digit, otherwise you could create a a number that is divisible by 2 or divisible by 5, (example: 523 is a prime, but because you could remove the 3 to get 52 which isn't, however, 23 is a prime and removing the 3 keeps it a prime). 5. As a result, a number cannot have BOTH have a 2 and a 5 in it, (they both can't be the left most digit, you'll always get 52 or 25 and neither are primes). From 1 and 2 the largest possible contender is 7532 and there are only 64 contenders. But when we apply rule 4 and 5 the largest contender becomes 573. So now we just follow the rules from 573 until we get our answer. Our largest contender: 573 is NOT a prime (it is divisible by 3) Next largest contender: 273 is NOT a prime (it is divisible by 3) Next largest contender: 73 Is a fully deletable prime. (73 = prime, 7 = prime, 3 = prime) Thus unfortunately there exists no fully deletable primes above 2 digits, and the largest fully deletable prime is thus 73.
A finite number of seeds will result in a finite number of branches if the growth of branches becomes scarce as it goes farther from the seed Basically only having 3 5 and 7 as the single digit prime results in a finite number of left-truncating primes, as primes in large numbers kinds dwindle a bit And once you break a chain it can not be repaired again, such as if the largest left truncating prime has a larger one but you need to truncate twice for the next one, well, that's a different kind.
Quick little code revealed a very short list of 'poppable' primes: 2, 3, 5, 7, 23, 37, 53, 73. Reasoning: all the two-digit possibilities must contain '3' as a digit. To expand to three-digit numbers would involve adding a number that is either a '3' or not a '3'. If the added digit not a '3', and you were to randomly pop off the '3' from the existing two-digit number, you would be left with a new two-digit number not in the list above. if the added digit is a '3', and you were to randomly pop off the number that is not a '3' from the existing two-digit number, you would be left with '33', which is not prime.
23, 37, 53 and 73 are the only randomly truncatable primes. Explanation: Since every digit can be the last surving one, all digits need to be prime number. So the only digits we can use are: 2, 3, 5, 7 2 and 5 can only be at the very beginning because otherwise we could always construct a number like "x5" or "x2" which will always be divisible by 5 and 2 respectively. So let's construct our numbers with the last digit first (which can only be 3 or 7) and lets let's start with 7. Possible numbers are: 27, 37, 57, 77 of which only 37 is a prime. So can we put another number in front of 37? No, we cannot use 3 or 7 again because we could construct 33 or 77 which is always be devisible by 11. We cannot use 2 because 27 is divisible by 3 and we cannot use 5 because 57 is also divisible by 3. So for numbers that end with 7, 37 is the only one. Now continue with 3: 23, 33, 53, 73 Those are all prime numbers except for 33. But since 2 and 5 can only occur in the beginning, we know that we cant extend 23 and 53 any further. So the only possible one would be 73 but we can use the exact same argument as for 37 (since the numbers are simply flipped) to show that we also cant extens this one further. And this leaves us with just 23, 37, 53 and 73
They talked about this in the video. Larger bases create larger trees, meaning larger primes. However, it is unknown whether or not they will extend to infinity at some point, though I would guess they don't as prime density goes down with higher numbers.
How far does the sequence go in each base. I didnt pause but for base ten, it looked like less than say 200 left truncateable primes. I guess in base two the sequence is 11=3 111=7 and that's all. Just these two since1111=15 which is not prime.
Reminds me of a geeky math game I came up with some years back based on truncatable primes where one person writes down a single digit number that is prime 2, 3, 5, 7 - then the other person writes a digit to the left or right to create another number that is still prime - and so on and so fourth - till someone can't continue or creates a composite non-prime number by mistake. As the numbers get larger, the easiest way to enter and accurately check the primes automatically is on a PC with the "openssl" command line program (one of the functions of which lets you check prime numbers). Linux PCs normally have openssl installed by default, but you can also find pre-compiled command line versions for Android, Mac OS, and MS Windows if you look around. To start play open a terminal and type "openssl prime" followed by a single digit prime: openssl prime 3 At which point openssl reports that 3 (after converting to hex for checking) is indeed prime. Then to continue play, the second player just hits the up-arrow key, which will make the terminal re-display the previous command for editing, then uses the left-right arrow keys to edit the 3 and add a number (for example 13, 43, 73, 31, or 37), for example. openssl prime 43 You can restrict the game rules as you see fit, only allowing addition of a single digit, or allowing multiple digits, on the right, left, or anywhere including the middle. This game gets really hard as the numbers gets larger.
You'd be adding digits, but you wouldn't always get a new number when you delte a digit. A 30-digit number with 28 zeroes would only contain two different prime numbers. And it would take a bit more work to test, since you could keep adding zeroes forever in the hope of hitting a prime number.
Trimon The fact that there are infinite numbers does matter because there is an infinite amount of numbers with n amount of 0s and 2 digits that are prime. All of these numbers are probably prime. Also the fact that it is proven to have finitely many of these numbers have been proven to have finitely many without any 0s.
All left-truncatable primes (in base 10) can be calculated in a fraction of a second with the following Mathematica/Wolfram language expression: Flatten@NestList[Apply[{#1 + 1, Select[PrimeQ]@Flatten@Table[10^#1 Range@9 + x, {x, #2}]} &], {0, {0}}, 24][[2 ;;, 2]]
Joshua Hillerup it probably wasn't all done by hand but a program could fairly easily brute force it. Especially if you gave it a list of known prime numbers to compare each result to. Basically say that if the result is on the list continue, if not then try the next path.
You couldn't determine that by brute force because there are infinitely many primes. There must be a proof but I don't have any clue how you would do that.
The only numbers from which you can remove both from left to right or right to left in base 10 and receive a prime number: 2, 3, 5, 7, 23, 37, 53, 73, 373.
The numbers in hexadecimal: 2, 3, 5, 7, B, D, 25, 2B, 35, 3B, 3D, 53, B3, B5, D3, 2B3, 3B3, D3D. This doesn't work at all in binary because the only single digit numbers are 1 and 0. (If we counted 1 as a prime, the highest we could go is 111, the binary representation of 7.) I'm not sure what the significance of several of the three digit numbers being palindromes is. There may be a reason for this trend but I'm not sure yet.
(An) answer to Brady's arbitrarily deletable prime: 35E7 in base 12 (6043 in base 10) I use the digit system 0123456789TE for base 12. 0 deletions: 35E7 (6043) 1 deletion: 35E (503), 357 (499), 3E7 (571), 5E7 (859) 2 deletions: 35 (41), 3E (47), 37 (43), 5E (71), 57 (67), E7 (139) 3 deletions: 3 (3), 5 (5), E (11), 7 (7) Primes of length 4 were the longest I found, and of them this was in the smallest base. Number bases through 140 were checked. There are some interesting trends to be observed, so I recommend anyone with some coding ability to explore this problem.
8:02 If we're willing to do a little bit of fudging and include the number "1," then 137 works ;) 137 can become 13 or 17 or 37 13 can become 1 or 3 17 can become 1 or 7 37 can become 3 or 7
Hold on everyone, we need to specify between deletable and any-digit-deletable. It satisfies one but not the other. They were pretty clear about that in the video.
Your _allowed_ to choose any digit (not being restricted to one of the ends) to create your chain of primes. There's no requirement that all choices must work. The existence of one sequence of choices is enough. I first had the same thought as you, but figured it out after a while. They could have phrased it a bit differently in the video. The existence of a 6 in the number is enough to throw the other interpretation out.
If you are allowed to remove a digit "in ANY order you want", what numbers are able to be prime after the removal of each digit down to just 1 digit. This is what he meant, not that all digits work all the time. Those numbers are 27, 37, and 57 and that's it none that are three digits or more.
Check out Brilliant (episode sponsor): brilliant.org/numberphile
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Numberphile Numberphile wearing at t shirt with a prime number on it is now a thought crime in the US. Although your chances of being detected are slim (how many members of the thought police even know what a prime is, and how would they explain their knowledge without implicating themselves?). The penalty is to be sent to a de-education camp (trailer park flooded with meth).
How much is pinning this in the comments worth? You can keep the number in GBP.
Joshua Jansen 50 pence?
Numberphile About 6 mins in; OH WOW.
Yeah.
Do you have to re-edit your videos if the company that's sponsoring them eventually stops the offer being offered?
That title tho... Good luck to anyone who searches for this video in the future.
try numberphile prime numbers
Thank goodness for the keywords/tags and the description, then.
Considering the importance of the number, all cultures will eventually find it.
what about the tom scott video written in inuktitut
Let's start searching for videos by just entering in truncatable primes in other bases, and see what comes up.
Even if you don't use base 10 by default, you can still find the video if you expand your search to other bases, assuming you at least use a similar number system.
I never knew how mutch I wanted a pencil with a giant prime number on it until now.
Same
Four two-digit deletable primes are "arbitrarily" deletable, i.e. removing either digit will return a prime number: 23, 37, 53 and 73. Any higher-digit arbitrarily deletable primes cannot, therefore, contain any numeral but 2, 3, 5 or 7, because to do so would introduce the possibility of reaching a two-digit number other than those four while deleting. But we can't go any further. Adding any of those numerals to any of the two-digit numbers introduces the possibility of reaching a different two-digit number. It can therefore be posited that the list of *all* arbitrarily deletable primes is finite, containing only 23, 37, 53 and 73. (And if empty string counts as a possible value, 2, 3, 5 and 7.)
Tantusar Just to further thid argument a bit: 2 and 5 must be the first digit because any number 2 digits or greater ending in 2 or 5 is non-prime. No digit can be repeated or you can end up divisible by 11. Since all the 2 digit numbers have a 3, you can't have another 3. You can't put a 7 on anything with a 7 and you can't put a 7 on anything starting with 2 or 5 or you'll get 27 or 57. You can't put a 2 or a 5 on anything with a 7 and you can't put a 2 or a 5 on anything with a 2 or a 5. That eliminates all the 3 digit numbers.
Yeah, I did some coding on Mathematica, those are the only ones in which you can keep deleting digits forever and only get primes.
but I did find several which are primes for at least one deletion of a random digit, here are the results among the first 1,000,000 primes:
23, 37, 53, 73, 113, 131, 137, 173, 179, 197, 311, 317, 431, 617, 719, 1013, 1031, 1097, 1499, 1997, 2239, 2293, 3137, 4019, 4919, 6173, 7019, 7433, 9677, 10193, 10613, 11093, 19973, 23833, 26833, 30011, 37019, 40013, 47933, 73331, 74177, 90011, 91733, 93491, 94397, 111731, 166931, 333911, 355933, 477797, 477977, 633317, 633377, 665293, 700199, 719333, 746099, 779699, 901499, 901997, 944777, 962233, 991733, 1367777, 1440731, 1799999, 2668999, 3304331, 3716633, 4437011, 5600239, 6666437, 6913337, 7333331, 7364471, 7391117, 13334117
Remove any digit form these and they remain prime.
Ok, Now do it for any base.
First I thought, "of course they EXIST -- 37 is one. The question is, do LARGE ones exist?" ("They" being "arbitrarily deletable" primes that remain "arbitrarily deletable" all the way down the chain to one digit.) Then I immediately realized they could have no repeated digits, since if they did, you could delete all the other digits and get down to something divisible by 11. They also can't have any digits but 2, 3, 5, and 7 (the one-digit primes), so you can't possibly get beyond 4 digits. And if they have a 2 or a 5, it has to be the first digit, since any number that ends with 2 or 5, other than 2 or 5 itself, is not prime. That leaves very few possibilities to check, and checking all the 3-digit possibilities quickly shows that you can't even get to 3 digits.
Just worked that out before going through the replies. Kind of a disappointment that there's nothing beyond 2 digits that satisfies the conditions.
7:00 James reminded me of a politician there, just outright changing the question.
Perfect misdirection!
That's an interesting statement, but what if we said this about Matt? It would be true then, right?
"One isn't a prime number, so I don't know why I've even mentioned it."
i just love how happy this guy is every time i watch him talk bout numbers :D
(which is almost every video)
Numbers? You mean Noobahs!
I believe all random digit remove primes are: 2, 3, 5, 7, 23, 37, 53, 73.
Considering the sum of digits must never be divisible by 3 (else the number is divisible by 3) and there can be no repeated numbers (else divisible by 11 if removing anything else), there can't be many of the any order digit (random) deletable primes.
You can't have any 0, 1, 4, 6, 8 or 9. (for 0, remove all after, result divisible by 5)
The 5 and 2 can only be in the beginning. (else remove all after, result divisible by 2 or 5)
Can't have 5 and 7, 2 and 7. (else remove all other, divisible by 3)
We only get 2, 3, 5, 7, 23, 37, 53, 73, larger is impossible.
Agreed, I used the same logic
wow, nice
That's true, I ran a program testing the first 1000 numbers.
i got the same answer for base 10. the list ends at 73 because no 3-digit numbers can give 3 2-digit numbers that are random digit remove primes which also chain reaction means no 4-digit numbers exist for this either. i want to see if i can go further in higher bases but cant find a list of primes in base 12 :/
In case you are interested, I made a little python program to calculate all random digit remove primes in different bases. I have not verified the code, but I am pretty confident it is correct. They are in their corresponding notation, using ABC for digits higher then 9. Interestingly, base 8 has a three digit number and base 12 has a four digit number!
2:
3: 2,
4: 2, 3, 23,
5: 2, 3, 23, 32,
6: 2, 3, 5, 25, 35,
7: 2, 3, 5, 23, 25, 32, 52,
8: 2, 3, 5, 7, 23, 27, 35, 37, 53, 57, 73, 75, 357, 573, 753,
9: 2, 3, 5, 7, 25, 32, 52,
10: 2, 3, 5, 7, 23, 37, 53, 73,
11: 2, 3, 5, 7, 27, 72,
12: 2, 3, 5, 7, B, 25, 27, 35, 37, 3B, 57, 5B, 75, B5, B7, 357, 35B, 375, 3B5, 3B7, 5B7, 35B7,
13: 2, 3, 5, 7, B, 23, 25, 2B, 32, 52,
14: 2, 3, 5, 7, B, D, 23, 2D, 35, 3B, 53, 5D, 73, 75, 7B, B3, BD, DB,
15: 2, 3, 5, 7, B, D, 27, 2B, 2D, 32, 72, B2, D2,
16: 2, 3, 5, 7, B, D, 25, 2B, 35, 3B, 3D, 53, B3, B5, D3.
17: 2, 3, 5, 7, B, D, 23, 27, 2D, 32, D2.
18: 2, 3, 5, 7, B, D, H, 25, 27, 2B, 2H, 35, 37, 3D, 3H, 57, 5B, 5D, 5H, 75, 7B, 7D, BD, D5, D7, DH, H5, H7, HB, 357, 375, 3D7, 3DH, 3H5, 57D, 5BD, 5D7, 5DH, 5H7, 75D, DH5, H75.
19: 2, 3, 5, 7, B, D, H, 23, 25, 32, 52, B2.
20: 2, 3, 5, 7, B, D, H, J, 23, 27, 2D, 2J, 37, 3B, 3D, 3J, 53, 57, 5D, 7B, 7H, B3, B7, BD, BJ, D3, DB, DH, H7, HD, HJ, J3, JH, 2D3, 3B7, 3BD, 3BJ, 3DB, BD3, BJ3.
21: 2, 3, 5, 7, B, D, H, J, 25, 2B, 2H, 2J, 52, 72, B2, H2, J2.
22: 2, 3, 5, 7, B, D, H, J, 23, 2H, 35, 37, 3D, 3H, 53, 5H, 73, 7D, 7J, D7, H5, J3, JD, 23H, 35H, 37D, 3H5, 53H, 73D.
23: 2, 3, 5, 7, B, D, H, J, 27, 2D, 32, 72, J2.
24: 2, 3, 5, 7, B, D, H, J, N, 25, 2B, 2D, 2J, 2N, 37, 3B, 3H, 57, 5B, 5H, 5J, 75, 7B, 7D, 7N, B5, B7, BD, BH, BJ, D5, DJ, HB, HD, HN, J5, J7, JB, JN, N5, NB, NH, NJ, 25B, 25J, 2BD, 2DJ, 2J5, 2JB, 2N5, 2NJ, 37B, 3B7, 5BJ, 5HB, 5J7, 5JB, 7D5, B5H, BD5, DJ5, HBD, J57, J75, JB5, NJ5, 2NJ5.
25: 2, 3, 5, 7, B, D, H, J, N, 23, 2B, 2H, 2N, 52, B2, N2.
Edit: Typo in the code, 15 is not a prime lol. Fixed now.
I just recently learned how big primes go. I always thought they didn't go very high, because it's hard to believe that a number on the billions doesn't have any single number that decided into it besides one and itself. If the prime number is on the billions, that means there's billions of numbers that could have divided into it and none of them do. That's pretty amazing to me.
Yep, the primes go on forever!
The easiest way to see this is to just take the product of all the primes you’ve found so far, and add one. This must always be a new prime!
More like the new number is either prime, or it divides into a prime thats larger than the largest prime in your list, either way youve shown that theres always a larger prime than what you originally thought was the largest prime
Hey, Brady!
I've always wondered why time isn't "metric", so I googled it and found "Decimal time". I totally fell in love with the concept. It would be nice to see a video in your format where you ask your friends what they think of it. Potential pros and cons, a bit of history and so on.
(Please upvote, so he can see this)
Pros and pros, it has no cons...
@@XenophonSoulis You do lose a lot of functionality if you're very strict about sticking to powers of 10. It's nice having highly composite divisions instead, so you can talk about quarters and thirds of an hour for instance.
@@fastpuppy2000 You don't need thirds of an hour if they aren't integer multiples of the second in the first place...
Let's just scrap hours and minutes and use kiloseconds instead. Sure, it'll mean a day on earth is 86.4 ks long, but there's nowhere else where a day comes out to a nice round number of seconds either.
@@Roxor128 there's no reason we'd have to keep using seconds though. Use a unit of time that's slightly shorter and have 100,000 of them in a day instead of 86,400.
Highly compressed format for storing certain lists of primes.
The ultimate "let's be honest, you didn't search for this" video
A sharpened pencil with that number would also be a prime piece of stake.
James Jumper - lol, Dad Jokes: Numberphile Edition
But it wouldn't make a prime product line.
*ba dum tss*
But if the pencil is unsharpened it's pointless
Every once in a while I like to enjoy the joy you guys get out of numbers. I am not gifted with such abstract numeric talent, but find it fascinating to watch. Cheers and thanks.
Unless you sharpen that pencil so that 8 becomes a 3 and now you no longer have a prime number :)
Maybe there exists *very* truncatable primes that verify this haha
Eliot any truncatable number that doesn't use 8 as a digit would work
Oh yeah you're right that's not fun :'(
You beat me to this comment! XD
Have you verified that? ;)
373 is the largest prime such that each sub-string is prime. So you can truncate both side *simultaneously*, or *left*, or *right*, and still get a prime. You however can't erase the 7 and keep the 3s. The others are 2, 3, 5, 7 (obviously), 23, 37, 53 and 73
1:30 "He is also a mathematician." Wow, could have knocked me over with a feather when I heard that.
I like how so many of these end up with "And we just don't know!" with a happy and curious expression.
8:02
I actually got to work. I have a list of all the primes where you can remove any digit, and no matter which one, it will still be a prime, all the way to 1 digit.
2, 3, 5, 7, 23, 37, 53, 73.
That's it.
The smart thing about this channel: there will always be enough "special number" videos because they're infinite.
A TH-camr uploads infinite videos on their channel. First video is a minute long, second one is two minutes long, third one is three minutes long. He puts all of those videos in a playlist. A viewer finished watching all of them 5 seconds before he made the playlist.
@@tiletapper4everis this real? What's the channel name
@@thoughtfulsoul3402no, it’s a joke about the sun of all natural numbers and -1/12. Just look up -1/12 and you’ll get several videos explaining it.
@@thoughtfulsoul3402it’s a
joke based on famous 1+2+3+4… = -1/12 result
Thats a contradiction
of course the "random digit deletable" primes exist, example: 23
Numberphile Deleting 7 would give 33 = 3*11 which isn't prime :-(
The rest are 37, 53 and 73.
The only possibilities are 23, 37, 53 and 73:.
3C Kitani 1 aint prime, yo
tommihommi1 Yeah, i know xd
In the long System this number would be spelled out like that :
357 trilliard 686 trillion 312 billiard 646 billion 216 milliard 567 million 629 thousand 137
This is how we spell in french. It feels so unnatural.
How does it feel unnatural? A billion is a million to the power of two, a trillion is a million to the power of three, and so on... This makes the long system feel much more natural than the short system (at least to me).
Cedros It has no point, if you really think of it. Why use one latin number every 2 10 powers?
YipHyGaming - Minecraft Agario Cytus and more! Because we started counting on the second group.
Counting "million, millionard" is unnatural. That's like counting "one, oneard, two, twoard, three, threeard".
The value millionard represents isn't a subset of million or anything, so why would it look like there's a much stronger relation? Also, I know that qunitillion is million to the power of 5 in short system, that's easy. I have no idea what that would be in the long system, to the power of ten? How does that make sense?
Always in our prime. I want this pencil
8:13 There exists such numbers 23, 37, 53, 73. I believe these are all of them that don't include 0, which I excluded because if 0 is last to be deleted, that's not prime.
My first question was if you would have more or less digits with a larger base. I thought more because there would be more possibilities with each step, but then I thought maybe less. Happy you answered that for me.
You GOT to be kidding!
357686312646216567629137 has always been my favorite number! What are the odds?
That's pretty funny because mine is 357686312646216567629138(!)
Fascinating beyond imagination! That’s why I love you so much Numberphile!
I checked on the anywhere-deletable primes. The complete list of anywhere-deletable primes (in base 10) is 2, 3, 5, 7, 23, 37, 53, 73. So there are no 3-digit anywhere-deletable primes in base 10. Though you may end up with more interesting results in other bases.
It seems there is no number with at least three digits that can be truncated anywhere.
It has to obey some rules:
-Each digit has to be prime
-2 or 5 can only appear in the first digit, because otherwise you would get a two digit number ending with 2 or 5 and that can't be prime.
-All digits other than the first can only be 3 or 7.
-No digits can appear twice, because otherwise you will get a number that is divisible by eleven after having deleted all other digits
That gives us quite a short list:
237
273
537
573
None of those works, because you get either 27 or 57, which are not prime.
==> There is no number with more than two digits that you an truncate in any possible way and always get primes.
Exactly my thinking except for the last step: All four three-digit numbers are not prime, because they are all divisible by 3.
weilgooglesproduktintegrationsstrategievoneinemexmicrosoftsuitandtiemanagerfritzengeleitetwurdeunddahervonvornhereinzumscheiternverurteiltwar
Cedros I think he was just listing the only numbers that fit the criteria, and he may have accidentally glossed over the fact that they aren't prime which is even more proof that it can't exist.
Yes, it felt easier for me, because I saw the 27 and 57. Checking if they are prime only turned out to be easy, as they are all divisible by 3. Otherwise the prime check would have been more complicated.
Basically 73 becomes the largest
James Prime is Bacc
Largest one I could find that is always prime with random truncation: 717 (but it has the potential of spitting out 1). Here I've made the start keep going James
0:43 you can see a metal CGP grey logo
1:35 that’s gotta be the nerdiest pencil I’ve ever seen, and I love it! 😂
If it's 739397, it's okay with an eraser.
huh?
An eraser to go with the pencil, with 739397 written on one side, and 'prime for writing wrongs in your life' on the other side.
or just for idiots like me that sharpen their pencils on both sides at once
@@onetwothreefour3957 you got three hands or something?
For numbers where it doesn't matter what digits are deleted you can not use any duplicate numbers, because the you can always get a number that can be divided by 11. You also can't use any non-prime digits because when you leave only one digit it still has to be prime, no matter what digit is left. So the maximum number of digits would be 4, since there are only 4 prime numbers with one digit (2, 3, 5 and 7). If you use 2 it must be the highest digit since if it isn't you can always make an even number, by removing numbers and leaving the 2 as the unit. However, if you use 5 it also has to be the highest digit, since any number ending in 5 (except 5) is not prime. So you can't use 5 and 2 in the same number, leaving the number with the maximum of 3 digits: a 3, a 7 and a 5 or a 2. That leaves us with four numbers with 3 digits: 237, 273, 537 and 573, but none of these are prime. Thus the only numbers that are prime and remain prime when any digits are removed have a maximum of 2 digits. (Btw they are 73, 37, 53 and 23)
Primes that always result in primes when you delete any digit are pretty few. First of all, all digits must be primes, as you can end up with each of them. Second, the last digit can't be 2 or 5 if you have more than one digit. Third, no prime can appear twice, as that would allow you to produce a multiple of 11. This is a comprehensive list of such primes: 2, 3, 5, 7, 23, 37, 53, 73. There are no three digit ones, as 237 and 537 are divisible by 3 and 237 can produce 27, which is not a prime. Any number with more digits would need to contain those, but as they already fail, no such numbers exist.
If you allow 1 to be prime, there are 20 numbers which work: 1, 2, 3, 5, 7, 11, 13, 17, 23, 31, 37, 53, 71, 73, 113, 131, 137, 173, 311, 317
Wait... what number is 27 divided by?
been a fan of this channel for years.... still can't wrap my brain around "infinite primes."
Without thinking too much, 73 for example is prime whichever digit you remove, and any such number is left (and right) trunkable, hence there are finitely many and at least one, there's a biggest.
After quick computations by hand, the list of such numbers is,
2, 3, 5, 7, 23, 37, 53, 73, 373
The largest such is 373
373 doesn't work if you remove the 7.
Yeah you're damn right, then 73
The restraint that makes this less interesting is that you can’t repeat any digit and you can’t have a 1 anywhere, nor an even number else you can end up with 1, or a factor of 2 or 11. That limits the whole thing to subsets and combinations of 3,5 and 7.
"hence there are finitely many and at least one, there's a biggest." , What makes you say there are finitely many of them ?
This is my favorite numberphile guy
11, 13, 17, 23, 31, 37, 53, 71, 73
113, 131, 137, 173, 311, 317
These are the "remove any and still prime" numbers I could find if you consider 1 being a prime
23, 37, 53, 73
Are the only real ones
1 is *NOT* a prime.
'completely out of left field' came completely out of left field ... nice deliver :)
“always in our prime” lol I love it
Always get hyped when James is in numberphile!
that pencil needs to be merch
You summed up why I watch your channels nicely at the end there. I've made a lot of different choices in my life since 2012 or so when I started watching you and Vsauce and et al. Maybe all the high maths and physics isn't being used in my daily life, but I think that I look at problems differently as a result of TH-cam.
What if you sharpen that pencil too much; there would be no prime!
Nargis Akter then it doesn't say 'always in your prime' anymore either ;)
+ Nargis Akter: That's how you know when it's time to get a new prime pencil!
Fred
at some point, it will say "ur prime".
James is such an inspiration to a person like me who wants to major in it.
I love prime numbers
I love refrigerators
Love your name
I like trains
Why doesn't anyone like Easy Bake Ovens?
jake me Too
Aw man, I have a lot to do, but now I'm really tempted to follow that delete-any-digit idea...
I love how unsuspicious some numbers look but they have such amazing unique qualities
Do you mean inauspicious?
We loooooooooooooooove Dr James Grime! 👍🤓
The Grimes man :D I love him
It seems like for the left-truncatable primes, you should be able to add a 0 to the left, as long as you ended up with a non-zero integer at the very left as a stopping point.
7:40, but you're not allowed to do it any way you want?
6 wouldn't be prime, neither would 46. What exactly is a "deletable prime"?
kiffe22 I’m confused too, 453 and 45 are divisible by 3 as well
kiffe22 The thing with the 'Deletable Primes' is that you can deliberately _choose_ which one you delete.
If the number that's deleted is random, we can quite quickly find out all of them. Let's call them Random Deletable Primes (RDP) or whatever.
Say a number is RDP. That means it is a Left Truncatable Prime, because deleting the numbers from left to right is a random possibility.
For the same reason it is also a Right Truncatable Prime. Therefore an RDP is in both the Left and Right sets (RDP is a subset of intersection(LTP, RTP), for those that like notation 😜). Both these sets are finite. Hence, the set of RDPs is also finite.
He must mean you can delete any digit, but only delete once.
But that's not the way the example played out.
He's not saying that any number you remove will leave a prime. He's saying that you can try deleting any number, and if one of them is prime, you can continue to chain.
Another way to say this is that at least one choice of digit can be removed and create another prime.
To answer Brady's question on numbers where removing any digit keeps it prime:
1. there are finitely many. Given that you cannot repeat digits (else there's a scenario where you're left with a number divisible by 11), this limits us to 4-digit numbers (using the digits 2,3,5,7).
2. However, We can't use 2 in any but the leftmost position, since otherwise we can end with an even number. Same story with 5, though, meaning we're now stuck at 3-digit numbers.
3. Any combination of 3, 7, and one of the others is divisible by 3. That means no 3-digit numbers starting with 5 or 2. It also means no 3-digit numbers containing both 3 and 7.
4. Given the above, it is impossible to construct a 3-digit number which does not break one of those rules. If we start with 5 or 2, we either have to put the other somewhere not in the leading space, or use both 3 and 7. If we start with either 3 or 7, we have to put either 2 or 5 (or both) in non-leading positions.
5. So, in short, we're left with 2-digit numbers. The largest of which is 73.
Sheldon Cooper will be happy to hear the news.
We need a pencil with the parker right truncatable prime.
There is no choice. We HAVE to have it. Put it in the shop, quickly, you're gonna make a LOT of money (maybe cause i'll buy a heck of a lot of them)
Smallest randomly deletable number: 37.
Delete 3, you get 7.
Delete 7, you get 3.
I may write a C# program to find more.
73
23?
Pencil primes is a nicer name than truncatible primes
Why pencil prime? That doesn't make sense. Truncatable makes sense because that's what you're doing, you're truncating it.
And if you're referring to the sharpening thing, that's definitely gonna cause more confusion for the people that doing get the reference. Besides that, that pencil was made after the logical name was already given.
Austin Bryan
I just think it's a nicer name because it functions the same way as a pencil- you take some off the top and it still works.
Henry Bownes if you remove part of a pencil, you're truncating the pencil
Truncation is a standard name for the operation they're doing to the number.
Wasn’t expecting a numberphile video today!
What about adding 0s to these. Like 103 -> 03 > 3?
Could including 0 lead to a potentially infinite number of these truncatable primes? For instance maybe 3000...trillions of 0s later...0007 is a prime so using the truncated method we would only need to check it and the number 7. Who knows how many of these types of numbers there are? You could then of course add any of the numbers found using the method in the video to the front of these other long numbers as well.
Zsemberi Dániel interesting, as the Video didn't include Zeros...
NJ S Yeah after writing my comment I gave it more thought and came to the same conclusion as you. Next time I'll think before I write :)
It would make for a perfect Parker's Truncatable Prime.
That did occur to me, but - at least from my point of view - it's a bit of a cheat. 7, 07 ,007 etc. are the same number, so you're effectively skipping a digit. A 30 digit number with 28 zeroes would only contain two different prime numbers.
To answer Brady's question at the end of the video (re: deletion in any order), the number would need to have digits consisting solely of single-digit primes (2, 3, 5, 7), since we can choose any digit to be the last one standing. The digits 2 and 5 could only appear as the first digit (otherwise deleting all the digits to their right would yield a composite). Furthermore, no digit can appear more than once (since an appropriate deletion would yield a number divisible by 11). Two-digit candidates for satisfying these conditions are 23, 27, 37, 53, 57, and 73, of which only 23, 37, 53, and 73 are prime. Since the order in which deletion occurs is arbitrary, we can get all the three-digit candidates by right-appending to these four two-digit primes, while adhering to the earlier conditions: 237 and 537. Both of these are divisible by 3, so the list is complete: 2, 3, 5, 7, 23, 37, 53, and 73.
7:46 Huh? But deleting the 7 would leave 6, which isn't prime?
45, 56, 57, 63, 456, 453, 567, 573, and 4563, also aren't prime. The point with that example is you don't have to remove digits from the ends, but you do still have to remove them in a specific order.
I was literally thinking about right truncatable primes in the shower this morning, then this 2-year-old video was in my feed.
Is a zero allowed? For example, could 503 be a left truncatable prime? If so, the list can be expanded.
2003
But then you can expect a number like 20000000......0000003 as a prime number, and then the number 357686312646216567629137 is no longer interesting.
One of the rules that they forgot to mention is that the number can't contain a zero for it to be considered left-truncatable, and you obviously can't have a zero in a right-truncatable prime because at some point you'd end up with a number divisible by whatever base you're using.
03 isn't a real number. You can't have 03 of something.
@@chrismanuel9768 03 = 3. It's just a matter of notation.
It is a very funny and interesting video, but what I liked the most was Brady's last words on why there has to be a point to study these numbers (or any field of study, for that matter).
"oh, me?". He's lovely
I’m more impressed on how he remembers which are prime and which aren’t for the larger numbers holy
I want that pencil now!! Where can I get one ?!!
For numbers for which there is some sequence of digits you can delete and always get a prime, I propose "startling primes", because there is a sequence in which you can delete letters from "startling" and always have an English word.
“Random digit deletable” primes can only include the digits 2, 3, 5, and 7 and must:
Include 2 at the beginning of the prime (if at all)
Include 5 at the beginning of the prime (if at all)
Not include two or more 3’s or 7’s
237, 273, 537, and 573 are all composite, so 73 is the greatest of this kind of prime.
Full list: 2, 3, 5, 7, 23, 37, 53, 73
*QED*
Roger Wang If you allow 1 to be prime, there are 20 numbers which work: 1, 2, 3, 5, 7, 11, 13, 17, 23, 31, 37, 53, 71, 73, 113, 131, 137, 173, 311, 317
He didnt say "random digit deleteable prime" he said deleteable prime. You can delete the digits in some order, not necessarily in any order.
A prime that will remain a prime if you delete any of its digits must:
1) Have only prime digits: 2,3,5,7
2) Not have 2 or 5 after the first digit because if then we right-trunkate it we will get a number ending in 2 or 5,which is not prime. _If_ it is in the first place it does *not* matter, because even if we right-trunkate it we will get 2 or 5 themselves, which are primes.
3) *Not* have repeating digits, because if we delete all other digits the result will be divisible by 11, 111, etc. Numbers that only include 1 as a repeating digit would be allowed, but they would not fulfill the first two criteria.
4) Pass the prime test.
The only numbers that fulfill the three first criteria are 2,3,5,7,23,27,37,53,57,73,237,273,537,573. Of these, 2,3,5,7,23,37,53,73 pass the first prime test and it so happens that all of them pass all required prime tests. So, the largest number that will always be a prime even if we delete all its digits is prime is 73.
James..make something on german tank problem
Love these videos that talk about one number and why it is cool, great video!
Lets find the largest "fully" deletable prime, this is a prime where you can delete any digit in any order among the chain and always get a prime.
First a few facts that make this problem easier:
1. All digits must be a prime, otherwise the digit by itself would be not a prime (and therefore it would be non-deletable)
2. There can be no replicated digits, if you replicate a digit then you can delete all the digits but the replicated one which would then be divisible by 11 and not a prime (except for 9 but 9 isn't a prime and so we won't use it via rule 1)
3. All subsets of digits within the deletable prime must also be deletable primes (this should be obvious why).
4. If the number has a 2 or a 5 in it, the 2 or 5 must be the left most digit, otherwise you could create a a number that is divisible by 2 or divisible by 5, (example: 523 is a prime, but because you could remove the 3 to get 52 which isn't, however, 23 is a prime and removing the 3 keeps it a prime).
5. As a result, a number cannot have BOTH have a 2 and a 5 in it, (they both can't be the left most digit, you'll always get 52 or 25 and neither are primes).
From 1 and 2 the largest possible contender is 7532 and there are only 64 contenders. But when we apply rule 4 and 5 the largest contender becomes 573. So now we just follow the rules from 573 until we get our answer.
Our largest contender: 573 is NOT a prime (it is divisible by 3)
Next largest contender: 273 is NOT a prime (it is divisible by 3)
Next largest contender: 73 Is a fully deletable prime. (73 = prime, 7 = prime, 3 = prime)
Thus unfortunately there exists no fully deletable primes above 2 digits, and the largest fully deletable prime is thus 73.
bleesev2
Good proof. The only (non-trivial) fully deletable primes are 23, 37, 53 and 73.
Why not 317?
S-N-A-IL PS4 Remove the 3 and 7 to get 1 which is not prime.
You can also just look at the intersection of the left-deletable and right-deletable lists, since it will necessarily be a subset of that.
A finite number of seeds will result in a finite number of branches if the growth of branches becomes scarce as it goes farther from the seed
Basically only having 3 5 and 7 as the single digit prime results in a finite number of left-truncating primes, as primes in large numbers kinds dwindle a bit
And once you break a chain it can not be repaired again, such as if the largest left truncating prime has a larger one but you need to truncate twice for the next one, well, that's a different kind.
3:47
imagine if he found a bigger one while doing this example
Quick little code revealed a very short list of 'poppable' primes: 2, 3, 5, 7, 23, 37, 53, 73.
Reasoning: all the two-digit possibilities must contain '3' as a digit. To expand to three-digit numbers would involve adding a number that is either a '3' or not a '3'.
If the added digit not a '3', and you were to randomly pop off the '3' from the existing two-digit number, you would be left with a new two-digit number not in the list above.
if the added digit is a '3', and you were to randomly pop off the number that is not a '3' from the existing two-digit number, you would be left with '33', which is not prime.
This number is bigger than the possibilities of the Rubik’s cube!
23, 37, 53 and 73 are the only randomly truncatable primes.
Explanation: Since every digit can be the last surving one, all digits need to be prime number. So the only digits we can use are:
2, 3, 5, 7
2 and 5 can only be at the very beginning because otherwise we could always construct a number like "x5" or "x2" which will always be divisible by 5 and 2 respectively. So let's construct our numbers with the last digit first (which can only be 3 or 7) and lets let's start with 7. Possible numbers are:
27, 37, 57, 77 of which only 37 is a prime.
So can we put another number in front of 37? No, we cannot use 3 or 7 again because we could construct 33 or 77 which is always be devisible by 11. We cannot use 2 because 27 is divisible by 3 and we cannot use 5 because 57 is also divisible by 3. So for numbers that end with 7, 37 is the only one. Now continue with 3:
23, 33, 53, 73
Those are all prime numbers except for 33. But since 2 and 5 can only occur in the beginning, we know that we cant extend 23 and 53 any further. So the only possible one would be 73 but we can use the exact same argument as for 37 (since the numbers are simply flipped) to show that we also cant extens this one further. And this leaves us with just 23, 37, 53 and 73
If you allow 1 to be prime, there are 20 numbers which work: 1, 2, 3, 5, 7, 11, 13, 17, 23, 31, 37, 53, 71, 73, 113, 131, 137, 173, 311, 317
well, this is only tractable primes for base 10, are there any bases that have longer, or even arbitrarily large tractable primes?
They talked about this in the video. Larger bases create larger trees, meaning larger primes. However, it is unknown whether or not they will extend to infinity at some point, though I would guess they don't as prime density goes down with higher numbers.
How far does the sequence go in each base. I didnt pause but for base ten, it looked like less than say 200 left truncateable primes. I guess in base two the sequence is
11=3
111=7
and that's all. Just these two since1111=15 which is not prime.
That begs the question what is the maximal length of the sequence of left-truncatable primes as a function of the base.
When you sum the digits you also get a prime. You get 113 (30th prime) lol.
You ve heard about digit 0, havent you? :)
I must have missed so many of numberphile's videos!
"It's quite a big number so I'm gonna write it out" Not that big then I guess
Reminds me of a geeky math game I came up with some years back based on truncatable primes where one person writes down a single digit number that is prime 2, 3, 5, 7 - then the other person writes a digit to the left or right to create another number that is still prime - and so on and so fourth - till someone can't continue or creates a composite non-prime number by mistake.
As the numbers get larger, the easiest way to enter and accurately check the primes automatically is on a PC with the "openssl" command line program (one of the functions of which lets you check prime numbers). Linux PCs normally have openssl installed by default, but you can also find pre-compiled command line versions for Android, Mac OS, and MS Windows if you look around.
To start play open a terminal and type "openssl prime" followed by a single digit prime:
openssl prime 3
At which point openssl reports that 3 (after converting to hex for checking) is indeed prime.
Then to continue play, the second player just hits the up-arrow key, which will make the terminal re-display the previous command for editing, then uses the left-right arrow keys to edit the 3 and add a number (for example 13, 43, 73, 31, or 37), for example.
openssl prime 43
You can restrict the game rules as you see fit, only allowing addition of a single digit, or allowing multiple digits, on the right, left, or anywhere including the middle. This game gets really hard as the numbers gets larger.
Why not use the 0 to make the chain longer ?
You'd be adding digits, but you wouldn't always get a new number when you delte a digit. A 30-digit number with 28 zeroes would only contain two different prime numbers. And it would take a bit more work to test, since you could keep adding zeroes forever in the hope of hitting a prime number.
Yep that would be the idea and this list might not be finit because of that :)
Patrick BORE but it's already been proven it's finite, the fact that there's infinite numbers doesn't matter
Trimon The fact that there are infinite numbers does matter because there is an infinite amount of numbers with n amount of 0s and 2 digits that are prime. All of these numbers are probably prime. Also the fact that it is proven to have finitely many of these numbers have been proven to have finitely many without any 0s.
I like the fact that the slide rule is just casually sitting next to the awesome prime action :D
Plot twist:
Graham's number+2 is a left truncatable prime 😯
Did you say +2 because Graham's number is an odd multiple of 3, so it can't be prime, and Graham's number +1 also can't be prime because it's even
All left-truncatable primes (in base 10) can be calculated in a fraction of a second with the following Mathematica/Wolfram language expression:
Flatten@NestList[Apply[{#1 + 1, Select[PrimeQ]@Flatten@Table[10^#1 Range@9 + x, {x, #2}]} &], {0, {0}}, 24][[2 ;;, 2]]
So, how do you know there's only finitely many end points? Brute force?
Joshua Hillerup it probably wasn't all done by hand but a program could fairly easily brute force it. Especially if you gave it a list of known prime numbers to compare each result to. Basically say that if the result is on the list continue, if not then try the next path.
You couldn't determine that by brute force because there are infinitely many primes. There must be a proof but I don't have any clue how you would do that.
adam poulter that's my assumption too, but I'm just wondering if that was done, or if there was a more elegant proof.
You wouldnt be able to prove a list is finite using brute force.
I'm pretty sure you can bruteforce that. Apply the same method he did with 7 => 47 => 947 => 3947. There aren't infinitely many paths.
The only numbers from which you can remove both from left to right or right to left in base 10 and receive a prime number: 2, 3, 5, 7, 23, 37, 53, 73, 373.
The numbers in hexadecimal: 2, 3, 5, 7, B, D, 25, 2B, 35, 3B, 3D, 53, B3, B5, D3, 2B3, 3B3, D3D.
This doesn't work at all in binary because the only single digit numbers are 1 and 0. (If we counted 1 as a prime, the highest we could go is 111, the binary representation of 7.)
I'm not sure what the significance of several of the three digit numbers being palindromes is. There may be a reason for this trend but I'm not sure yet.
(An) answer to Brady's arbitrarily deletable prime: 35E7 in base 12 (6043 in base 10)
I use the digit system 0123456789TE for base 12.
0 deletions: 35E7 (6043)
1 deletion: 35E (503), 357 (499), 3E7 (571), 5E7 (859)
2 deletions: 35 (41), 3E (47), 37 (43), 5E (71), 57 (67), E7 (139)
3 deletions: 3 (3), 5 (5), E (11), 7 (7)
Primes of length 4 were the longest I found, and of them this was in the smallest base. Number bases through 140 were checked.
There are some interesting trends to be observed, so I recommend anyone with some coding ability to explore this problem.
I have found a truly remarkable number that is both left and right truncatably prime, and this comment field is just large enough to contain it:
2.
Jeff Irwin 313
I assume this was your last comment.
I believe that PRIME numbers should be renamed to GRIME numbers because of this man.
So, when he says "and that's a prime number", did he know before hand or is his IQ above 420?
Jorge C. M. His IQ is prime.
And truncatable from any side as well as randomly.
so at most 73? That's pretty low....
At most? Are you crazy? How about 373?
373 is not truncatable randomly. If you truncate the 7, you get 33, which is not prime.
"I can remove the digit!" - said while covering it up with his own digit
6:21 Is that the Parker right-truncatable prime?
8:02 If we're willing to do a little bit of fudging and include the number "1," then 137 works ;)
137 can become 13 or 17 or 37
13 can become 1 or 3
17 can become 1 or 7
37 can become 3 or 7
Oops. 415673 does not fit your requirements as you could select an order of deletion that leave you with a 16, 6, or a 1. All not prime.
T Perm, "In ANY order you want". Not one predefined order.
Hold on everyone, we need to specify between deletable and any-digit-deletable. It satisfies one but not the other. They were pretty clear about that in the video.
Chad Eichhorn, Ok, reconcile that with "In ANY order you want" for me.
Your _allowed_ to choose any digit (not being restricted to one of the ends) to create your chain of primes. There's no requirement that all choices must work. The existence of one sequence of choices is enough.
I first had the same thought as you, but figured it out after a while. They could have phrased it a bit differently in the video.
The existence of a 6 in the number is enough to throw the other interpretation out.
If you are allowed to remove a digit "in ANY order you want", what numbers are able to be prime after the removal of each digit down to just 1 digit. This is what he meant, not that all digits work all the time. Those numbers are 27, 37, and 57 and that's it none that are three digits or more.
That's a good looking nail and gear on the desk in the background.