Are there infinitely many (you meant to say) palindromic prime numbers? There are about c*10^n/n primes of length n (where 1/c is log 10). There are about c*10^(n/2)/n palindromic primes of length n. So yes, but that's not a proof.
i would back you up, but you don't believe in the imaginary. If only 3, 5, 7, and 9 left, then your opponent would be back at square one. Try to be careful, though, I think 3, 5, and 7 are in their prime. Here's a tip, you might be able to create a division between 9 and 3.
Every now and then I find a new video on this channel that takes a simple concept and uses it to blow my mind in the most satisfying way. Today is one of those days.
Actually, a lot of the palindromic numbers do reduce to another three palindromes. Pretty much all the ones I randomly tried did, until I got to three-digit ones.
I never comment, but here is my first exception on UA-cam, with the hope that James or the other guy reads it. In Spanish we have a special word for PALINDROME NUMBERS. It is called CAPICÚA. We don't say "números palíndromos" ("palíndromo" being the word for, of course, your guessed, "palindrome"), but we are mathematically conscious. We say: ¡capicúa! I think it's a nice word that everyone should learn because it saves time. So: "todo número es la suma de tres capicúas." I hope it catches up.
I read somewhere that "capicúa" is originally Catalan. It literally means "head and tail" and was coined specifically for palindromic lottery numbers, which were believed to be lucky.
Did some mucking around in Excel using Gaussian's birthday, and found another solution: 30041777 = 29955992 + 85058 + 727 I would expect that the larger a number is, the more solutions it has.
I'm very glad you put links in for the original paper and also for the website (for when I lose patience with the paper). I'm going to have to try this at work if it gets slow...
By the way, I went to the website that does it for you and entered 1 000 000 006 (that is, a billion six... I did not put the spaces in, those are just for the reader's convenience). It actually came up with a nifty set of palindromes!
3:19 "You can do this in any base" That's what I was waiting to hear! Before 3:19, I was thinking "You do know that base 10 decimal form is not the One And Only True Form of a number, right?"
at 7:40 there is no need to assume you have a carry you definitely do as the column to the right already has a value greater than the target i.e there is already a 4 so to reach a 1 a sum of 11 must be reached hence there is definitely a carry
Perhaps the paper calculates the numbers in a way that makes it only necessary to carry over one. Maybe that's what happens every time? Then again, there could just be twos and they happened to not be in the examples.
That would work for this particular number but that may not always be the case. The algorithm James is using was designed to work for all numbers in any base, so it has to be made robust enough to deal with all cases.
palp1880 Algorithms are designed to require no thinking at all when you are running them. You are correct, but it is either faster or simpler (probably both) to not think about it and correct it at the end using the algorithm.
He's right, you do have a choice when it comes to the second digit of the second palindrome! He put 2, as per the algorithm, but I put 1 just to see, and it worked out to: 2210122 + 915519 + 15951. Then I tried it with 0, and I got 2220222 + 904409 + 16961. It doesn't work with 3, though. ...And then if you change the second digit of the first palindrome, you really open up the possibilities! 2159512 + 971179 + 10901 2163612 + 966669 + 11311 2143412 + 990099 + 18081 I'm sure you could get a dozen more possibilities as well. Once you have the first and last bit of each palindrome sorted, solving the rest really does have the same sort of feel as doing a sudoku. It's pretty fun, I high recommend it as a way to procrastinate from writing your master's thesis while still feeling smart.
7:33 You already know that column 4 is going to generate a carry because the 4 from the third number is greater than the target sum of 1. No need to "assume" a carry.
Sooo... if this works in every base, does this mean you could encrypt a binary file into 3 palyndromes that can be stored separately, and that you would need all three parts to reconstitute the message? Also, because they are palyndromes, you can also compress the information by a flat 50% per number of palyndromes (multiplicatively) before standard compression, since you only need to store half of it. In addition to all this the message can be reconstructed by two or three simple additions! This is pretty significant cutting edge maths! I'm gonna try to do this in C# with BigInteger type or a bitmap...
@@sumdumbmick Although it won't make the file smaller, it might make it more secure as you have to have all three palindromes before being able to tell what the original file is.
For binary it actually takes 4 palindromes. That means it's no better than the trivial algorithm of writing it as an XOR of two things (i.e. a "one time pad"), which is already optimally secure.
i can think of an easier way than the methods in the video: first you identify the type of the number and fill in the first and last digits, then you subtract the 3 numbers you get from this step. For example, for 3141592 in the video, you subtract it by 2000002, 900009 and 10001 and get 231580. you then delete the zero from the end and get 23158. this number will follow the same property as it can be put into 3 palindromes, which are the middle part of the answer for 3141592. then you can identify the type of number this is and carry on.
Aaron Cruz It isn't true for all bases. Dr. Grimes misspoke. The authors give 10110000 as an example of a binary number that requires 4 palindromes. The problem is that, with the exception of 0, all binary palindromes are odd. An even binary number is the sum of three palindromes only if it is the sum of two palindromes.
7:30 - in this case you know the carry is a 1 without assuming it, because you've got a 4 in the column that needs to end up with a 1, and you can't get a carry of 2 from that row, so it has to be 4 + 0 + (7 or 6 with a carry) = 11, carry the one.
At 7:40, you don't need to assume. You can see that the next column already has 4 and 0 and it should add up to 1 so obviously it would leave a remainder of 1.
Psykopate Both statements are true. Why attack others and call them names? Take a break from math for a while and consider the most important question for you to ask, “What is the payoff to me of putting others down? Comfort? Safety? Stability? Importance? Getting laid?...you get the idea.”
Why don't you take a break from internet, take a deep breath and re-read my comment with some distance and not literally ? PS: i like how you subtly attack me and call me names while calling me on attacking others and calling them names
I see lots of comments here wondering about how this holds for small numbers, but to me, the surprising thing is that it works for *large* numbers. Palindromes get sparser as numbers get larger, so I would've expected more palindromes to be required to make larger numbers.
If you consider 1000 digit palindromes, there are about 10^500 of them, and roughly 10^1500 triplets of such palindromes. These triplets far outnumber the roughly 10^1000 numbers that they might add up to.
So does that mean then that every palindrome number can be written as the sum of 3 palindrome numbers? 22+44+55 = 121 is all well and good, but of course all these can then be written as palindromes 9+7+6 = 22, 1+2+3 = 6, 1+0+0 = 1. From this we can determine that 0 is a palindrome, but if that's the case, suddenly it become quite easy to find these triplets for any palindrome number, simply by adding 0 twice to the original number.
If you require the two additional numbers to have more than one digit unless the beginning number is less than 19, then you could probably calculate the reduction of many numbers and graph it out into a tree to see if it forms any significant shape.
Assuming that zero is a palindrome, then adding zero twice only works if the other number is a palindrome. The premise was that every positive integer can be written this way, in which case zero just doesn't work, and other single-digit "palindromes" are only useful up to maybe two- or three-digit numbers.
try it in my base-14,414,400 number system. it's a nested sequence of base-52, base-77, base-60, base-60, where base-52 is itself composed of 4 base-13 cycles which are themselves made of 2 base-6 cycles (add one and you get the 13), the base-77 is composed of 7 base-11 cycles composed of 2 base-5 cycles (add one and you get 11), and the base-60 is composed of 3 base-20 cycles composed of 4 base-5 cycles (this level is identical to ancient Mayan numerals). this results in the system only requiring 11 symbols, despite being base-14,414,400. one great cycle is built up as base-60,base-60,base-77.base52: a comma follows a base-60 tier a period follows a base-77 tier a colon follows a base-52 tier in base-52 you count as follows: _: = 0 `: = 1 >: = 2 3: = 3 g: = 4 /: = 5 (: = 6 `(: = 7 >(: = 8 3(: = 9 g(: = 10 /(: = 11 ((: = 12 _`: = 13 _>: = 26 _3: = 39 ((3: = 51 52 itself would be a one in the base-77 tier: `._: = 52 or for brevity just `. in base-77 you count as follows: `. = 1x52 >. = 2x52 3. = 3x52 g. = 4x52 /. = 5x52 `/. = 6x52 >/. = 7x52 3/. = 8x52 g/. = 9x52 X. = 10x52 _`. = 11x52 _>. = 22x52 _3. = 33x52 _g. = 44x52 _/. = 55x52 _(. = 66x52 X(. = 76x52 77x52 is then a 1 in the first base-60 tier: `,_._: = `, = 77x52 in base-60 you count as follows: `, = 1xn (n can be either 77x52 or 60x77x52) >, = 2xn 3, = 3xn g, = 4xn /, = 5xn `/, = 6xn >/, = 7xn 3/, = 8xn g/, = 9xn X, = 10xn `X, = 11xn >X, = 12xn 3X, = 13xn gX, = 14xn N, = 15xn `N, = 16xn >N, = 17xn 3N, = 18xn gN, = 19xn _`, = 20xn _>, = 40xn gN>, = 59xn so 14,414,399 is rendered as follows: gN>,gN>,X(.((3: add 1 to this and you get 14,414,400: `:_,_,_._: or for short this can be written as `:: I didn't show a use of the 11th symbol, but it would simply serve to indicate the 'ones' position, acting basically like a decimal point. and using this you can then freely switch around which tier you're using at the level of precision that you're interested in. like, say for instance you're interested in powers of 11, well in that case it would be nicest to set the base-77 tier as your lowest level instead of base-52.
I can do better than that. Every positive integer is a sum of ONE palindrome. In base 1-higher-than-itself. Because it's a 1-digit number in that base, you see, so it's a palindrome.
I know this video is three years old, but I've been hard at work in the last three years -- and have finally found success. I have developed an algorithm that, when given the target sum and two of the three palendrome numbers, will compute the third palendrome number to complete the set. Amazingly, it does this without any knowledge of the algorithms used to find the first two numbers! I would share the algorithm here, but this comments section is to small to contain it.
@7:35 You don't need to assume the carry 1. because in the column right next to it you need to add something to a 4 to get 1. Obviously that's not possible without a carry over. So yeah....no guesswork or assumption needed. We HAVE to carry over 1 to make the next culumn work :)
I found a glitch on the site: type the number (i.e.) 11021 and the site gives 4 numbers to add together, although one of them is zero ("00" actually) and can obviously be assumed to be unnecessary.
That's certainly an interesting glitch. It seems to be unique to that one site though, because I tried 11021 on a different site, and it didn't generate "00".
i wrote a program that just goes through all the combinations of 3 palindrome numbers. the palindrome numbers are just generated up to the number you want, and then checking if they're the same backwards. They have a lot of possible combinations. 3141592 has 3112 (which is funnily also a palindrome) 3 number palindrome sums
Irrelevant but I’ll prove why my birthday is the best. 16/04/02 Every number is the square root of the one before it (other than 16 since there are no numbers before it)
Asking the site for 10241024 resulted in the display being incorrect, but the answer being right. The first number (9900099) was offset by 1 to the left placing the ones digit 9 into the tens digit spot, but when summed the numbers did come out right.
Every number larger than some MINIMUM number is the sum of three palindromes! The smallest number that is the sum of three palindromic numbers in base ten is 33, 11 + 11 + 11. I'm not sure what the maximum number is in base ten that is not the sum of three palindromes, but a quick computer program I wrote suggested that it is 212. Clearly, in this video, they're working with numbers which are much larger than 212. But there is a lower bound.
Your definition of 'palindrome' might be stricter than it needs to be. 1 is a palindrome, so even if you don't allow 0, 3 = 1+1+1 is the sum of 3 palindromes.
It's a coincidence, it doesn't happen for other numbers. Maybe it always happens if the beginning number is a palindrome as well? I wonder how you would prove that...
At 6:19 you're assuming a carry of 0 in the second column contrary to later instructions. Is this something else in the algorithm not fully explained in the video or did you knowingly cheat the algorithm so you didn't have to do the adjustment at the end?
*_...so what's the relation of palindrome-triplets and prime-triplets...how does the count-of alternative-palindrome-triplets compare to the count-of alternative-prime-triplets, for each number..._*
Some guy: Man I with their was a way I could figure out the three palindromes that summed produce this number. Some other guy: there's an app for that... Also why?
6:40 how did you know that you had to put a 2 there and not carry, ehen you counld have put a 1 in second place of the first number and do carry a 1 from the column right to it?
13311331 + 4404044 + 55055 would make a "proper" version of 17770430 - the digits of the ISO 8601 format. (Or 11022011 + 6360636 + 387783 if you use the algorithm)
If you like palindromes maybe you'll appreciate my discovery (at least I think it's mine), monodromes. A palindrome is when successively inner pairs of digits subtract to 0, or the absolute value of their differences equal 0. In 937739 9-9=0, 3-3=0, 7-7=0. If the number of digits is odd so there's a single digit, say 5, in the middle, then we subtract it from itself, 5-5=0. A monodrome is when such pairs subtract to 1, so 93612548 is a monodrome, 9-8=1, 4-3=1, etc. Monodromes can only have an even number of digits. Of course you can also have bidromes, and generally ndromes. I wonder if any number is the sum of a certain number of monodromes?
In that third column you could've predicted the carry of at least 1 as your 4 will become a 1 unless you add at least 7 with increments of 10 afterwards. You'd see you only need a 1 there so the rest solves itself again.
At 6:20, does the algorithm specify a 2 (or just specify that there's no carry in that column, or something)? Or is it similar to later on when he says "You have a choice here and if it's wrong you'll correct it later?"
How about 101+131+151=383? That’s 3 different Palindromic Prime Numbers added up to give you another Palindromic Prime Number.
and I here thinking 42 was the most special number...
Nice
Are there infinite palindromic prime numbers
Are there infinitely many (you meant to say) palindromic prime numbers? There are about c*10^n/n primes of length n (where 1/c is log 10). There are about c*10^(n/2)/n palindromic primes of length n. So yes, but that's not a proof.
OH NOOOO
Mad Maths Beyond Palindrome
Three numbers enter. One number leaves.
James Jumper this is one of the best puns I seen in a long while
I actually don't get it :(
Mad Max Beyond Thunderdome the film
Trust me Aleph. One of the memorable lines in the movie is about Thunderdome itself. “Two men enter. One man leaves.”
Ah, never heard of that movie but it seems many people have.
I got into a fight with 1, 3, 5, 7 and 9.
The odds were against me.
So funny I forgot to laugh
Even against those odds ... You should be OK.
If it was with pi, e, and square root of 2, then that'd be an irrational fight.
i would back you up, but you don't believe in the imaginary.
If only 3, 5, 7, and 9 left, then your opponent would be back at square one.
Try to be careful, though, I think 3, 5, and 7 are in their prime.
Here's a tip, you might be able to create a division between 9 and 3.
I think this is getting pretty complex.
Every now and then I find a new video on this channel that takes a simple concept and uses it to blow my mind in the most satisfying way. Today is one of those days.
Could be a module in "keep talking and nobody explodes"
Increases the Manual by 40 pages, just for one module. lol
Yesss
One might have to rename that: Call your loved ones, because you're going to explode.
I played that game yesterday, that'd be insane hahaha
Maybe just have a few different types of numbers so it's not as extensive
So, if every number is sum of 3 palindromes, every palindrome is sum of 3 palindromes too...
Palindrome + 0 + 0 = Palindrome.
Yes, I know it's disappointing.
I believe every single-digit number counts as a palindrome.
So it would just be 0+0+number
It is less cool than it sounds. n+0+0
Actually, a lot of the palindromic numbers do reduce to another three palindromes. Pretty much all the ones I randomly tried did, until I got to three-digit ones.
Aside from the trivial ones in the other comments, if the first digit is more than 2 you can split each digit eg 5639365 = 1111111 + 2222222 + 2306032
Any quality math/logic content:
* exists *
Brilliant:
*IT'S FREE REAL ESTATE*
So, kinda like... Gödels incompletenes Theorem?
I KNOW RIGHT? I THOUGHT I WAS THE ONLY ONE GETTING ANNOYED
I am annoyed by the sponsorship.
Marek Šťastný
*any quality -math/logic- content
*Super large numbers exist*
Yeah i don't think so
Love the Gaussian blur at 1:10... wonder how many others caught that reference!
I don't think I get it
0:00 "I've got a cute little mathematical fact for you" - instant like :D
I never comment, but here is my first exception on UA-cam, with the hope that James or the other guy reads it. In Spanish we have a special word for PALINDROME NUMBERS. It is called CAPICÚA. We don't say "números palíndromos" ("palíndromo" being the word for, of course, your guessed, "palindrome"), but we are mathematically conscious. We say: ¡capicúa! I think it's a nice word that everyone should learn because it saves time. So: "todo número es la suma de tres capicúas." I hope it catches up.
I read somewhere that "capicúa" is originally Catalan. It literally means "head and tail" and was coined specifically for palindromic lottery numbers, which were believed to be lucky.
@@chrisg3030 Yes cap i cua
Was ready to ask "Does this work outside of base 10?" and 3:17 comes around and smacks me. Great work to whoever discovered this.
Maybe should watch the entire video before asking questions that you think will make you sound smarter then those in the video.
@@bvbinsane1vanity projecting much?
I am impressed again and again in how people come op with that kind of theorem
They actually get paid for it!
High
Must've been really bored
@@jaakkonyh gold
They are nerds. I’m man enough to admit that I’m jealous of their superior intellect, though.
Every night I wake in a cold sweat assaulted by my thoughts asking me “how, how did somebody discover this”
And how many nights have you known this? The video is 1 day old.
@@BaddeJimme - the paper was published more than a day before this video
A new episode with James... what a treat!
As interesting as this is... I'm curious as to how many unique solutions of 3 palindromes you can have for certain numbers.
Did some mucking around in Excel using Gaussian's birthday, and found another solution: 30041777 = 29955992 + 85058 + 727
I would expect that the larger a number is, the more solutions it has.
I'm very glad you put links in for the original paper and also for the website (for when I lose patience with the paper). I'm going to have to try this at work if it gets slow...
By the way, I went to the website that does it for you and entered 1 000 000 006 (that is, a billion six... I did not put the spaces in, those are just for the reader's convenience). It actually came up with a nifty set of palindromes!
I liked how the carry line in the Pi example was also palindromic, but was a bit sad to see that this isn't always the case.
"It's like a Sir Dooku!"
Excuse me, it's _count_ Dooku.
Very underrated
Sudoku, not Sir Dooku... Don't laugh at Dr Grime's accent :/
@@arnavj.3927 Agreed
“...But your gonna have to memorize 40 pages of algorithms...”
Matt Parker can be heard distantly screaming
Meanwhile Speedcubers: So I did that
@@fisch37
57 oll and 21 pll is just the start
3:19 "You can do this in any base" That's what I was waiting to hear! Before 3:19, I was thinking "You do know that base 10 decimal form is not the One And Only True Form of a number, right?"
Mathematical proof of Brady's conversion
The site by Chritian Lawson is just .. ...AMAZING
at 7:40 there is no need to assume you have a carry you definitely do as the column to the right already has a value greater than the target i.e there is already a 4 so to reach a 1 a sum of 11 must be reached hence there is definitely a carry
He's following the algorithm exactly. Presumably the algorithm is simpler if you leave checks like that till the end.
Also there can be carries of 2 when adding 3 numbers...
Perhaps the paper calculates the numbers in a way that makes it only necessary to carry over one. Maybe that's what happens every time? Then again, there could just be twos and they happened to not be in the examples.
That would work for this particular number but that may not always be the case. The algorithm James is using was designed to work for all numbers in any base, so it has to be made robust enough to deal with all cases.
palp1880 Algorithms are designed to require no thinking at all when you are running them. You are correct, but it is either faster or simpler (probably both) to not think about it and correct it at the end using the algorithm.
I just studied maths for an hour and thought I needed to relax a little, and what did I do? Started to watch a Numberphile video.
He's right, you do have a choice when it comes to the second digit of the second palindrome! He put 2, as per the algorithm, but I put 1 just to see, and it worked out to: 2210122 + 915519 + 15951. Then I tried it with 0, and I got 2220222 + 904409 + 16961. It doesn't work with 3, though.
...And then if you change the second digit of the first palindrome, you really open up the possibilities!
2159512 + 971179 + 10901
2163612 + 966669 + 11311
2143412 + 990099 + 18081
I'm sure you could get a dozen more possibilities as well.
Once you have the first and last bit of each palindrome sorted, solving the rest really does have the same sort of feel as doing a sudoku. It's pretty fun, I high recommend it as a way to procrastinate from writing your master's thesis while still feeling smart.
7:40
"ashume"
???
Smerg the Dargon
You’re right, that is the correct pronunciation of “assume”
Pretty sure that's perfectly correct for Brits. Plenty of words are correctly pronounced differently in different countries.
It's a regional dialect. It's not very common in the UK.
The completely crazy (and potentially coincidental) thing is that the carry row contains a palindrome number as well for the first digits of π: 101101
I guess if you sum palindrome numbers and the result is palindrome you have this behaviour
7:33 You already know that column 4 is going to generate a carry because the 4 from the third number is greater than the target sum of 1. No need to "assume" a carry.
Sooo... if this works in every base, does this mean you could encrypt a binary file into 3 palyndromes that can be stored separately, and that you would need all three parts to reconstitute the message? Also, because they are palyndromes, you can also compress the information by a flat 50% per number of palyndromes (multiplicatively) before standard compression, since you only need to store half of it. In addition to all this the message can be reconstructed by two or three simple additions! This is pretty significant cutting edge maths!
I'm gonna try to do this in C# with BigInteger type or a bitmap...
the palindromes are all close in size to the original number, so this compression will actually take more room than the original number.
@@sumdumbmick Although it won't make the file smaller, it might make it more secure as you have to have all three palindromes before being able to tell what the original file is.
For binary it actually takes 4 palindromes. That means it's no better than the trivial algorithm of writing it as an XOR of two things (i.e. a "one time pad"), which is already optimally secure.
i can think of an easier way than the methods in the video: first you identify the type of the number and fill in the first and last digits, then you subtract the 3 numbers you get from this step. For example, for 3141592 in the video, you subtract it by 2000002, 900009 and 10001 and get 231580. you then delete the zero from the end and get 23158. this number will follow the same property as it can be put into 3 palindromes, which are the middle part of the answer for 3141592. then you can identify the type of number this is and carry on.
Even with the algorithm, it's like solving Minesweeper
5:52 "It's like you're doing a sudoku"
lol
I was flabbergasted when I learnt that its true for all bases...😯😯
Why did it upset you? It's amazing.
Aaron Cruz It isn't true for all bases. Dr. Grimes misspoke. The authors give 10110000 as an example of a binary number that requires 4 palindromes. The problem is that, with the exception of 0, all binary palindromes are odd. An even binary number is the sum of three palindromes only if it is the sum of two palindromes.
No, he said it was proven to work for all bases 5 or greater, and that someone else worked on the base 2, 3, 4 numbers.
Did you not read the first Abstract he showed?
It is only valid for bases >= 5
I am not a native English speaker so I had to look up Flabbergast. I thought it meant "loud farting" or something.
7:30 - in this case you know the carry is a 1 without assuming it, because you've got a 4 in the column that needs to end up with a 1, and you can't get a carry of 2 from that row, so it has to be 4 + 0 + (7 or 6 with a carry) = 11, carry the one.
S U D O O K O O
Count Dooku
@@schizophrenicenthusiast hello there
my old friend
it *is* a bit like a sudookoo, yeah!
also, *ASHOOM.* the paper tells you to ashoom.
At 7:40, you don't need to assume. You can see that the next column already has 4 and 0 and it should add up to 1 so obviously it would leave a remainder of 1.
I am having an existential crisis right now because of VSauce and Numberphile.
For some reason, this felt really natural fir me. Cheerio, James!
I see james grime,
*I CLICC*
Gotta hand it to them, I'm super impressed! Top demonstration too, as always on this channel.
It's also easily demonstrable that many numbers have more than one solution. 9 for example can be 9+0+0 or 8+1+0 or 7+1+1 or ...
Yes, I went on that website and easily found alternative solutions myself for 5-digit numbers even.
Mind blown. Glad I wasn't the one who had to write the code that captured that algorithm. Lot of work.
For me the most remarkable piece of information from this video is that there exists someone whose name is Christian Lawson-Perfect
for me its your username.
James Jumper took my word... this was MAD! Felt like doing soduko or solving the Rubik's cube...
Random uncivilized animal: "This is useless"
Us: "Shut up this is art"
Psykopate Both statements are true. Why attack others and call them names?
Take a break from math for a while and consider the most important question for you to ask, “What is the payoff to me of putting others down? Comfort? Safety? Stability? Importance? Getting laid?...you get the idea.”
Why don't you take a break from internet, take a deep breath and re-read my comment with some distance and not literally ?
PS: i like how you subtly attack me and call me names while calling me on attacking others and calling them names
Psykopate Sure.
false.
Gaussian blur on Gauss' portrait--subtle, but quite neat at 1:06
I see lots of comments here wondering about how this holds for small numbers, but to me, the surprising thing is that it works for *large* numbers. Palindromes get sparser as numbers get larger, so I would've expected more palindromes to be required to make larger numbers.
exactly my thoughts
If you consider 1000 digit palindromes, there are about 10^500 of them, and roughly 10^1500 triplets of such palindromes. These triplets far outnumber the roughly 10^1000 numbers that they might add up to.
Yeah, that makes sense.
Proving the folklore true is worthy of pixelated sunglasses (for sure).
Anybody else’s spine hurt when he drags the sharpie across the rough paper?
because, like you, they are weirdoes
Me starting to write: “ does this work in other bases?” at 3:07.
James at 3:17: “...you can do this in any base...”
I’ll just shut up and listen.
So does that mean then that every palindrome number can be written as the sum of 3 palindrome numbers? 22+44+55 = 121 is all well and good, but of course all these can then be written as palindromes 9+7+6 = 22, 1+2+3 = 6, 1+0+0 = 1. From this we can determine that 0 is a palindrome, but if that's the case, suddenly it become quite easy to find these triplets for any palindrome number, simply by adding 0 twice to the original number.
If you require the two additional numbers to have more than one digit unless the beginning number is less than 19, then you could probably calculate the reduction of many numbers and graph it out into a tree to see if it forms any significant shape.
entering any number of 3's into the algorithm seems to agree with you
Great breakdown, you show that this theorem is not significant math.
Assuming you are replying to me, I don't know that you're right, but I do think clarification is needed.
Assuming that zero is a palindrome, then adding zero twice only works if the other number is a palindrome. The premise was that every positive integer can be written this way, in which case zero just doesn't work, and other single-digit "palindromes" are only useful up to maybe two- or three-digit numbers.
try it in my base-14,414,400 number system.
it's a nested sequence of base-52, base-77, base-60, base-60, where base-52 is itself composed of 4 base-13 cycles which are themselves made of 2 base-6 cycles (add one and you get the 13), the base-77 is composed of 7 base-11 cycles composed of 2 base-5 cycles (add one and you get 11), and the base-60 is composed of 3 base-20 cycles composed of 4 base-5 cycles (this level is identical to ancient Mayan numerals). this results in the system only requiring 11 symbols, despite being base-14,414,400.
one great cycle is built up as base-60,base-60,base-77.base52:
a comma follows a base-60 tier
a period follows a base-77 tier
a colon follows a base-52 tier
in base-52 you count as follows:
_: = 0
`: = 1
>: = 2
3: = 3
g: = 4
/: = 5
(: = 6
`(: = 7
>(: = 8
3(: = 9
g(: = 10
/(: = 11
((: = 12
_`: = 13
_>: = 26
_3: = 39
((3: = 51
52 itself would be a one in the base-77 tier:
`._: = 52
or for brevity just `.
in base-77 you count as follows:
`. = 1x52
>. = 2x52
3. = 3x52
g. = 4x52
/. = 5x52
`/. = 6x52
>/. = 7x52
3/. = 8x52
g/. = 9x52
X. = 10x52
_`. = 11x52
_>. = 22x52
_3. = 33x52
_g. = 44x52
_/. = 55x52
_(. = 66x52
X(. = 76x52
77x52 is then a 1 in the first base-60 tier:
`,_._: = `, = 77x52
in base-60 you count as follows:
`, = 1xn (n can be either 77x52 or 60x77x52)
>, = 2xn
3, = 3xn
g, = 4xn
/, = 5xn
`/, = 6xn
>/, = 7xn
3/, = 8xn
g/, = 9xn
X, = 10xn
`X, = 11xn
>X, = 12xn
3X, = 13xn
gX, = 14xn
N, = 15xn
`N, = 16xn
>N, = 17xn
3N, = 18xn
gN, = 19xn
_`, = 20xn
_>, = 40xn
gN>, = 59xn
so 14,414,399 is rendered as follows:
gN>,gN>,X(.((3:
add 1 to this and you get 14,414,400:
`:_,_,_._:
or for short this can be written as `::
I didn't show a use of the 11th symbol, but it would simply serve to indicate the 'ones' position, acting basically like a decimal point. and using this you can then freely switch around which tier you're using at the level of precision that you're interested in. like, say for instance you're interested in powers of 11, well in that case it would be nicest to set the base-77 tier as your lowest level instead of base-52.
I can do better than that. Every positive integer is a sum of ONE palindrome. In base 1-higher-than-itself. Because it's a 1-digit number in that base, you see, so it's a palindrome.
Every positive integer n > 2 is a palindrome in base (n-1): 11 base (n-1) = 1(n-1) + 1
Congratulations youv'e just discovered how to count
Your profile picture is evil
I was today years old when I realized that tally marks are base 1
There is no need to assume at 7:43. If we see in the column next to it...it's clear that to get a 1 from 4 and 0 we have to have a carry.
I want to see a 3 Parker palindrome.
300=191+99+010
It's not technically right because 10 isn't a palindrome, but it works because of the leading zero.
excellent Gaussian blur gag 1:07
Was it a car or a cat I saw?
+Sebastian Elytron.
_Was it a car or a rat I saw?_
A rat. Tara!
Ask Schrodinger.
It was a racecar
A man, a plan, a canal. Panama!
Notable gel baton?
I know this video is three years old, but I've been hard at work in the last three years -- and have finally found success. I have developed an algorithm that, when given the target sum and two of the three palendrome numbers, will compute the third palendrome number to complete the set. Amazingly, it does this without any knowledge of the algorithms used to find the first two numbers!
I would share the algorithm here, but this comments section is to small to contain it.
cool, now I have something to do in my spare time, program this in python XD
That's gonna be a loooong script judging by that paper they showed
excellent, I'm going to have fun for days XD
Please share us your program when you complete it
All the best
Adrés Firte
sure (o.o)b
@7:35 You don't need to assume the carry 1. because in the column right next to it you need to add something to a 4 to get 1. Obviously that's not possible without a carry over.
So yeah....no guesswork or assumption needed. We HAVE to carry over 1 to make the next culumn work :)
Pure mathematics: fun with no foreseeable application
Number theory is used in cryptography
@@neonKow but that's such a small part of everything mathematicians do
It's not the solution, but the path to it, that furthers understanding.
The same used to be said about fractals.
@Hootkins Fractals are used in computer animation.
I think the only application for palindromic numbers is wizardry.
; D
Great stuff hear! Thanks for the video, I could watch this all day!
Nice video!
I found a glitch on the site: type the number (i.e.) 11021 and the site gives 4 numbers to add together, although one of them is zero ("00" actually) and can obviously be assumed to be unnecessary.
That's certainly an interesting glitch. It seems to be unique to that one site though, because I tried 11021 on a different site, and it didn't generate "00".
Numberphile is meant to be about numbers :) Videos like these are refreshing. No magic tricks, no physics. Just number theory.
i wrote a program that just goes through all the combinations of 3 palindrome numbers. the palindrome numbers are just generated up to the number you want, and then checking if they're the same backwards. They have a lot of possible combinations. 3141592 has 3112 (which is funnily also a palindrome) 3 number palindrome sums
3112 is not a palindrome
@@thewhitefalcon8539 that's how many combinations there are ...
Irrelevant but I’ll prove why my birthday is the best.
16/04/02
Every number is the square root of the one before it (other than 16 since there are no numbers before it)
Unless you were born on the 256th minute of that day.
Four days late, but congrats on your birthday.
I've stopped watching Numberphile videos, but when it's James Grime it's an insta-click for me
ASHume
Gesundheit.
Would love to see a video where Cliff Stoll talks about sphericons!
Fark I love this stuff but got lost at 0:02
I love it when the video really is just about numbers. Classic style!
Amazing
Asking the site for 10241024 resulted in the display being incorrect, but the answer being right. The first number (9900099) was offset by 1 to the left placing the ones digit 9 into the tens digit spot, but when summed the numbers did come out right.
Yeah. It seems confused by 10xxxxxxxx seems to confuse it if the x's are smallish.
Are palindrome representations unique?
No. 2=2+0+0=1+1+0 for example.
T L Ak, I should have thought of that. Shame really. If they were unique representations palindromic numbers would suddenly be really mysterious.
Also, palindromes like 999 have lots of different solutions.
333+333+333
616+232+151
787+111+101
Nobody published a proof for positive integers in base 1. Here it is: every number is a palindrome when written in base 1.
8:36
Anyone else notice that the carry is also a palindrome! :D
The videast Michael Launay, from the channel Micmaths (French), gave me the will to start watching your cool videos again :)
1. IS THE LONLIEST NUMBER THAT YOU'LL EVER DO.
Two can be as bad as one.
It's the loneliest number since the number one.
Every number larger than some MINIMUM number is the sum of three palindromes! The smallest number that is the sum of three palindromic numbers in base ten is 33, 11 + 11 + 11. I'm not sure what the maximum number is in base ten that is not the sum of three palindromes, but a quick computer program I wrote suggested that it is 212. Clearly, in this video, they're working with numbers which are much larger than 212. But there is a lower bound.
1= 1+0+0
Your definition of 'palindrome' might be stricter than it needs to be. 1 is a palindrome, so even if you don't allow 0, 3 = 1+1+1 is the sum of 3 palindromes.
Lagerregal
Regallager
Reggerregger
Is it crazy how saying sentences backwards creates backwards sentences saying how crazy it is?
Dude put your art up. It’s been there for like a year.
Love the vintage microphone at the end.
The carry numbers also formed a palindrome... coincidence?
Or something more sinister?
the universe is a -hologram- *palindrome*
I would assume so, otherwise this information would likely be used in finding the palindromes
I THINK NOT
It's a coincidence, it doesn't happen for other numbers. Maybe it always happens if the beginning number is a palindrome as well? I wonder how you would prove that...
At 6:19 you're assuming a carry of 0 in the second column contrary to later instructions. Is this something else in the algorithm not fully explained in the video or did you knowingly cheat the algorithm so you didn't have to do the adjustment at the end?
They found the algorithm and proved that it's always possible, I'm more interested in WHY you can do that.
What about small numbers? If we allow zero it seems silly because x = x + 0 + 0
In base 1, every sum is entirely made up of palindromes.
*_...so what's the relation of palindrome-triplets and prime-triplets...how does the count-of alternative-palindrome-triplets compare to the count-of alternative-prime-triplets, for each number..._*
Some guy: Man I with their was a way I could figure out the three palindromes that summed produce this number.
Some other guy: there's an app for that... Also why?
I just love how excited this guy is about math
That is the most useless thing I learned all week.
6:40 how did you know that you had to put a 2 there and not carry, ehen you counld have put a 1 in second place of the first number and do carry a 1 from the column right to it?
Do better with the headlines! "Every Number" is not the same as "Every Positive Integer"!
Come one thats obvious. How should this be possible for non periodic irrational numbers.... think b4 flame.
Headlines are meant to be catchy...not overly wordy.
It was clearly explained once you watched the video so...get over it.
Mathematicians often just say "number" when it's clear from context that they are talking about natural numbers.
One of the more amazing videos.
And if you write that date *properly* the three palindrome numbers are 4001004 , 281182 , and 19591.
Well, not properly, it's just Americans write their dates backwards.
13311331 + 4404044 + 55055 would make a "proper" version of 17770430 - the digits of the ISO 8601 format. (Or 11022011 + 6360636 + 387783 if you use the algorithm)
We need a sequence on oeis. a(n) = # of distinct palindromic triples which sum to n (in whatever base).
If you like palindromes maybe you'll appreciate my discovery (at least I think it's mine), monodromes. A palindrome is when successively inner pairs of digits subtract to 0, or the absolute value of their differences equal 0. In 937739 9-9=0, 3-3=0, 7-7=0. If the number of digits is odd so there's a single digit, say 5, in the middle, then we subtract it from itself, 5-5=0. A monodrome is when such pairs subtract to 1, so 93612548 is a monodrome, 9-8=1, 4-3=1, etc. Monodromes can only have an even number of digits. Of course you can also have bidromes, and generally ndromes. I wonder if any number is the sum of a certain number of monodromes?
In that third column you could've predicted the carry of at least 1 as your 4 will become a 1 unless you add at least 7 with increments of 10 afterwards. You'd see you only need a 1 there so the rest solves itself again.
At 6:20, does the algorithm specify a 2 (or just specify that there's no carry in that column, or something)? Or is it similar to later on when he says "You have a choice here and if it's wrong you'll correct it later?"