What's special about 277777788888899? - Numberphile 

Can't you just ask what numbers give you 277777788888899 when multiplied, and then that number would have a persistense of 12 by default.

@@mrJety89 Or really you could look to multiply out to any permutation of the digits in 277777788888899. Maybe it's easier to end up with an even number like 997777778888882.

can you make a downloadable program for those codes?

interesting seeing python in a Numberphile video, as I just started learning python a couple days ago

@@ErixTheRed Yeah, I was thinking abou that too. Something tells me that they've already tried both of those ideas.

preach

He is obviously joking

Nope

This is actually super accurate. U stop using numbers at some point when u do math, so u lose basic skills

@@mmeister8582 What do you mean you stop using numbers?

ahh uhm i personally like 12

@@qwertyuiop-cy5en i could go for an exponential today.

@@theterribleanimator1793 It's been the number 91 again for the last few weeks.

32 is a number I find inherently attractive for no apparent reason (other than being a power of 2). I don’t find 16 as attractive but I do love it being the only number that can be described as x^y and y^x where x and y are differing positive integers (2 and 4, for the record). Overall I just love powers of 2. Numbers like 12 and 24 are cute and all, but that factor of 3 feels like an imperfection.

3, take it or leave it

Definitely, I barely coded ever and haven't coded since the 80s, but even I spotted that that was the problem.

I knew he would simply remove 1 from the total number, because the code gave him +1 because of the entra last step.

It's a Parker Program: almost right.

What got me is print statements in the first place. Python2 is strong with him.

With a little bit of extra interfacing it could make logical sense. Like the final number gives an additional string "Final entry: X". But yeah, he said he "could just" remove it, he never said he would.

Never trust a perfect person

Except Matt isn't perfect, he makes tons of mistakes all the time, he's just secure that his mistakes don't reflect on his actual abilities. Otherwise we wouldn't have the brilliant "Parker X" banter that makes his channel and Numberphile's channel so entertaining.

@@efulmer8675 I think that was the point. A person who claims to be perfect is likely to be fake, and you can’t trust someone like that. A person who doesn’t mind getting caught making mistakes, because he never pretended to be perfect in the first place, is more likely to be someone you can trust.

You can only use 1s

1 11 111 1111 11111 Now play the game backwards..

Both 1 and 0 are the BOOM 🤣🤣🤣

nowonmetube It‘s always 1 step whatever the number is 111111111111111-> 1*1*...*1=1

@@samodelkini yea i think they knew that thanks for clarifying tho

Stuka don't forget to double check it!

It was me in engineering school... I was double checking simple multiplications because I was more stressed to get it wrong because of them than with the reasoning of the math!

He's a mathematician, he does it in a sequence, it doesn't matter how simple is an action. It's the order inside your mind. Because the moment he says "nah, I don't have to do it, it's obvious" he starts to make mistakes. From small to bigger and for math these are all the same. He's learned the hard way to be patient and correct. So it's now a reflex. =)

I had a weird habit of refusing to accept a calculator when I took math tests in junior high school in the late 90s. My math teacher was always like "shouldn't you have a calculator, though?" and I said NO, and insisted on calculating everything by hand. Not sure why, but maybe I wanted to prove to myself that I could do it without a calculator or something.

Well he wanted to multiply all the digits together, so he started with 2 x 7. Sure he could have typed 14 straight away and saved two button presses, but he didn't.

That's something we would have called a *Parker Code*

Every programmer writes perfect code while they are staring at it. Nothing can ever go wrong.

2777788888899?

Yep, I think so.

@@numberphile nitrogen-infused coffee ☕

He said "Primes" (plural), so yes.

that was obvious to me immediately and I'm an idiot

It's a parker prime

I entirely agree. I deeply admire and envy that about others, while I despise my own fear.

these numberphile regulars are really great.

He makes mistakes live on the internets for breckfist and doesn't give 2 Fs! XD

Mik he was talking about the individual digits being prime

Parker square

Googology has entered the chat

Just came here to see if anyone said the same thing.

timestamp?

12:53

You get points for "doing the work". Kudos!

Did exactly the same. Ended up without bugs.

5 points to Gryffindor

Cool i m not in my room or i would have also coded it

@@kryo2k lol

i would think base 10 is definitely the reason for 11 being the maximum probably a proof if you generalize a base 10 number as a sum of multiples of powers of 10 where you can manage to pull out a "10" after 10 steps and then you will get the 11th step always resulting in 0 and there being no way to continue

@@douggggggg I don't think it's that clean, quite, but I do think the attrition of possible branches gets too fast

111 - the number of the star-gate?

Best punchline ever

Richard Yan LOL

Yeah that was super passive aggressive xD

@@krystofdayne Nah, it's just cheeky banter.

Coincidence?

you just can't comprehend his high IQ and thinking skills

he was using python 2 as well

He forgot to add "Hello World"

I know it's a joke. But let me educate others. Matt is a mathematician, not a coder. He is not expected to know the tiny nuances of coding. What matters is whether his logic is correct. So returning "DONE" is perfectly fine. Also, using Python2 is also fine. It's not like this code is some multi-client multi-threaded huge application which needs to maintained for a long time. It's just a simple script to crunch some numbers. No need to bring Python3 here as he is comfortable with Python2. But it's encouraged to shift to it. (I myself shifted to Python3 a month back)

Sumitro Bhaumik how is returning done fine? It makes no sense

Egor Strakhov 😂

He's trying to take over the tri-state area

Freaky Fred vibes

The Weeknd

Goku hairstyle

That's a buzz cut, not a Mohawk.

@@RonJohn63 Look at the shadow on the door.

@@WrenAkula I choose to not understand!

Matt is secretly an anime protagonist with a fancy transformation sequence.

me: "Matt with a moh.... ?" _spots the joke_ ".... omg..." _facepalm_

woah

It’s higher than the original, and just has the same sequence of numbers. Bit underwhelming.

It's literally the same sequence of numbers lol, just in a slightly different order.

If you add 900 to yours, it also works (or 990, 9990, 99990....)

any permutation of these digits have the exact same persistence and it's pretty obvious. changing two digits around and calling it adding 90 is a little weird

Clever

I noticed that, too.

Can I learn these Cscience things all by myself?

@@randomdude9135 yea, fortunately there are tons of programming lessons on the internet

Just say no to python.

@@anthonyhoffmann why so?

How about subscribing to computerphile?

I had high hopes... and then we got 15 seconds into the video.

Someone explain?

The Parker persistence of "277,777,778,888,899" is still 11.

PARKER SQUAREEEAAAAAA

@@Sodiumitis no, he said some numbers reached the same length

Honestly it's no more risky than doing mathematics in front of a class. Everyone makes mistakes at the board.

I always coded in front of my class. Hours and hours per day

You can make it much bigger, if the original number is a String, and you just multiply the digits.

@@roguechlnchllla6564 Even better, notice how the result of the multiplication is always of the form 2^a × 3^b × 7^c, since you're just multiplying lots of digits together. The 4's, 8's and 9's can be broken down into multiple 2's or 3's, and the 5 never appears with any 2's without instantly giving a multiple of 10 and therefore a zero at the start. Now, the problem just boils down to finding three integers, a, b and c, such that 2^a × 3^b × 7^c is a number with a multiplication persistence of 11.

I have written a simple one in python . Upto 6-7 its fast, but after that the time it takes grows exponentially

i got them all to the last one as well

@@defectus1769 You beat me to it. I too was thinking about that. I also thought of reverse engineering the next number. Ex. 2, 12, 34, it ends at 34, since it needs 2*17. 17 is a 2 digit number and a prime. Ex. 2, 12, 43. Ex. 2, 12, 223 or 232 or 322. Some of which can go further. Ex. 2, 21, 37. Ex. 2, 21, 73. I tried 6, and got myself a little tree of options to start from. They all end with 6. But I don't know how to write a program that looks for reversed engineer all options. If someone makes that, you could use that 11 parts number as a start. All possible answers would roll out. And perhaps even more. Since you are right about 2, 3 and 7. You could easily weed wrong answers out for the next step.

I tried 3927 and got five (378, 168, 48, 32, 6) steps.

@MilesTisserand , if you mean any number has an equal chance of being chosen, I have a feeling the odds would lean heavily to 1 and 2 steps because of how often 0 comes up. It would be interesting to see the odds if you automatically exclude certain numbers that you know will drag down the average (like any that has 0 from the start).

@@Ascordigan Then 679 (2x3) would have the same steps. Now you know after watching the rest of the video, that the aim is find the lowest number.

@MilesTisserand Interesting idea. If I wasn't a busy undergrad, I would definitely try it out and graph the probability distribution. It'd be crazy to find a hidden poisson or something haha

I got six steps with 888888888

Step 1. Run complex problems that nobody has answered! Step 2. Wait Step 3. Wait ... Step 9. Wait Step A. Wait Step B. Wait ... Step FC. Wait Step FD. Wait Step FE. Wait Step FF. Wait Step 100. Boom! You got an answer and a fire!

Are you running a 90s processor without a cooler? Lol

Made my day.

@@whatisthis2809 Brilliant .

Play it on today's google doodle

Soon on 4chan, someone asks a question about an anime and someone else mathematically analyzes it to find a 15 -step number.

It does indeed seem like 11 is the limit. I calculated up to 1500 digits and there aren't any solutions. Similarly, I computed limits for bases 2-10, Base 2: 1 Base 3: 3 Base 4: 3 Base 5: 5 Base 6: 5 Base 7: 8 Base 8: 6 Base 9: 7 Base 10: 11 They all reach this "cliff" after which there's no answers as far as I computed, so it's probably fair to conjecture they all reach limits.

@@ehsan_kia Base 11: 13 23777777777777777778888888999999aaaaaaaaaaaaaaaaaaaa

@@ehsan_kia Any chance you could do this for bases up to 36? I have a conjecture I'd like to test and I cannot code. :(

Vihart already did that

fact, but what string of digits multiplied will get you that number.... figure that one out

@@dhycee8215 thats the problem lul

Nah, no knives

​@@bijeshshrestha2450 et tu, bijeshsrestha, et tu

this dude wrote all of that only to get 3 likes

​@@onehotseat You... missed the point. Ethan started on a random single-digit number, and created a sequence that would go to it (28 -> 16 -> 6) by factoring it. It's transparent how this method works - you find single-digit factors of a number, and string them together to create a new number (factors of 16 are 1,16, 2,8, 4,4, so 28 and 44 both go to 16 which goes to 6). Using the same method we continue - single-digit factors of 28 are 4,7. 47 is prime, and 74 has no single-digit factors. However, 4 is 2x2, so we also have 227, 272, 722. And none of them have single-digit factors, so we can't continue this sequence past length 3 (47 -> 28 -> 16 -> 6) Adding 1 to the _starting_ number is pointless, but we don't _have_ a starting number. However, a 1 in an intermediate step is not harmful, and only expands the number of possibilities for reversing a sequence. We can't continue the sequence from 23 because 23 is prime - nothing will multiply to it. However, by inserting 1s, we could end up with a number that contains only a 2, a 3, and a ton of 1s, with single-digit factors. From these, we can construct a number that multiplies the number we found with 2, 3, and 1s, and that number will multiply to 16. At the extreme, there's the number from the video. That one ends at 0, and there's infinitely many numbers that can lead to 0. Picking one at random, 20, gives us more factors to play with. 20 factors to 4,5, giving us 225, 252, 522, 45, and 54. Continuing in this fashion, we get to 438939648, which has factors 2^12, 3^7, 7^2, all single-digit. However, you may notice that there's no way to arrange and combine those factors that gives us another number with single-digit factors, _unless_ if you slot in a 1 and end up with the number 4996238671872. That number _does_ have single-digit factors - a 2, 6 7s, 6 8s, and 4 3s. I'm just not sure whether this number theory method is faster than brute forcing. There's already a plethora of ways to rearrange and combine factors of large numbers, and you have to find the ones with single-digit factors. You then have to account for the fact that every combination could have any number of 1s put into it, in any position, and factorise those too, all of which sounds computationally more expensive than just multiplying down the digits of most numbers from 1 to 10^320

Toying with this for a while, I think I found a way to get to higher numbers. If you check the steps on each number, they always repeat the lower steps. Like, 277777788888899, results first is 4996238671872, then later is 438939648. While the one beneath it, 3778888999, the next step is 438939648. This could mean that a way to find the 12th number, maybe the second step of it is the first step of the 11th one. Not a rule, since the smaller ones seem to break it, but bigger ones a very consistent.

Have you gone insane yet? Sitting around your town recording the numbers of car plates as they pass, guarding a shopping bag filled with Easter eggs?

It's not just a string of digits that multiple out to that number, but any combination of the digits in that number.

and you can add as many 1s as you want anywhere in the number

True. That would potentially make the search easier.

@@kevinr.9733 I agree with that, but couldn't it also exponentially increase the number of numbers that would have to be checked?

@@Stefan-ls3pb Oh i just realised :( you destroyed my life

I found one, but the RU-vid comment section is too small to contain it.

@@sobeeaton5693 isn't Numberphile gonna do a vid on it? :P

@@sobeeaton5693 Nice reference haha

@@sobeeaton5693 Get Outta here Fermat!

He's got 9 *whole* months before python2 is EOL ;-) [Yes! *stop* using python2; maybe try golang?]

I came into the comments to say this!

Lol

What's the difference though?

Or maybe use a decent programming language.

Would you pay for this on onlyfans?

Judging from the videos in this channel, anything and everything is a mathematical curiosity when the simplest problems can be scaled up, and up, and up, until it isn't trivial anymore. There are quite a few other videos like this where the game is pretty simple, but then when you continue to play it, it gets more difficult. Then they whip out the brown paper

So rightly, the Guarracino Number of 277777788888899 is 11... etc... 😊

Nothing wrong with giving it a go

Interesting...

Very lol

It's all marketing for Matt's new book about math mistakes... Available now: bit.ly/Humble_Pi

Numberphile Can confirm. All mistakes (for the rest of my career) are now officially deliberate.

@@standupmaths You could say it's a Parker Mistake

Cristian Baldi dudes aging fast

Julian Pinn Thanks! I plan to take a screwdriver to it soon.

Did it myself recently and I was surprised at how easy it really was.

Just go to the genius bar and get them to fix it practically for free (probably less than 2000 pounds at least)

@@Zolbat It's 7 pounds for a cable, over 100 for the genius bar.

@@gartinmarrix8484 so practically for free, which is amazing since the problem isn't even apples fault. Obviously all the users don't know how to treat their hardware... Opening and closing a laptop multiple times is nothing short of careless.

That's 2 steps. 1,524,382,530,048 0

You misspelled Stack Overflow.

@@Attlanttizz you can't put whitespaces in URLs. Even the logo is stack*overflow* as one word:.

@@wWvwvVUse your favorite search engine, search for Stack Overflow, you'll see.

@@Attlanttizz If this was a matter of you being able to find your page on Google, the original comment was already fine.

@@MrUwU-dj7js Well obviously: no. Stack Overflow is the name of the site and it's written with two separate words, not in one word. Even the term stack overflow as is used in programming is written with two separate words. Just wanted to point that out, nothing more :)

Also needs moar reduce()

@@BlueRaja iterative programming is easier for most newbies to follow along with, I have a suspicion that Matt would reduce x,y: x*y on his own.

Lua is better

@@rahimeozsoy4244 For select purposes, yes, just like python. "better" is subjective and intensely context sensitive. Please don't try to start flame wars.

@@crashmatrix This is not programming, this is coding. Programming requires skill.

Python 2 will not suddenly stop working in 2020. There are gazillion companies still using Python 2.x codebases, and they won't migrate everything to Python 3. Heck, forty-year old COBOL code is still used, millions and millions of lines of code.

@@PaulaJBean You're right. Also, Pluto is a planet.

@@Art-fn7ns well, it is, a dwarf one.

@@Art-fn7ns Now this is important!

@@leandrometfan smaller than our moon? it's an ateroid, not a dwarf planet. its diameter is smaller than the width of australia cmon

It kinda annoyed me that ALL they had to do was remove the very first print(n) statement to remove the duplication. I'm guessing if it wasn't being recorded live, Matt may have noted this. But yeah, why have the first print(n) when the previous call had already printed result.

You can also calculate the thing backwards (i.e. start from the smallest number, then work your way upwards), which would probably make the thing much faster.

@@Luxalpa I thought some more about it, and I think that any numbers beating it will have to start with 2,3,4,6,7,8,9 or 26 before we hit the wall of 789s (and obviously kept in digital order). If we cheat and count 26 as one digit for the purpose of recording length (obviously we multiply by 12 still), then we only need to check 210680 numbers to check all numbers of length 230, and even that is double counting some numbers.

@@manudude02 I would like to know how you got to that idea. I myself found a way to get up to length 1600 with fairly realistic effort (without requiring a super computer that is). Basically, the realization is that all chain numbers (i.e. all numbers except for the starter) must be N = 2^x * 3^y * 7^z. So you just need to go through all the x, y and z, find a number that doesn't contain 0 or 5, and then see if the multiplied digits result in any other of those numbers. With x, y, z < 1000 you only need to check 1 billion numbers. Storing them all on a hard disk would take some 1.6 TB of storage, however I think if you exclude all the numbers with 0 and 5 in it, the storage requirement will probably be much lower (I'll have to double check my math on that one or just learn Python and give it a try).

I tried to make it all the way around, the program ask the number of steps and then calculate all the numbers... didn't succeed...

@@Luxalpa Obviously we are not going to have zeros or one and removing 5s (not checked, but I doubt you can avoid even numbers for long), but look at every 2-digit other case. I ignore any starts with a 7,8 or 9 in it since we are already in the "wall" in that case 22 -> replace with 4 for 1 less digit 23-> replace with 6 for 1 less digit 24-> replace with an 8 for 1 less digit 26-> unable to find a fault in it 33-> replace with a 9 for 1 less digit 34-> using 26 gives a smaller number with the same product 36-> using 2 and a 9 gives a smaller number with the same product 44-> using 2 and a 8 gives a smaller number with the same product 66-> using 4 and a 9 gives a smaller number with the same product I started up a program (in java) about 10 minutes ago to check, so far it is has checked every candidate under 120 digits, but at least it scales by an order of N^2, while checking 2^x * 3^y * 7^z is scaling by an order of N^3. edit: Actually, I think your way may be more efficient if you don't have memory concerns, going from k=1000 to k=1001 gives you another 3 million-ish numbers to check, while going from 1600 to 1601 digits is another 9 million-ish numbers to check. You do have the downside in that you'd need to actually convert the x/y/z values into an actual number, but that's a small price to pay

pretty simple

That's definitely less stress on your processor. There are times I wish I knew languages other than Perl.

len(str(n)) is what's directly defined ("if there's only one digit left"). n

If you compile the code, str(n) is faster because you're going to call str(n) later on line 6 so the compiler will optimize it to compute it once and keep the result. Of course you can get rid of str(n) completely by using "% 10" (i.e. modulus 10) and "/10" in a loop instead to calculate "result", which I'm guessing is even faster.

Mattia Conti hes a mathematician. cannot do common sense. only complicated abstract concepts.

3.15 now :(

As a python programmer, i dont appreciate the continuation of python 2.

len(str(n))==1 instead of just n

print n return "DONE" How can appreciate this?

pithon, ok

@@kwinzman I know right completely unnecessary use of the length function. But its alright since he is not a programmer

Parker Python?

That throws up an error if the original number is only 1 digit, but that can easily be ignored or corrected

no no keep it! this way it is Parker code.

the method is recursive, so you only need one single print(n) on the first line

anoderone comments on that vid say they probably fell down the hole of "what counts as a flip"

If I knew how to write code so that it does this: Take the number 277,777,788,888,899, replace some of the 8s with 2 4 or 2 2 2, and some of the 9s with 3 3, then add an arbitrary number of 1s at the end, then rearrange ALL the digits in that number, to get another new number, which is an even number The digits in this number will still multiply to the same value and still have a MP of 11. Now check if that number can be expressed as 2^a × 3^b × 7^c. If not, repeat the process Surely by pure luck, the computer will find such a number that is in the form 2^a × 3^b × 7^c Which will be the digits of a number with an MP of 12.

Tha's atually 4. He counts the zero when he counts the 11 steps

@@Anders1314 I counted the 0 as well. I didn't count the 799 I started with though.

@@MEver316 You're right. My mistake

Mine was 7879

Nice