A 10,958 Solution - Numberphile

[@blacxthornE]

I'll give it to him. This isn't another Parker Square. He found the solution, fair and... uh... you know.

[@brokenwave6125]

Ersen Akçay Fair and cube?

[@emmanuelm07536]

Ersen Akçay 😂😂😂😂😂

[@thomasbjerrebojsen60]

HAHA! brilliantly done, sir.

[@workhardism]

Nice!

[@TedMan55]

this is between a parker square and a legitimate solution, we need a term for this. unlike the parker square, it’s a solution that does actually work, but a solution for a problem just slightly tweaked from the actual problem

[@shadout]

Accuracy : 10/10 Elegance : 4/10 Cheek : 10/10

[@EnderPig]

shadout IKR

[@TheNefariousness]

cheeki breeki

[@karl-erlendmikalsen5159]

I thought that use of concatenation in his math was elegant. Not what the one stating the problem wanted, but more elegant than that.

[@joshuaholtgreive3323]

shadout I think it's perfectly elegant.

[@dlevi67]

Why "not what the one stating the problem wanted"? If because the result is not an integer (but it rounds or floors to the right integer), then I agree. If because of his use of concatenation, I would invite you to look at the original paper on arXiv: it uses some form of concatenation in ~20,000 cases out of 22,000...

[@johnsadena1043]

"Do you know how many showed up to my shows wearing Parker square t-shirts?" I'd say 10,958.4.

[@bengineer8]

a few of the shirts were home made or torn?

[@pomtubes1205]

A serial killer came on the show.

[@johnsadena1043]

No some people just came with 0.4 of a shirt.

[@TheTombot]

Zinger!

[@pleaseenteraname4824]

3051

[@renxula]

"You could take these two videos that involve concatenation, and just play them one after the other." Comedy gold :D

[@tomonetruth]

Thanks, I'd completely missed that one!

[@bobwilson7684]

that was the best of the video/s

[@missellenmartin4152]

Ohh snap

[@mingyang4835]

Concatenate them together

[@alextaunton3099]

Look up Meshuggah - Concatenation

[@jonathanperryman1217]

For 1447 in ascending order Taneja has 1447 = 1^2 × 3 × 456 + 7 + 8 × 9, but I was able to find that 1447 = (12345 + 678)/9. Seems like there are simpler answers to some of these than are on the list.

[@chrisness]

Maybe he prefers solutions without parenthesis

[@Алекс-ц3т]

Again, if we count concatenation as an operation - yours has 8 operations and parenthesis, and Taneja's has as well 8 operations but without parenthesis

[@h-Films]

@@Алекс-ц3т it seems this person's answer was a true parker square

[@_catzee]

@@h-Films I disagree. The Parker Square is not a solution to the problem it concerns, while (12345+678)/9 is a solution to the problem it concerns. They are not comparable.

[@spudhead169]

@@Алекс-ц3тActually no, because the order of operations used in the original solution is left to right ordering. Thus if we use the same ordering on Johnathan's solution, the parenthesis become unnecessary, so his solution also has none. Also if you count concatenation as an operation, all solutions are going to have 8 operations. So both solutions are equally simple.

[@XMachete]

I like this solution, he's spot on that the concatenation operation being restricted to setting up the problem is arbitrary. He's opened a can of worms for the paper's author, however, because if he accepts this solution he must now revisit the existing solutions with this new information.

[@rich8037]

Not necessarily - all existing solutions are still valid and no one was ruling out the possible existence of an indefinite number of others.

[@maxinesenior596]

Concatenation in this case seems to go right after brackets, so there will be no case where calculations in previous sections have a concatenation with a number next to it like what he did with (120)||7. They happen before all other non-bracket calculations

[@h-Films]

@@maxinesenior596 they meant trying to find shorter solutions

[@Zephle]

exactly

[@swapyzzlevisionneur9145]

I Love this comment

[@NoSugarAllowed]

Now we need a shirt that just has 10,958.4 written on it

[@FirstnameLastname-zc6ym]

And a Parker square

[@nikolastoyanov1318]

Yup

[@TheBehm08]

Yesssss

[@Triantalex]

false.

[@stan4143]

"Do you know how many people show up to my shows wearing Parker Square T-shirts?" "Not enough" Priceless

[@mercurywoodrose]

hopefully no fractional people show up. or people wearing fractional shirts

[@alextaunton3099]

No' eenuff

[@alextaunton3099]

also it's "proiceless" not priceless

[@asheep7797]

*$27.48 AUD

[@SebvdBergh]

It's probably going to be at least 10958 of them!

[@jameshowell7178]

The paper's author missed a chance at alliteration when you said he loves plus product potentiation and brackets. he could fix that with "Parentheses"

[@siarles]

Parker nomenclature

[@ianmoore5502]

Parkers plus product pronunciation and parenthetical notation

[@zevenforch]

Basic examples of Concatenation: (Considering a, b, c... to be single digit integers) a||b = 10a + b a||b||c = 100a + 10b + c (10a + b)||c = 100a + 10b + c (10a + b)||(10c + d) = 1000a + 100b + 10c + d .....etc You get the idea... The one used in this video: 120||7 = 1207 (100*1 + 10*2 + 1*0)||1*7 = 1207 Which is of the form (100a + 10b + c)||d = 1000a + 100b + 10c + d Concatenation needs more recognition as a proper function guys...

[@diegonals]

|| = Best function

[@eFse7en]

i think the formula could be simplified to always have 2 parameters a||b = 10a + b, but allow it to be chained, so a||b||c = (10a + b)||c = 10*(10a + b)+c and 120||7 = 10*120+7 that way you don't need to define what a||b||c||... is

[@Raminagrobisfr]

I don't think it can be called an operation, because it's base dependant. If you use another base than 10, the result changes

[@leong108]

I think the simpler way to think of this concatenation is that he wrote ab, which means that you do not write anything for concatenation, its really writing the set of operators there between a and b as "empty set". So we know what it means, but the word "write" implies the example is correct, that we can write "ab" (a empty set b ) but we cannot write a|b (because thats a new operator " |" not empty set. the empty set operator allowed by the non-mathematical process of "write". ) . That is, if set A is the operations written, set A union "empty set" is still just set A.

[@tomdekler9280]

@@Raminagrobisfr The entire exercise is more or less base dependant. Even the numbers that are reached without concatenation would still be entirely unsatisfying if you had to be like "OoOoOh I found this quinary number using all numbers from 1-14", that number range just feels arbitrary.

[@hydrosamedh]

his accuracy has always been to the nearest Parker square order of magnitude

[@ConstantlyDamaged]

±1-1/(parker square of infinity)

[@ShinySwalot]

I'd use a square as a symbol for concatenation. Because it's not a "real" mathematical operation, it just gives a go at being one. Kind of like a Parker Square

[@RomeForWar]

D'alambert operator. Maths pretty much used every single freaking symbol you can think of. We gotta move to hieroglyph or japanese characters ahaha

[@snatchngrab8262]

palmomki No, concatenation is not a mathematical operation. Operations work values. Concatenation works on digits, just the numerals, just the symbils used to represent them. Concatenation is just putting strings together.

[@kiefac]

Shiny Swalot but it still generates a new number

[@pbpbpbpbpbpbpbpbpb]

But you could say that a||b = 10a + b which makes it a function, just with multiple arguments​ (EDIT: Wait, that doesn't work. Nevermind.)

[@snatchngrab8262]

kiefac Randomly picking digits also generates a new number. Randomly picking digits is not an operation. Generating a new number is not what defines a mathematical operation. In fact, a mathematical operation expresses the same number expressed differently. That is what is meant by "2x3=6". "2x3" is "6". Different ways of expressing the SAME value.

[@VideosRunescape4u]

Love whenever Matt Parker is in a Numberphile video!

[@OlafDoschke]

That's called Parkerphile

[@pr3nzlb3rg3r]

Only now did I realize, this is on numberphile, not his channel.

[@googlespieonsomeoneelse4898]

Ikr, he's my favourite almost-mathematician

[@Danish_raven]

Google Spie On Someone Else he is a parker mathematician

[@mb00278]

+Google Spie On Someone Else That's just disrespectful man. He's most certainly a mathematician, he's just a silly person.

[@mellowords]

*watches video* *looks at comments full of mathematical wizards* I don't belong here...but that was fascinating.

[@johnchessant3012]

10,958 is the exact number of days in three decades!

[@VACATETHE48]

10957, but close enough I guess.

[@tonyhakston536]

You guys are both wrong, it's 10957.5

[@SilentBudgie]

Thirty is not divisible by 4, so it depends on the number of leap years in there.

[@Fater4511]

also leap years don't always occur every 4 years. they occur every 4 years unless that number is divisible by 100 then they wouldn't occur unless that number is also divisible by 400. so 2000 had a leap day but 1900 would not.

[@АлександрСуходольский-л6б]

значит только через 30 лет мы узнаем ответ :)

[@dlevi67]

Out of interest (and spurred about a comment on concatenation being "allowed"), I analysed the arXiv paper, and found that it contains a number of errors - 16 of the formulas do not evaluate to the correct number. Some are clearly typesetting errors/interpretation of negation operators, and I have been able to recover the correct formula, but for others the matter may be more difficult to resolve. Here is the list (including the six "corrections" I found): Increasing Order Digits: 292: 1 + 2 × 3 + 4 + 56 × 7 + 89 (computes to 492) 312: 12 + 34 × 5 + 6 + 7 + 8 + 9 (computes to 212) 1548: 1 + 2 × 34 × (5 + 6) + 789 (computes to 1538) 4498: (1 + 2 + 3^4 + 7 × (5 × 6)) × (8 + 9) (computes to 4998) 6374: −1^2 × 34 + (5 + 67 ) × 89 (may compute to 6442, unambiguous as −(1^2) × 34 + (5 + 67 ) × 89 = 6374) 7622: −1^2 + (3 + 4 + 56 × (7 + 8)) × 9 (may compute to 7624, unambiguous as −(1^2) + (3 + 4 + 56 × (7 + 8)) × 9) = 7622) 9055: 1 + 2 × 345 × (6 + 7) + 8 × 9 (computes to 9043) 9070: −(1 + 2^3)^4 + 5^6 + 7 + 8 − 9 (may compute to 22192, unambiguous as: −[(1 + 2^3)^4 − 5^6 − 7 − 8 + 9] = 9070) 9940: 1 + 2 × 3 × 4 × 5 × (6 + 7 × 8 + 9) (computes to 8521) 11093: −1^2 + (3^4 + 5) × (−6 + (7 + 8) × 9) (may compute to 11095, unambiguous as −(1^2) + (3^4 + 5) × (−6 + (7 + 8) × 9) = 11093) Decreasing Order Digits: 289: 98 + 7 + 65 × 4 + 3 + 21 (computes to 389) 7683: (9 × 8 × 7 + 6) × 5 + 4 × 3 + 21 (computes to 2583) 8580: 9 + 8 × 7 + 65 × 43 × (2 + 1) (computes to 8450) 8989: 9 − 8 + 7 × (6 + 5) × 4 × 321 (computes to 98869, correct as 9 − 8 + 7 × (6 − 5) × 4 × 321 = 8989) 9069: 9 × 8 × 7 × 6 × (5 + 4) × 3 − 2 − 1 (computes to 81645, correct as 9 × 8 × 7 × 6 × (5 - 4) × 3 − 2 − 1 = 9069) 10535: 9 + 87 × (6 + 5)(4 × 3)^2 − 1 (computes to 137816)

[@Darkstar.....]

dlevi67 is using a subtraction symbol as a minus even allowed? You should use subtraction to get a negative number not use it as an actual number in the nagative. I guess its not anything to do with subtraction. A negative number is still a number. Ignore my comment after i thought it through better. Numbers haven't been my thing for a while. 😉

[@kerlunkfish]

For 292, how about: (((-1 x (23+4)) + 56) x 7) + 89? More brackets than strictly necessary but makes it completely unambiguous!

[@kerlunkfish]

Another method for 292: ((-1 + 2) x (3 + 4) x 5 x 6) -7 + 89

[@kerlunkfish]

For 312: 1 + 2 + 3 + (4 x 56) -7 + 89

[@kerlunkfish]

For 289, descending: 9 + 8 + 7 - 6 + (54 x (3 + 2)) + 1

[@MatthiasCorvinu]

Be sure to let us know what the guy says if he responds.

[@furrehIzzy]

Can we get a T-Shirt that says 10,958.4 on it?

[@IntergalacticPotato]

Ye

[@ky-gp4sz]

Ask him to sign your Parker square shirt 10,958.4

[@agar0285]

666 likes

[@audrey-chan6290]

my poor CS brain trying not to read the contatenation symbol as the OR operator

[@hexa3389]

Same here.

[@hikarii8758]

pretty much the same.

[@cindydong798]

so true

[@milind8541]

Well your CS brain should also see the CONCAT as command

[@thesmart4128]

it's rather it looks like the parallel sign for me

[@Jaymac720]

There isn’t a “concatenation” button on my calculator. I can’t accept this

[@UstazFarhanBD]

Lol

[@Cream147player]

Secret that calculator companies would HATE for you to learn! Unbelievable shortcut I found that works to get concatenate on *all* calculators in ONLY 4 BUTTONS!!! Where the concatenate function should be press the following buttons: x 1 0 +

[@SoI-]

Just do the two numbers like 4||5 to 45

[@nanamacapagal8342]

@@Cream147player 17||29 = 1729. 17 * 10 + 29 = 199. 1729 ≠ 199. Disproven.

[@_catzee]

Please tell me you're joking. PLEASE.

[@DailyDrumLesson]

Matt: "Do you know how many people show up to my show wearing Parker Square T-Shirts?" Brady in the Background: "not enough" That's where I cracked up!

[@lkmlmlioj]

I ALMOST DIED LOOOOOOOOOOOOOLZ

[@california_wang8078]

Cervix Crusader it’s about a Parker square amount of t-shirts

[@K_amonger]

666th like

[@stetsongray1975]

This very comment has an upvote count very nearly double the count of main video downvotes. Somewhere in this data is a positive proof of the healing power of comedy in academia.

[@kw3494]

I love this guy. He motivates me to try problems I couldn't dream of solving. Thx Parker.

[@standupmaths]

Kim W You're welcome!

[@Yntec]

Hey Matt, I wish to ask you if you were actually annoyed about the whole Parker Square thing, and if you regret/regretted it at some point, because you seemed upset at that point, but I'm not sure.

[@SlimThrull]

I can't speak for Matt. But he does humor as a profession. Most people who do that have incredibly tough skins. I suspect he finds it mildly annoying but far more entertaining. Again, this is just a guess on my part.

[@guyzavaro958]

sorry for the trolling: 0^1+2×(3+4×(5+678))+9=10958 0 can be added to all of the solutions, both ascending and descending

[@TheStoicBrowser]

Guy Zavaro why not just multiply it all by 1?

[@Wordsnwood]

"Do you know how many people show up to my shows wearing parker square t-shirts!?" --- Hah! Matt, I hope you're getting a cut on those sales! But more importantly, I think you should be keeping track of the numbers of people showing up wearing those shirts. I'm sure some fun math humour could come out of that!

[@kcwidman]

Wordsnwood (Art Mulder) he's probably keep track of it in his recreational spreadsheet.

[@googlespieonsomeoneelse4898]

He should make a Parker Square with the numbers of people that turn up wearing Parker Square t-shirts

[@livedandletdie]

If I ever became a city planner, I'd make a parallelogram with 89º and 91º corners, and name it the Parker Square. As an homage to the Parker Square and that time he tried to measure the most square square in New York.

[@WilliametcCook]

Try to calculate pi with the ratio of people wearing Parker Square t-shirts to people who didn't

[@tasherratt]

Make it tau.

[@invidious07]

This is absolutely a valid use of concatenation per premise of the original paper. Well done.

[@zacharypower7674]

I consider this to be thinking outside the Square

[@letstalkpoliticsBDG]

nooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

[@letstalkpoliticsBDG]

NO NO.

[@gingermatchstick]

#ParkerConcatenation

[@Gremlins422]

concatenation*

[@igt3928]

#ParkerSpelling

[@AlucardNoir]

I'm so using this.

[@qwertyquazo673]

Matt is a gap-filler.

[@2Cerealbox]

Is that a solution that almost doesn't work, but you have to include it anyway for completeness?

I really want to know Inder J. Taneja's opinion about this.

[@slook7094]

Apparently, Matt Parker did email him and Taneja has decided it's "correct".

[@eugeneeugene2511]

Julia Lilienstein the pdf stiil says still unavailable

[@lfvdb1]

He cheated

[@MrKrimson]

@@lfvdb1 How does one cheat maths?

[@melglb123]

@@MrKrimson well its not a mathematical operation so you have to go on what the rules for the booklet say, which did not include it as a valid operation

[@pvanukoff]

His first solution is a definite #ParkerSquare. The second one is ingenious though.

[@adamschlinker972]

As a programmer who uses concatenation often, this was satisfying.

[@jacobpeters5458]

in what way do u use it? I'm curious

Oh, I see what you did there, Sir. Well played. Well played indeed. You are awarded the Kobayashi-Maru ribbon for this!

[@adamecek64]

I smell a Star Trek fan here

[@AnimeLord512]

I only know about Kobayashi Maru from its reference on Young Sheldon but know nothing about Star trek. Is there a reason its name sounds Japanese?

[@AbudBakri]

I got 99 problems but 10958 ain't one

[@vendetta7221134]

curious, were you able to solve it Dr?

[@Vedvart1]

I got 23 problems and Fermat's Last Theorem ain't one

[@CalvinHikes]

Dr.StickFigure that would also be a great t-shirt

[@kamoroso94]

Why am I suddenly seeing you everywhere?

[@AbudBakri]

Kyle Amoroso because I am everywhere

[@SimonFoster63]

"Classic Parker Square" LOL

[@martinshoosterman]

I completely agree. I dont like concatenation as a solution, but if its used, use it properly to it's fullest. This video isnt a parker square.

[@achu11th]

martinshoosterman you are absolutely right. It is the complete opposite of a parker square. IT IS A PARKER SQUARE ROOT. What a wonderful way to celebrate the fact, that the parker square meme is officaly one year old now.

[@achu11th]

SoulSilver Snorlax #ParkerSquareRoot indeed.

[@wingracer1614]

I suspect that if you allow it, the whole list would have to be redone because there would be shorter solutions to many of the other numbers.

[@prplnau]

The 5 digit ones only use each of the 1-9 sequence once anyway. So unless you are also counting in the number of operations there wouldn't be shorter ones. For the 3 digit one they only mentioned the least amount of digits I believe, which to me would only mean the amount of numbers.

[@Tekay37]

achu11th: actually it's a Parker Square of a Parker Square, so we should call it a #ParkerSquareSquare

[@lachlach3449]

6:13 - Most people in Maths Exams

[@amritaanshbhandari6199]

Way too true

[@JamesTheFoxeArt]

Those 17 seconds are very relatable

[@eddiejohnston1853]

Yessssss! 😄

[@annabago8621]

I laughed out loud

[@SwagnerCountsThings]

I love Matt, and this has quickly become a top two numberphile video. I've watched the two videos 3 times since it came out

[@aquawoelfly]

B: in what reality do you think im not going to put up a square? *puts up a square* M: just as long as you dont try yo sell the t shirts *puts up tshirt* M: do you know how many people show up to my talks wearing the shirt? *puts link in description* B: not enough. Bradey, breaking math, one mathematician at a time.

[@anon8109]

The next obvious step: revisit all of the results while allowing full use of concatenation.

[@FavoO]

I hope there will be a followup with the answer to his solution.

[@jpchevron]

Likewise.

[@sab611]

Agreed, we need to know if it's accepted.

[@snatchngrab8262]

It's not valid. Using concatenation as part of a series of operations is not mathematically sound, which is why the 10958 problem rules do not specifically state it can.t be done.

[@finthegeek]

Snatch n Grab looks sound to me. Also I use it in real life applications frequently

[@Lexivor]

But lots of the official solution did use concatenation, just as the first step, It's inconsistent to allow concatenation for the first steps but not later steps.

[@XManium]

I don't know how this made it on my recommended list, but it's a welcome break from political commentary.

[@dedouluk]

Same here! And I have just been following a link to watch Fatboy Slim's "Right Here, Right Now"..

[@nocturnalsunlight3639]

@@dedouluk Such a great song, though!

[@Dylan-rc2cg]

Heres one without concatenation (1+2^3)^4+5-6!+7!+8*9

[@johnkeefer8760]

Dit Is Dylan that does get the correct solution but I believe factorials are not allowed

[@Dylan-rc2cg]

@@johnkeefer8760 welp, I tried

[@pamanes7]

still impressive!

[@ibperson7765]

The way the op did it is also not allowed.

[@xadielplasencia3674]

@@ibperson7765 why?

[@Eggmasstree]

6:06 dat "Really :D ?!?" is so fun :D Like a child (no offense) getting his parents' approbation for doing something silly

[@jerryto803]

As the philosopher Jagger said "You don't always get what you want, but if you try sometimes, you just might find you get what you need."

[@adamplace1414]

Ok Dr. House. 👍

[@GameKevLP]

Matt trying to stop this Parker Square thing is just another Parker Square. He tries to end it, but it's just not gonna work. Classic #ParkerSquare

[@abdulmuhaimin9780]

iCake he didnt embrace it though

[@brokenwave6125]

iCake He isn't trying to end it. He allowed the first video to air after all. and mentions it regularly now. he was just saying this isn't a Parker Square...and it's definitely not.

[@schadenfreudebuddha]

we should rename the Streisand effect to the Parker effect.

[@bengineer8]

even better: Integral from -2 to infinity of 1/n! dn = a ParkEurler constant for a less parker square result, start from 0.5970060971725153 instead of -2

[@andreasjohansson1952]

It's 4:30 am and I don't know anything about mathematics. I watched the video and I was right all along, I don't know anything about mathematics. :)

[@whatisthis2809]

4:28 am... what

[@shepilepsyy5572]

i had so much fun reading these comments. maths geeks.....ASSSEMMBLLLEEEEE!!!!!! love you all x

[@rosiefay7283]

1 + (2-3^(4*5/6/7))^-8 + 9 = 10958 (copied from my comment to the 1st video) OK, I admit, it's a Parker formula. But it's close. Even closer than 10958.4. Anyone got anything closer?

[@oschi2537]

Rosie Fay Nice work. The funny thing is , that if you take to the power of 8+9 at the end, it almost equals 1

[@AyrtonTwigg]

WTF how do you even come up with stuff like that?

[@Vorgu]

Genius :-O

[@lukevacca1537]

equal to 10958.00206 close enough lets stop there

[@kamoroso94]

I didn't know you could use negation as one of the operators. They never mentioned it.

[@seanehle8323]

He should publish the section with just the 4's and call it "A Complete List of Fourier Transforms."

[@vara202]

Only the fouriest transforms

[@lowercaserho]

I sometime try to prove the Colatz Conjecture or the Twin Prime Conjecture. Usually just by thinking about them in the bath. I know the probability of me actually succeeding is basically zero, but I'm fine with that.

[@brokenwave6125]

lowercaserho Hey, we all appreciate math here...we can acknowledge that the odds of you solving it are in fact greater than zero...so don't feel bad ;) Some things are impossible...but that isn't.

[@lowercaserho]

"Basically zero" was my shorthand way of saying "not actually zero, but extremely close to it". It is probably slightly higher than an infinitesimal as well. Probably. :D

[@NoriMori1992]

lowercaserho I'm the same way, dude. You're not alone.

[@GabyGrecu]

If an infinite number of men would do that then for sure one of them will find the solution

[@kamoroso94]

Gaby Grecu Maybe more than one :P

[@thenoeticskeptic5819]

+Numberphile The one thing you did not mention was that Dr. Inder J. Taneja did have a solution to 10,958, he just was not sure that adding additional representations (square roots and factorials for only one solution was within the parameters of his work [in the same way you concatenate (4x5x6)||7]. 10958=12×3+√4+5!×(67+8×√9)=(9+8×7×65+4)×3−2+1 Also, if you read his introduction, he talks about larger than single digit numbers, i.e., ab or 23, he did not say that concatenation was a viable solution.

[@delroland]

Technically the square root function is the same as raising to the power of 1/2 and so I think fails the spirit of the exercise.

[@johansundman8221]

@@delroland Just how devision with the second digit would be multiplying with 1/2 so I think it's already failed in that case.

[@jobigoud]

@@delroland log and root are the reciprocal operations of power so if we can use divide for multiply, then log and root should be game. And not just square root, any root and any log.

[@delroland]

@@jobigoud Except it still requires the use of numbers, so unless we were raising to the power of 2/3 or 3/4 etc, it would be a rules violation imo. Like 1^(2/3) would be valid but (root)1 + 2 + 3 would not as the (root) function is shorthand for ^1/2

[@JasonKatsanis]

Did you ever get a response from Taneja about your proposed solution?

[@campbellstarky2144]

After an exhaustive computer search I am pleased to report the following: -There are 2220 near solutions (near solution meaning -0.5

[@itisALWAYSR.A.]

"Do you have any idea how many people come to my shows wearing Parker Square t-shirts??" "....not enough." A+ savagery

[@sethx1138]

A related question is, "What is the maximum integer that can be represented using the digits and operators from the video?". Perhaps it is 12^(3^(4^(5^(6^(7^(8^9)))))). Call this integer "M". Then one might say that from 10958 to M there are many integers that can't be represented using the digits and operators. I would also venture that these un-representable integers occur with increasing frequency as you move from 10958 to M.

[@slook7094]

They'd just need to use a factorial.

[@Bippah]

"Thats as close as I can get.." yeah as close to the most perfect example of a Parker Square next to the original! xD

[@Silverizael]

But then the second answer he gave does work. The first is totally a Parker Square, but the second is a real solution. The prior non-use of concatenation to the fullest was completely arbitrary.

[@TheArachnid]

@Bippah xD Also: ayyye a death note fan!!! sup?

[@emiraee]

"You know how many people show up to my shows wearing parker square t-shirts?" "Not enough."

[@craigdawson9391]

I was recently trying this problem again, and I got an answer today. It seems Mr Taneja wants to avoid square roots and factorials. I got an answer without concatenation but did use 1 square root. ((1+2)^(3+4))X5+6-7+(8Xsrt9) If you want to use this in a video, you may do so.When looking through previous emails, it seems that I found a web page saying a computer program found no solution using only basic operations.

[@stevethecatcouch6532]

+numberphile There is a potential problem with allowing run time concatenation. If it is allowed when representing numbers using 9 digits, there is no obvious reason not to allow it when representing numbers using a single digit. That would result in at least one of Professor Taneja's representations to be non-minimal. His representation of 15 using only 6's uses six copies: 6 + 6 + 6 × 6/(6 + 6). Using run time concatenation, that can be reduced to five: ( (6/6) || 6) - (6/6). I doubt that is the only representation that would be affected. edited to change "on the fly" to "run time"

[@brokenwave6125]

Steve's Mathy Stuff So a new solution being more efficient is a "problem"? Damn...I'm glad you weren't in charge of mathematics throughout history. Just because someone already wrote stuff done doesn't mean something that would change it is wrong.

[@thishandleistaken1011]

+Broken Wave The fact that he was able to do all those other numbers without run time concatenation is pretty amazing. Run time concatenation makes it much easier, and sort of defeats the purpose.

[@the1exnay]

Terran Loyalist Does it really makes it that much easier?

[@jakubbajdak4634]

+Broken Wave More efficient solution is not a problem. He just mentioned that to suggest, that prof Taneja hasn't used concatenation when it would give better results, therefore it might not be allowed to use it (it is not allowed, if prof Taneja hasn't just missed it, but hasn't used it intentionally).

[@mkolnay]

let's allow floor function and we are done

[@RedsBoneStuff]

This is one of the best videos on KZbin. You solved an interesting problem and made a great moral out of it. Three times hooray for Matt!

[@pelgervampireduck]

he didn't solve it, he cheated!.

[@jackscrivens9520]

3:06 careful matt, language

[@jaylashTV]

Here's another solution: 1-2x3+(4^5)||6+7||(8+9).

[@Zoova]

what. Wait which comes first? Concatenation or addition?

[@FakeMichau]

writing multiplication using "x" bother me a lot!

[@SoI-]

@@FakeMichau what about ×

[@SoI-]

Or ÷

[@trdi]

I'm pretty sure Matt Parker has proven Riemann hypothesis, he is just waiting for the right moment to present it in a Numberphile video. At the moment they are looking for a way to shorten the 15-minute proof into a 10-minute version. Current idea how to achieve the sub 10-minute mark is to present the lemma that rabbits have long ears in a separate video on the second channel.

[@pauldraper1736]

Spoiler: it involves concatentation

[@nienke7713]

I agree with matt, either concatenation should be completely allowed, including how he used it, or it shouldn't be allowed at all, which would invalidate many of the other solutions and leave much more gaps

[@1224chrisng]

most certainly, yes

[@DehimVerveen]

Don't listen to the people saying it's a parker square. I would agree with them on your first solution for this problem, but I think your second solution is very elegant.

[@magicmulder]

Gah, I was expecting an attempt to explain why 10958 is the only number where it isn‘t possible.

[@JMartinez69261]

When doing things like this, things happen. Sort of like prime numbers

[@gehirnschmelze]

My closest approach is 1 ^ 2 + (3 / 4) ^ (5 - (6 * 7 * 8 / 9)) ≈ 10957.85254 (Error ≈ 0.15) when: - negative one is not allowed - concatenation with things in brackets is not allowed - operations other than +, -, *, /, ^, || are not allowed Found this by brute force with Java. I think it is the closest number you can get with this set of rules, but I can't prove it is. Beat me by getting closer (with the given rules)

[@1enaic]

Beat "me"

[@deldarel]

That first solution is just some parkering around, but that second solution is some seriously impressive out-of-the-box thinking

[@MrEngineeringGuy]

I'd give him that! That is a clever and ingenious solution.

[@IceMetalPunk]

Thanks, guys; I just screwed up a job interview, and I needed some cheering up. All this Parker Square talk had me laughing out loud :) Also, here's my Parker Attempt, from before I watched this second video: 10958 = 1 * 2 * (3 + 4 * (5 : 6) + 7 + 8 + 9) Where a : b is an operation defined such that 5 : 6 = 1363 Definitely Parker in nature, right? :P

[@100nodog]

Can you define the operation, plox?

[@JacobBanerjee2821]

it just makes the result equal to 1363

[@Dracopol]

That feels like cheating, because he hasn't concatenated digits, but concatenated a digit with the PRODUCT of other digits. Taneja's solutions imply an order of operation where concatenation has always come first.

[@newsupermariobros.uworldla5626]

Parentheses negate order of operations though. That's like saying using parentheses is cheating.

[@ahmetsezginn]

@@newsupermariobros.uworldla5626 Thats not his point and he is right.

[@natural2112science]

I thought the same thing. This is not allowed. All the other concatenation were numbers in order like 123, 4567. His is 1207. If it was that obvious there would have already been an answer.

[@Dracopol]

@@newsupermariobros.uworldla5626 Okay, I'll accept that. I'm no mathematician but the rules of precedence must be clearly stipulated in the Great Book of Goofy Arithmetic Operations or whatever.

[@jesusthroughmary]

Of course it's cheating. Matt knows it too, it's kind of the point of the video.

[@g00dvibes47]

I readily accept your answer due to your charisma, creative problem solving, bracket placement, and respect/observance to the original mantra of mathematical minimalism.

[@TDiff]

I've done coding my c++ program. Let's wait and see if i can get closer. :^)

[@TDiff]

(1*234-5/6)*(7*8-9)) = 10958.83333...

[@TDiff]

12*(3-(4^5*(6-7)*8/9)) = 10958.66667

[@TDiff]

1*(2*3)^4/(5/6)*7+8*9 = 10958.4!!! Can i make the 10958 without cheating with concatenation?

[@TDiff]

1+(2^3-4/5)^6*7/89) = 10958.28637

[@punchb]

((1+2/3+4)^5*6-7)^(8/9) = 10958.1156

[@LaatiMafia]

I have discovered a truly remarkable connotationless proof for this problem which this comment box is too small to contain.

[@firstlast8858]

We're waiting!

[@erebusvonmori8050]

Fermat is that you?

[@yoloswaggins2161]

Stop bullying poor parker.

[@SireSteckdose]

Square

[@hasch5756]

It's not bullying, it's just an _approximation_ of bullying

[@javierlim4873]

Yolo Swaggins 29 1 47 41 37 1 23 41 29 #PARKERSQUARE

[@Hyuji1111]

You can actually define concatenation as a function quite easily. Say there's some function dig(x) that counts the digits of a number (there is probably one that actually exists but I don't know of it). We can define this as dig(x)≡{p: floor(x/(10^p)=0}. Then, we can define concatenation as a||b≡10^(dig(b))*a+b. This was fun, thank you. Maybe this should be my next research topic haha.

[@MasterofBeats]

lemme try?

[@punchb]

The closest I got with no negation: ((1+2/3+4)^5*6-7)^(8/9) = 10958.1156

[@achu11th]

Happy parker birthday, matt. The video is exactly one year old now.

[@YT7mc]

achu11th no it's not...

[@achu11th]

PaulTheSuperstar it is. Look at the upload date of the parker square video (18.04.2016). It is exactly a year since parker square started to become a thing.

[@YT7mc]

achu11th two days ago was the release date, it's not a year old.

[@YT7mc]

achu11th oh ok I thought you meant this video I must be blind and bad at reading

[@achu11th]

PaulTheSuperstar no problem. It is a time to clelebrate anyways. The first ever mathematical meme, I have witnessed growing up. Our conversation almost worked out but quite right, just like the honorable parker square.

[@BernhardTittelbach]

I honestly think concatenation should be disallowed alltogether. It's not an operation on numbers but an operation on the representation of numbers. The value or semantic of the concatenated number changes depending on your choosen representation base which is not true for "true" mathematical operations. Would be interested in a counter argument though.

[@sf21787]

Agreed, but a formula could be written to describe concatenation. ***DISCLAIMER: Only works with positive digits*** Ex. 1||7 = 1(10^(n-p))+7(10^(n-p)) Whereas, n = number of digits in concatenation p = position of digit in concatenation written left to right 1||7 = 1(10^(2-1))+7(10^(2-2)) 1||7 = 1(10^(1))+7(10^(0)) 1||7 = 1(10)+7(1) 1||7 = 10+7 1||7 = 17 4||5||6 = 4(10^(n-p))+5(10^(n-p))+6(10^(n-p)) 4||5||6 = 4(10^(3-1))+5(10^(3-2))+6(10^(3-3)) 4||5||6 = 4(10^2)+5(10^1)+6(10^0) 4||5||6 = 4(100)+5(10)+6(1) 4||5||6 = 400+50+6 4||5||6 = 456 This is about the mathematical equivalent to making up sounds to rhyme in poetry. This was written on a phone, I hope there aren't too many mistakes I made.

[@EvelynV123]

This feels slightly overly complicated to me. a||b = (a*10)+b a||b||c = ((a*10)+b)||c = ((a*10)+b)*10)+c Treating it as non-associative resolves any need to increase the power of 10. All of these fall apart if we start using decimals in any way, assuming we want to append the number at the end of the decimal. The desired solution to 1.7||5 is 1.75, but following any of the above equations gives 22. Since the amount of digits after the decimal can be arbitrary, there's no obvious & elegant solution.

[@sf21787]

@@EvelynV123 I wasn't aware of use of concatenation in decimal numbers. Lol this really puts a stick in my wheel. From now on I should add a disclaimer that my formula only works for positive digits and is still highly unnecessary.

[@zerochan2915]

exactly. If concatenation is truly an operation, it should work for all real numbers. Think about pi || 3. It just doesn’t work. Unless you invent new number theory to allow manipulations after an infinite amount of digits.

[@asdasd-ho3mm]

@@zerochan2915 that's also a bad argument. Is exponentiation an operation? Then (-1)^(-1/2). Even worse, is division not an operation since 1/0 isn't defined? Operations can be operations even if they don't apply universally. Anybody who's studied fields and rings in abstract algebra could tell you that binary operations don't always need to be defined to be useful and valid.

[@riko8606]

Zero dislikes... classic parker square

[@AnonCaptainAwesome]

Excel uses the ampersand symbol for concatenation. 1&2&3&4 If you use the INT function, you can round the number down to a decimal place.

[@tomonetruth]

Not only that, but if you right click in a cell you can change the background colour, too! Put that in your pipe and smoke it, Taneja..

[@OJB42]

As a programmer, I totally agree that concat should be treated that way. Great solution!

[@1Darco1]

Concatenation is just a mathematical function like any other. It is just not an operator, more like two: a || b = concat(a, b) = a*10 + b Use that as an argument for either side, I think this video is more like a #ParkerSquareParkerSquare. It's almost a ParkerSquare, but not really.

[@1Darco1]

Btw, since concatenating is, of course, not commutative, it has to be given an order of precedence from left to right: a || b || c = (a || b) || c = (a*10 + b)*10 + c = a*100 + b*10 + c (which has been done implicitly the whole time anyway).

[@1Darco1]

#ParkerSquare^2 #ParkerSquareSquared

[@1Darco1]

Alright, then lemme just concat : Z x {1, 2, 3, 4, 5, 6, 7, 8, 9} --> Z there we go, well-defined domain and co-domain :P

[@blablablerg]

what if b is two digits?

[@1Darco1]

That is ruled out by the statement concat : Z x {1, 2, 3, 4, 5, 6, 7, 8, 9} --> Z. It means that concat is a function which maps two inputs, the first from the set of all whole numbers, and the second from the set containing the digits 1 thru 9, onto another whole number as a result. Thus, b has to be an element of {1, 2, 3, 4, 5, 6, 7, 8, 9} and can't be greater than 9.

[@Kolyanes]

*Немного жульничает, но все равно очень круто!* *В оригинальном труде нельзя дописывать цифру к результату действий* *Либо составляется число из цифр подряд, либо математические операции* _Но повторюсь, что сделана крутая работа в поиске решения._

[@ДонАйс]

Хитрые янки

[@oleksandrhorskyi8442]

Вроде в оригинальном труде не было об этом сказано конкретно, что этого делать НЕЛЬЗЯ. Сказано просто что можно a+b, a-b, [...] ab. Имелось ли ввиду составление числ из цифр подряд как ab либо конкатенация a||b?

[@maksimvoylokov681]

@@oleksandrhorskyi8442 если в других примерах этого нет, то подразумевалось, что нет

[@Suvorovgold]

Как же автор труда записал число 123 или 1234? как 1||2||3 и 1||2||3||4? Просто недочет слов, как обычно любят использовать в судебной системе. Автором конечно же данное действие не подразумевалось.

[@notd4335]

@@maksimvoylokov681, ты же не не покупаешь виноград домой только потому что из твоей огромной семьи его ест одна только бабушка? Она такой же член семьи как и все остальные, просто со своими особенностями. Тут такой же принцип. Какое-то число не использует вычитание, какое-то скобки: у каждого свой способ решения. Так что твой аргумент тут не катит.

[@bengt-goranpersson5125]

So what did Inder J. Taneja respond to getting this submission?

[@kyleduvall8818]

To finish off the problem, you could use the floor function to round down the entire answer to 10958. Simply, use the function that rounds down to the closest integer less than or equal to the given value.

[@jamesl8640]

3:20 to matt, particularly funny when you turned up to citation needed and someone in the front row had a Parker square t-shirt. It was brilliant.

[@Basscoach]

You look so genuinely happy. I'm happy for you :-)

[@robknightfilms]

Here's my solution that *doesn't* use esoteric concatenation, although it does cheat with notation a bit... (1+23)*456+7+8-.9... That last .9... is point 9 repeating, or 1. This indeed evaluates to 10,958 (with a hint of imagination ;) ).

[@Mohammed8778]

which makes your sequence 12345678099999999999999999999999999999999999999999999999999999999999.... not sure thats intented

[@thethinker2688]

That's cool even if it's cheeky

[@SD-io3hq]

@@Mohammed8778 0.9 repeating equals one but go off

[@gaius_enceladus]

The answer to the ultimate question of Life, the Universe and Everything is...... "Parker Square"....... I think I'll build a starship with a "Parker Square drive"...... :)

[@kabbaage]

As a programmer, I have to accept this because if I'm writing a program that takes in the nine digits as an input, one of the functions I'll write will involve concatenation and there is no rule stating that that function can only be run on the base digits

[@ANelis-z5d]

You can write Entier(10,958.4) to obtain 10,958, where Entier(x) is the greatest integer which is smaller than or equal to x. For example, Entier(2.0) = 2: Entier(1.9) = 1; Entier (2.1) = 2.

[@abhibhatta1]

Found another number 21212 which can neither be written in increasing order nor decreasing order using the operations permitted

[@TheHereticAnthem20]

So closeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee 1 * 23 * 456 + (7 * 8 - 9)*10

[@TheHereticAnthem20]

So close againnnnnnnnnn 1 + (23 - 45) * (6 - 7 * 8 * 9)

[@TheHereticAnthem20]

aaaaaaaaaaaaaaaaaaaaaaaaaaa 12 + 3 + (456 * (7+8+9))

[@AmikaofMan]

​@@TheHereticAnthem20 Sleep first, then come back and do it my friend

[@tylerabernathy8942]

but the first one works???

[@jkk1337]

@@tylerabernathy8942 the 10 is not allowed

[@Statalyzer]

4:17 - the entire thing is arbitrary (or arbitry).

[@Asterlibra]

My solution within 20 minutes: 1x2x((3+4)x(5+6)x7+8)||9 Concatenation is used as well :) This result shows that there are many solutions.

[@emmettfarrar9594]

Bro, you are such a hero, you should have been on e of the cameos in deadpool

[@marcobonera838]

Is there an algorithm? (also, the fact that we are using concatenation with an arbitrary base is baffling me)

[@luciojb]

I thought about an algorithm too, doesn't makes sense, this guy would have no life otherwise

[@nekodahaltom2443]

I'm in geometry right now so every time he writes a concatenation sign I automatically think it means parallel

[@kadenvanciel9335]

That Brazilian professor should see this.

[@xFuaZe]

console.log( 1 + Math.floor(Math.pow(Math.pow(2,3)-4/5,6)*7/89) ); If you put that in the chrome developer console (ctrl+shift+i), it'll return 10958, using only "Mathematical functions" (in programming context).

[@rolfmohme6188]

I confirm the result of the original author. I could not find a "normal" solution even though I tried potentiating and allowed complex intermediate results. The best approximation is: 1+(2-3^4^(5/6/7))^-8+9 = 10958.00205791.