6174 - Numberphile

6174 is also known as Kaprekar's Constant.
More links & stuff in full description below ↓↓↓
This video features University of Nottingham physics professor Roger Bowley, who is also featured on our Sixty Symbols channel at / sixtysymbols
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[@carlton6953]

02:57 damn this professor just ended this man's life and didn't even stutter 😂😂

[@kkycble47]

@SuperSatandevil666I tried it and it reached 6174.

[@WantoTo-it5if]

Ajarin aq cara menghitung nnya boss ku

[@CasualTrainspotter]

Starting from 0001 to 9998 there are 705 unique combinations limited by the rules. 20 of them take 1 step to converge 34 - 2 steps 140 - 3 steps 129 - 4 steps 113 - 5 steps 153 - 6 steps 116 - 7 steps

[@MdRaza-iw3cm]

Hj

[@gautam6956]

4 digit does not mean 0001 it will be counted as a 1 digit number

[@jakobandrews2096]

@@gautam6956 its exactly the same as using 1000 though so doesnt matter

[@brighthades5968]

@@MdRaza-iw3cm aitch jay

[@silkan7]

Very clever now explain why instead of what

[@carlmarsh8824]

Kaprekar's Constant actually has practical application. 6174 is a useful pedagogic tool. Since the Constant works with (almost) all four-digit numbers, it makes the great basis for a game.Have your young math students drill on fast subtraction by racing to the answer. The competition takes the boredom out of practicing subtraction, and Kaprekar's Constant is unforgiving: if you make a mistake in subtraction, you won't reach 6174 in the minimum number of steps. 7 iterations is all you should require (at the most) to reach the answer. It's a great way, in my opinion, to drill on subtracting rapidly and accurately, because the Constant always detects mistakes. Having four or five people race to calculate the answer is a fun way to hone working fast, but accurately. You can't cheat in this game, and you become fluent in working with numbers.

[@Emil-yd1ge]

That's a great idea. It's not as unforgiving as you think though, even if you make some mistakes you'll still always reach 6174- it might just take more steps.

[@paul55604]

@@Emil-yd1ge It might also take fewer steps. If you start on a number carefully chosen to take 7 steps and make a random mistake, maybe you would find yourself on a number that only takes 2 steps (like the examples in the video).

[@RajeshSingh-yh5qs]

NM

[@nithumb2483]

Loo o

[@nithumb2483]

Loo o oo o

[@Daealis]

Had to do a little numbercrunch on this to find out the iterations and how they are distributed. So in order to reach 6174, this is how many numbers reach it with said amount of iterations : 1 iteration : 384 numbers 2 iterations: 567 3 iterations: 2400 4 iterations: 1272 5 iterations: 1517 6 iterations: 1656 7 iterations: 2185

[@brotolanolysingle8164]

MaLcolm lll l l l l 'll LN l ll Lolo l ppl" l l l l l l l ll l lll 'll l l l L l 'll l ' l l l pl a l. L 'll l l l l ll l p l l l l l a l LaHood ll l l l a l l l l ll lll ll l L lL l ll L l 'l l l L l p ll l p lol Ppl l l 'lla a l l ll l l l l l ll l L L luLl l" l'll lbe Pulaski 'll l l l 'll l l ll a ll l l ll l ll l l l l a l l l l L l 'll p l l l P L 'll Ll lL L LL l l ploppeddown l last ll luLl l l 'll L ll l 'll l l ll l 'lll l l l l l l llllll L l l l l l a ll l L l L l L l p a l l la l l l l l l l l l a 'll L ll 'll l L l Men lol l l ll ll l l ll

[@Sattaking-uf1ge]

👈👈👈₈₈₇₅₈₄₀₁₃₇.....👈👈j

[@ajitkumarrouth9712]

Yr

[@stagdragon3978]

Is there any kind of pattern that forms in the iterations based on the number increased? for example if I were to have 2197 and then 2198 etc while of course skipping ones with the same number. Would I start to see a pattern form in the number of iterations?

[@suhadibewok7468]

Gttg yg t Rp

[@TooMuchDad]

"Not everything has to be useful to be appealing and fun" Too true! :D

[@thec-m]

For the people saying that numbers such as 1000 do not work, they actually do! 1000-0001=0999 (It has to be a 4 digit number) 9990-0999=8991 9981-1899=8082 8820-0288=8532 8532-2358=6174! [Edit: I'm probably one of the many people who have commented this, not that it really matters though. This post is old, almost 3 years old as I write this, I'm not sure why it has been getting a lot of attention recently... Admittedly, I should have mentioned everyone in one comment, but I scrolled through the comment section to find anyone who reported otherwise. Too late now, since I couldn't know who is who.]

[@muzyshayk8006]

Do you realise your comment is 10 years old? 😅

[@CazTanto]

So glad I finally got to checking these older vids. This is now one of my favourites especially as (at 21 and a half years old) its forced me to finally get to grips with Carry over Subtraction so I could properly try this out :)

[@manjunathchivatagundi5826]

JOtJ ObSR,

[@Melpheos1er]

This works with 5 numbers too, you will always end up with a rotation starting at 82962 There are probably other series with 5 numbers

[@maxwell8866]

Maybe it's always true no matter the number of digits. I wonder if this can appear in anything other than base10...

[@error.418]

@@maxwell8866 Yeah, afaik that hasn't been proven yet, just a conjecture.

[@error.418]

Yep, also works for 3 digits, again barring repdigits

[@Yottifferent]

And instead of it being 6174 it’s 61974 this time

[@williampatrick8814]

​@@Yottifferent tending to golden ratio ?

[@aesth1cc]

One of the last non clickbaiting titles on KZbin

[@fabulator2779]

Yup

[@unreal-the-ethan]

i dunno man, something about 6174 just seems so clickable to me

[@kisikabaapbhinhikaatsaktat3170]

8750678505))) singal Jodi for wats app

[@basavarajshetgar532]

@@unreal-the-ethan +

[@venkateshamankam4588]

Q

[@JosephDuffy94]

Another great episode, loving the series so far!

[@headshotgamer9232]

L loom o

[@disawarok7150]

⁹⁸⁹⁶⁰⁷²⁶⁰⁴....

[@gurdeepsinghchahal655]

Ok

[@gravityfallsguru5964]

I just realized something: when you take any number, multiply it by zero, and add 7644, it will akways come out to be 7644! Worsk every time!

[@Plus2Cuber]

wow it does

[@okie9025]

why do you spell words like that

[@tjreynolds685]

what about 1/0?

[@DSN.001]

NO. IT DOES NOT BECOME 7644 FACTORIAL IT BECOMES 7644

[@EulyDerg]

n∈ℕ; n*0 = 0 n*0 + 7644 = 0 + 7644 = 7644 Correct.

[@denfarc]

This is great. It's fun and very interesting to know. Amazing numbers. 😊👍👍

[@Noqism]

A new channel and more professor Bowley? Oh my, is it Christmas already?

[@garetr]

I've seen this done with 3 digit numbers as well. If one does the same thing with a 3 digit number (in which not all of the digits are the same), they always end up with 495. (And, as with 6174, if one puts, 495 through this process, it comes out as 495 again.)

[@JustinK0]

@@Shambagai checked, pretty much no, with 5 digits it goes in a cycle 63954... 61974... 82962... 75933... Then back to 63954 with 6 digits its like this 851742... 750843... 840852... 860832... 862632... 642654... 420876... then back to 851742...

[@sanderhfl]

Wow! I love this 1!

[@catakuri6678]

short and straight to the point :)

[@user-he4dk4sb7h]

My favorite professor, PROFESSOR ROGER BOWLEY

[@numberphile]

@RobertSeattle I'm not sure... I'm pretty sure there is a three digit version... Worth looking into hey?

[@neerudahel2498]

8222094434 yah mera WhatsApp number hai

[@sayedizaanahmad]

It always converge to 495

[@f4u5tus]

This is the video that got me hooked on numberphile

[@carlosm5711]

Interestingly, if you look at the position of 6174 between the first 10000 numbers, a very close approximation to the golden ratio pops up: 10000/6174 = 1.6196954972465

[@Irondragon1945]

That actually seems more like a coincidence. ... right?

[@tormentor2285]

@@Irondragon1945 there is golden ratio even in a certain type of electronic circuits, under certain conditions

[@Irondragon1945]

@@tormentor2285 still no, any number with leading digits 618 and around will, when inversed, give you the golden ratio. You're gonna need a better explanation

[@DeJay7]

@@Irondragon1945 but why were the leading digits 618 (or 617 here)? Technically Kaprekar's constant could be any number

[@tormentor2285]

@@Irondragon1945 consider a very long circuit made as a ladder, powered by single voltage source, where you ad two resistors parallel to a previous one each time. basically L structures one after another with resistors with variable of R1 and R2. by the property of constant resistance the source "sees" resistance R0 before each L. If you consider the structure at the very end it will have an R0 parallel to R2. to find R0, you can put R0=R1+(R2//R0)=R1+(R2R0/(R2+R0). if it is also true that R1=R2=R, by solving this you get R0=Rφ

[@hemanthpattem9949]

That moment after 3:00 when the 'wig' joke goes over the top of the head.

[@kisikabaapbhinhikaatsaktat3170]

8750678505))))) singal Jodi for wats app

[@Dhadkanmadhu]

Opan 6...0

[@emailjwr]

am i the only person who always opens up excel and programs something to prove this stuff? love this channel!

[@sbrunner69]

No you are not. Was thinking to knock together a c# example. I did the pi one with fractions was cool.

[@aiocafea]

replying to a 10 year old comment i see

[@joshua.ryan.powell]

Replying to a 0.5 year old comment I see.

[@stargazeronesixseven]

Amazing! Thank You So Much for this Fun Maths tutorial! 🕯🌷🌿

[@hezatulolaoli4313]

Thank you my friend

[@flowerofsilver]

Hahahah grand! Love love love this channel!

[@MrOldprof]

@Kargoneth Yes, I'm still here, doing research and making videos. Any requests will be carefully considered, provided they can be turned into a video.

[@fruity4820]

The first thought i had when he described the procedure was "i want to make a script for that"

[@stevefrandsen7897]

Thanks Roger! More number magic for my repertoire.

[@rajeshsaini7820]

फरीदाबाद। 15।सही। गलत

[@dickoamadou4372]

Bonjour

[@dickoamadou4372]

@@rajeshsaini7820 bonjour

[@AniruddhaKale-ys2bq]

Thanks to kaprekar

[@numberphile]

@Jim777PS3 thank you... do spread the word for us (or should that be number.. hmmm?)

[@RahulRahul-ft4do]

Hi

[@JordanSommer]

FINALLY! Someone else who writes 7 with the bar!

[@PhillipAlcock]

Me too! Comes from have to write coding forms (?) as a student / young engineer, that we’re then converted into punched cards I think.

[@Puleczech]

Also, the step before the last one is 61974. It's that Kaprekar number with 9 stuck in the middle :)

[@Zazzauser]

"You haven't got anything on the top of your head, but that's a wig, isn't it?" I was expecting a "Ba-dum-TSS" afterwards

[@abbasmia3566]

Sar you no thai lothario tips

[@NickWaffleson]

cant wait to show this to my friends

[@pkf1r3pk]

wow this is really awesome. 6174 is currently my favourite number now! >:)

[@SaveSoilSaveSoil]

Very interesting! Thank you for sharing!

[@mdhhh6066]

5r

[@abir_existz7325]

This also works with 3-digit numbers where you will get 495, 5-digit numbers where you will get a rotation of the numbers 53955, 59994, 61974, 62964, 63954, 71973, 74943, 75933, 82962, 83952 and 2-digit numbers where you will keep rolling out multiples of nine.

[@abir_existz7325]

edit: grammatical mistake

[@alxjones]

Are there versions of Kaprekar's Constant in other number systems, such as binary, hexadecimal, etc.?

[@janaki3829]

I made a spreadsheet to do the calculations for me, and found some interesting stuff It seems like any base 5*2^n has a constant, but some of them have cycles as well Some of the constants for smaller bases are 0111 and 1001 for base 2 3021 for base four 3032 for base 5 6174 for base 10 9:2:11:6 for base 15 12:3:15:8 for base 20 There's probably sampling bias in there though, because I really need to study for an exam so just looked into what's right in front of me

[@DamianReloaded]

I wonder if there is any relation between the original number and the number of iterations need to get 6174. Like if you add the original digits the result is the number of iterations, or something cool like that :)

[@portalsrule1239]

0, 0, 495, 6174, 0 The ending of this process for differently-lengthed numbers. (from 1 to 5) (except for some anomalies where it ends at 0)

[@giorgigusakov719]

Investigated by the program all valid numbers from 1 to 9998. The result was that the Maximum iterations = 7 (2184 numbers) The most biggest number of numbers - 2400 - come to the Kaprekar's Constant in 3 iterations. Minimum number of numbers - 384 - come to the Kaprekar's Constant in 1 iteration. The apparent pattern nowhere. You can find the hidden.)

[@DepFromDiscord]

Георгий Гусаков no need to comment the same thing twice

[@simonmasters3295]

@@DepFromDiscord that's harsh...

[@gandlebot]

I LOVE THAT GAME. My friend played that game with me once. I couldn't figure out out, so I asked him all the numbers from 1 to 20 and graphed it. He laughed at me. I deserved it.

[@karwos1992]

i did write a program to check the theory shown in video and it works! You need to take at maximum 7 steps to get number 6174 from any 4 digit number (exept 1111 etc.)!

[@jgtb0pl]

What about 1000?

[@jgtb0pl]

Cause I also have written this program and it's breaking on 1000

[@CorderoDeDios86]

This made me write a program and run through the first 10,000 numbers. It works for all and the maximum number of iterations happen to be 8 (for quite a few numbers)

[@steffen5121]

Which ones?

[@waseemahmed9365]

Very nice work

[@numberphile]

@SuperLaugh20 you are!

[@neerudahel2498]

Hi

[@praetorian85]

Tried it, seems I can't subtract any more. Thanks school...

[@surajvkothari]

Funniest part of the video (2:57): "You haven't got anything on the top of your head. That's a wig, isn't it?" 😂

[@ARBB1]

Brutal!

[@ajeydevadiga6652]

just amazing......

[@ilikevines]

Kaprekar really did have a lot of time on his hands.

[@trunkszetto]

glad to see some more of Dr. Bowley after his retirement. :)

[@DadiSpriydi]

8o al hasil

[@Incepter.]

6174 can be an unusual number in Kaprekar’s constant as the old man demonstrated here with Numberphile.

[@jaypi6423]

This reminds me of Langton's ant. I love these videos.

[@audiblemagician6751]

The five digit version eventually repeats int to a pattern n,n,n,n....and these numbers (in any order) 82962,75933,63954,61974 (in any digit order)

[@zixuan1630]

use 96715. that doesn't work with these.

[@aizennickoadante5572]

97651 - 15679 = 81972 98721 - 12789 = 85932 98532 - 23589 = 74943 97443 - 34479 = 62964 96642 - 24669 = 71973 hmm

[@chrisg3030]

Add 6174 to its reverse 4716 and you get 10890, which is also what you get when you do the same to the famous 1089.

[@eddiemaiden2012]

thank you nice work

[@spyderkam9225]

“Don’t have anything off the top of your head, that’s a wig in it?”

[@emailjwr]

I just did it in Excel, and this is the pattern of solutions (6174 reached) after x iterations 1 iteration: 384 numbers solved 2: 576 3: 2,400 4: 1,272 5: 1,518 6: 1,656 7: 2,184 these add up to 9990 of course :p

[@sheepslayer649]

“That’s a wig, isn’t it?”

[@sudamadeshlahre7321]

Very wonderful number sir

[@peterbaader334]

Brilliantly interesting!

[@DamianReloaded]

@SchumiUCD True. You are right. I was kind of hoping that there could be some deeper mathematical reason why the operation leads to 6174 :) Also, the sumatories of the multiples of nine at 1, 12, 112, 1112, (and so on) are 9, 594, 55944 and on, which are also multiples of nine. If you divide those numbers by nine you get another very pretty symmetric number. Probably not related to this, but fun, nonetheless. lol

[@error.418]

Also 6174 can be written as the sum of the first three degrees of 18. Also 6 + 1 + 7 + 4 = 18. Also the sum of squares of the prime factors of 6174 is a square.

[@DaviddosantosDavid]

Thank you mister

[@latifnaqshbandi3030]

Nice work

[@SeaFan1310]

Is there a number like this for other amounts, if not all amounts of digits? I have found that for 2 digits there is a loop 9 -> 81 -> 63 -> 27 -> 45 ->9, with every number entering the loop by the 2nd implementation, and on the first ether enter the loop or get a reverse of a loop component. and for 3 digits the number is 495, with all numbers reaching it by the 6th implementation.

[@faboo93]

Been through ten iterations from "mope" in base-26 (a = zero, b = one, etc.). So far I've got "vmle," and I think I'll have to go through a lot more paper, if I even decide to keep going.

[@error.418]

Try coding instead...

[@MiyuKuhaku]

20th April 2021 AD-Julian calendar, 2:56 AM (GMT+1) The day I felt in love with math after getting really interested in it 2 days ago.

[@sbsheikh6151]

Very nice video

[@BangaviSlamovi]

Very interesting. I tried multiple times and it worked. :) I like your videos.

[@anilgoutham1334]

Paused and laughed at that wig dig for about 2 minutes 😂❤️

[@KasabianFan44]

+Austin Ghelli Yes, the second number in the second line should be 0378, not 0387.

[@bornnaked2928]

Awsome party trick :D I´m going to try that at the next family event :P

[@joebykaeby]

"Gonna make this up off the top of my head, uh..." "You haven't got anything on the top of your head" XDDDD

[@kalinbhaiya3709]

*KALYAN FIX 2 JODI OR 2 PANEL JIS KISI BHAI KO BHI CHAHIYE ho JO LOSS ME HO WO BHAI WHATSAPP KARE 9685656907*

[@daneru]

Tried with 5 digits. It does not get to a definite number but repeats the same numbers every 4th iterations after a while.

[@daneru]

Works with 3 digit also. It will be always 495 at the nth iteration.

[@makingconnections777]

Cool also a 9

[@baddog1309]

0:50 This is the lottery formula and it cannot be denied

[@greasyclean]

is there any known relationship between the number you start with and the number of iterations necessary before reaching the constant?

[@adibakarlen]

my question too

[@TyphoidBryan]

This is fun, thank you!

[@ddborze5930]

Thank you ....

[@nhungmai8077]

cam on Thay nhieu lam Thay chot so qua chuan.chuc Thay luon chief Thang a

[@luismark3038]

I've chosen 7263. 7632-2367=5265 6552-2556=3996 9963-3699=6264 6642-2466=4176 7641-1467=6174

[@ferosekhan5649]

👍😁

[@DamianReloaded]

@richardathome65 Heh, I'm having quite a lot of fun with this. Watch this: There are 1112 multiples of 9 between 9 and 9999 (0,9,18,...,9999). The number 6174 is the 686th multiple of nine among those. 6174 / 686 = 9;

[@MrMebigfatguy]

The obvious question not answered is what's the longest chain?

[@patrickwienhoft7987]

7, but it appears a lot (2184 times)

[@todorsostar7974]

try this trick with number 2364

[@nielsdenbesten1852]

11 (2111)

[@michaelempeigne3519]

6432 - 2346 = 4086 8640 - 0468 = 8172 8721 - 1278 = 7443 7443 - 3447 = 3996 9963 - 3699 = 6264 6642 - 2466 = 4176 7641 - 1467 = 6174

[@oromain]

@@nielsdenbesten1852 2111 2111 - 1112 = 0999 9990 - 0999 = 8991 9981 - 1899 = 8082 8820 - 0288 = 8532 8532 - 2358 = 6174 5 Steps. Not 11.

[@FSMonster]

Interesting, a lot of comments with calculations feature the last calculation as 8532-2358. For example, I chose '6661' and of course if worked and ended with the same 8532-2358. Perhaps Dr. Gabor can explain how many out of 9990 end up with this as the last calculation?

[@tyleringram7883]

There are 2 numbers that you will get when your about to get 6174. Either 8532-2358 or 7641-1467. Will always end up like that

[@PrimusProductions]

Could you do a video on the order of operations and why they are that way; brackets, indices, multiplication and division, addition and subtraction

[@FiveToedSloth1]

That's numberwang!

[@arunghoderao1596]

Very nice sir

[@singlespies]

Cool, I wonder what number requires the most steps to get to 6174...

[@alirezasampour]

Most step is 7, numbers like 1004, 9985 and some more requires 7 step to become 6174

[@tribektiagustinus157]

Ku sms bulik&ibu.di pagi hari.kenudian ku nulis bpk.pada malam hari.prnahananya semua kena.sms.ku.mktn.salam lotry.ok?.

[@tribektiagustinus157]

Coba cek hpnya bpk sama ibu.khan ada kemiripan cuman beda.waktunya&jamnya.mktn.salam lotry.ok?.

[@LoseControlForDeath]

this calmed me after a fight with my father like nothing could, mathematics is really magic

[@apoorvsrivastava3816]

Mista: Something off in here, I can feel it

[@mdawladhasan8316]

I'm a big fan of yours

[@jsnadrian]

I was looking for a formal proof, which can be found in "The determination of all decadic Kaprekar constants," Fibonacci Quart (1981). Apparently 6178 and 495 are the only ones, but in modulus notation there are others, as shown by Walden, "Searching for Kaprekar's Constants: algorithms and results."

[@simonmasters3295]

Great addition...is the source available online?

[@itsaduck7633]

I was just checking out a 3 digit version on a calculator, and found the 495. I was also checking out a 5 digit version, and it SEEMS to have a broader loop. A start number ending with a zero (possibly sometimes not ending with a zero) seems to produce a variation of the digits 8 3 9 5 2. A start number not ending with a zero (possibly sometimes ending with a zero) seems to produce a variation of the digits 7 3 9 5 3.

[@pnutbutrncrackers]

How anyone would ever discover this 'constant' is beyond me.

[@pardeepgarg2640]

Beyond us

[@dhrvukatara943]

Nice video

[@MKder2te]

Amazing

[@Mojosbigstick]

D'oh! I used to do this as child, sitting for hours writing out numbers, reversing them and adding them up. I'd fill whole notebooks with it. If I'd only taken away instead of adding, that would have been called the Smith number :-(

[@3dplanet100]

Thumbs up if you grab a piece of paper and a pencil to try that.

[@danbuck11]

"but there isn't anything on the top of you head. that's a wig." GET OWNED BRADY

[@praveenkumardixit4650]

thanks

[@CazTanto]

Yes loop, as I too discovered, seems a better term than constant. The process for any number of digits always enters one of these loops of which all the outcome numbers will be multiples of 9, checkable by adding its digits together and repeating until you get a more recognisable multiple of 9 or just back to 9 itself. 3 digit and 4 digit numbers just happen to have a single number loop. Infact try 2 digits; you get a very familiar 5 number loop. With their flipped counterparts its even clearer

[@AdamHecht]

did it with 3124, worked after 5 cycles! I can't wait to show my son!