6174 - Numberphile
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- čas přidán 11. 09. 2024
- 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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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.
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.
@@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).
NM
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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
Hj
4 digit does not mean 0001 it will be counted as a 1 digit number
@@gautam6956 its exactly the same as using 1000 though so doesnt matter
@@MdRaza-iw3cm aitch jay
Very clever now explain why instead of what
02:57 damn this professor just ended this man's life and didn't even stutter 😂😂
@SuperSatandevil666I tried it and it reached 6174.
Ajarin aq cara menghitung nnya boss ku
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
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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?
Gttg yg t Rp
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.]
Do you realise your comment is 10 years old? 😅
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!
wow it does
why do you spell words like that
what about 1/0?
NO. IT DOES NOT BECOME 7644 FACTORIAL IT BECOMES 7644
n∈ℕ;
n*0 = 0
n*0 + 7644
= 0 + 7644
= 7644
Correct.
"Not everything has to be useful to be appealing and fun"
Too true! :D
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
That actually seems more like a coincidence. ... right?
@@Irondragon1945 there is golden ratio even in a certain type of electronic circuits, under certain conditions
@@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
@@Irondragon1945 but why were the leading digits 618 (or 617 here)? Technically Kaprekar's constant could be any number
@@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
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
Maybe it's always true no matter the number of digits. I wonder if this can appear in anything other than base10...
@@maxwell8866 Yeah, afaik that hasn't been proven yet, just a conjecture.
Yep, also works for 3 digits, again barring repdigits
And instead of it being 6174 it’s 61974 this time
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 :)
JOtJ
ObSR,
One of the last non clickbaiting titles on CZcams
Yup
i dunno man, something about 6174 just seems so clickable to me
8750678505))) singal Jodi for wats app
@@unreal-the-ethan +
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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.)
Георгий Гусаков no need to comment the same thing twice
@@DepFromDiscord that's harsh...
That moment after 3:00 when the 'wig' joke goes over the top of the head.
8750678505))))) singal Jodi for wats app
Opan 6...0
0:50 This is the lottery formula and it cannot be denied
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.)
@@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...
851742 is an anagram of 142857
Another great episode, loving the series so far!
L loom o
⁹⁸⁹⁶⁰⁷²⁶⁰⁴....
Ok
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.
edit: grammatical mistake
am i the only person who always opens up excel and programs something to prove this stuff? love this channel!
No you are not. Was thinking to knock together a c# example. I did the pi one with fractions was cool.
replying to a 10 year old comment i see
Replying to a 0.5 year old comment I see.
"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
Sar you no thai lothario tips
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
I've chosen 7263.
7632-2367=5265
6552-2556=3996
9963-3699=6264
6642-2466=4176
7641-1467=6174
👍😁
Are there versions of Kaprekar's Constant in other number systems, such as binary, hexadecimal, etc.?
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
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)
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)
use 96715. that doesn't work with these.
97651 - 15679 = 81972
98721 - 12789 = 85932
98532 - 23589 = 74943
97443 - 34479 = 62964
96642 - 24669 = 71973
hmm
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.)!
What about 1000?
Cause I also have written this program and it's breaking on 1000
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.
The first thought i had when he described the procedure was "i want to make a script for that"
@RobertSeattle I'm not sure... I'm pretty sure there is a three digit version... Worth looking into hey?
8222094434 yah mera WhatsApp number hai
It always converge to 495
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)
Which ones?
FINALLY! Someone else who writes 7 with the bar!
Me too! Comes from have to write coding forms (?) as a student / young engineer, that we’re then converted into punched cards I think.
this calmed me after a fight with my father like nothing could, mathematics is really magic
A new channel and more professor Bowley? Oh my, is it Christmas already?
Also, the step before the last one is 61974. It's that Kaprekar number with 9 stuck in the middle :)
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.
I took "5369"
9653 - 3569 = 6084
8640 - 0468 = 8172
8721 - 1278 = 7443
7443 - 3447 = 3996
9963 - 3699 = 6264
6642 - 2466 = 4176
7641 - 1467 = 6174
Wonder what the max/min amount of calculations is when doing this.
+Dashed
6642-2466=4176
7641-1467=6174
+Dashed Minimum?
7641-1467=6174
.
+( ͡° ͜ʖ ͡° )TheNoobyGamer 6174 would be the minimum, actually
Jose GomezFranco
what did I just post
Works with 3 digit also. It will be always 495 at the nth iteration.
Cool also a 9
for 5 digit numbers its 82962, but it doesn't work in the same way. Everything leads to 82962, but 82962 leads to 75933 which goes to 63954 which goes to 61974 which goes back to 82962, so there's a larger loop. I don't know for 6 and 7 digits, but you can easily find out. Start with 654321-123456, order the digits, and continue iterating until you get a loop.
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
It does.
You have to add zeroes to make it a 4 digit number after you subtract.
9998-8999 = 999
9990-0999 = 8991
9981-1899 = 8082
8820-0288 = 8532
8532-2358 = 6174
Ah THANK you!! I tried 2111 to keep below 6 even after subtracting, so got 2111-1112=999. I now know I got 0999.
Tried with 5 digits. It does not get to a definite number but repeats the same numbers every 4th iterations after a while.
“Don’t have anything off the top of your head, that’s a wig in it?”
The obvious question not answered is what's the longest chain?
7, but it appears a lot (2184 times)
try this trick with number 2364
11 (2111)
6432 - 2346 = 4086
8640 - 0468 = 8172
8721 - 1278 = 7443
7443 - 3447 = 3996
9963 - 3699 = 6264
6642 - 2466 = 4176
7641 - 1467 = 6174
@@nielsdenbesten1852
2111
2111 - 1112 = 0999
9990 - 0999 = 8991
9981 - 1899 = 8082
8820 - 0288 = 8532
8532 - 2358 = 6174
5 Steps. Not 11.
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?" 😂
Brutal!
Tried it, seems I can't subtract any more. Thanks school...
I don't know why i did that, but look:
for all numbers you need 0 - 7 iterations (subtractions) to reach 6174.
also:
Need only 0 iteration for 1 number (it is 6174)
Need only 1 iteration for 356 numbers
Need only 2 iterations for 519 numbers
Need only 3 iterations for 2124 numbers
Need only 4 iterations for 1124 numbers
Need only 5 iterations for 1379 numbers
Need only 6 iterations for 1508 numbers
Need only 7 iterations for 1980 numbers
Kaprekar really did have a lot of time on his hands.
So, I wrote some code and it seems like no matter which four-digit number you choose (except for the exceptions) you always end up with doing no more than 7 operations until you hit 6174.
Some output:
Number of operations: 1; Occurrences: 383
Number of operations: 2; Occurrences: 576
Number of operations: 3; Occurrences: 2400
Number of operations: 4; Occurrences: 1272
Number of operations: 5; Occurrences: 1518
Number of operations: 6; Occurrences: 1656
Number of operations: 7; Occurrences: 2184
And every result along the calculation is a multiple of 9 or generally the highest digit value of the current number system.
"Gonna make this up off the top of my head, uh..."
"You haven't got anything on the top of your head"
XDDDD
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I would be curious to know if similar constants occur for other n-digit numbers, or if this is the only one.
So i yeeted it at my computer.
Numbers going to zero is because i didn't filter out things like they did in the video
like i allowed
1111→0 and 2111 → 889→99→0 for example
Length 1: all numbers turn to 0
Length 2: all numbers turn to 0
Length 3: all numbers turn to 495 or 0
Length 4: all numbers turn to 6174 or 0
Past here i haven't checked every single number, but i used random sampling.
Length 5: numbers go to one of these loops:
(0,)
(53955, 59994) gets about 1/30
(61974, 82962, 75933, 63954) gets about 14/30
(62964, 71973, 83952, 74943) gets about 1/2
Length 6:
(0,) about 0.01%
(420876, 851742, 750843, 840852, 860832, 862632, 642654), about 93%
(549945,) about 0.1%
(631764,) about 6%
I'm not gonna keep saying exactly what loops there are because there are so many, just what lengths of loops exist. (ignoring going to 0)
7: 8 (also there is 1 loop all numbers go to)
8: 1,3,7
9: 1,14
10: 1,3,7
11: 1,5,8 this is the last anomalous one
Between 12 and 18, evens have loops 1,3,7, odds have loops 1,2,5
It seems that length≥19 evens have 1,3,5,7 and odds have 1,2,3,5
Also powers of 2 ≥16 always having loops of length 2.
Tho for large lengths the proportions get more extreme, so the numbers appear to vanish (i don't think they actually do tho), in the order 2,7,5,1.
So lengths 1 and 2 don't work, lengths 3 and 4 have one fixed point all things go to, lengths 5 and 7 only get into loops, and everything else gets either into a loop or a fixed value.
I also tried it in other bases, and it seems to get into a pattern eventually, for base N often at around N or N^2.
Sometimes the pattern is a loop, sometimes its a loop but new things are added every now and then.
Checking for patterns takes longer so i haven't checked much, but so far it seems that powers of 2 stop increasing. The only other numbers I've found that stop increasing are 10 and possibly 20.
And I can't be sure any of these do stop increasing, I can't properly check and large bases lead to potentially hiding some loops like how 2,7,5 and 1eventually become too rare to find in base 10.
@@d.l.7416 2111 does not go to 0. You're not following the algorithm correctly. You're not supposed to leave off leading zeros. 2111-1112=0999, 9990-0999=8991, 9981-1899=8082, 8820-0288=8532, 8532-2358=6174
@@M0z1ng0 in the video they stated that repeated digits weren't allowed right? probs because of this situation.
you're way is a valid way to extend the rule but i don't like leading zeros so i didn't do it like that. its really a matter of choice.
EDIT:
my main reason is i prefer it as a function on numbers not strings of digits
like 123 = 0123.00 in numbers
but "123" ≠ "0123" in strings of digits
@@d.l.7416 All digits being the same isn't allowed, it would cause an instant 0. Mattison's way is correct.
💯💥👈
My favorite professor, PROFESSOR ROGER BOWLEY
If you're wondering how many combinations of numbers there are limited by the rules, eg. 9998-0001, there are 705. I had to break it down into "Triples" such as those, "Doubles" which I broke down into double "starters" (9987) "Middles" (9887) "Enders" (9877) and "double twins" (9988.) Then there were just "Singles" where no digit was ever used twice (9876, 9875... 9430... 3210.)
do you play the lotto
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 :-(
9:90-9=81-18=63-36=27,72-27=45,54-45=9; 74943:97443-34479=62964;96642-24669=71973;97731-13779=83952;98532-23589=74943 this begs a few questions; 1. can math(s) explain why there isn't a single answer to the other sequences? 2. will all other sets make similar sequences or will some have specific answers? 3. if some will have specific answers, might there be a function that can help find or define said number? (another video in the makeing?)
Musahid
9612056288
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 :)
I studied the K-constant now more than 30 years ago and succeeded to demonstrate that stability in 6174 is reached at most in 8 cycles for the 4-digit non-uniform set of numbers. Besides the practical way, by constructing an algorithm that tries out every such number, I got a theoretical proof by studying the properties of those numbers. Now I would need some help to conduct research of K-constants in other numerical systems other than the decimal system. Do they exist in all bases? Sole vertices or cycles? After all, it is a numerical or algebraic property? Thanks for any help or hint.
+Argentarii Homini , yes I found also the multiconvergence pattern but could not estabilish any sensible rule yet. My main concern was to discover if the k-constant phenomena and others alike are base-dependent or not, that means if they are arithmetic or algebraic (or mixed?) rules.
+Argentarii Homini
Well, the first time I learnt about the K-constant, it was from an ancient Science or SciAm review, where clearly stated in the form of a chalange that the longest iteration was of 8 cycles. By the way, I tried by hand all the 9990 eligible 4-digit numbers (excepting 0000, 1111, .... 9999).
+Argentarii Homini Yes, in those times there were almost no handy computers (IBM /360 and Fortran IV were the latest cry), besides the handy work allowed to discover some very beautiful properties of the number chains. Of course the search was not linear but followed the 'number-trees' or chains, by families. This colored pattern idea you suggested seems to be valuable to examine further. My ultimate goal was to find a unique algebraic description of the K-const that explains its nice property.
+Argentarii Homini I've been having the same experience and am beginning to wonder if there is in fact a single constant in base 16 like 6174 is in base 10. What seems to be special about that number is that if you subtract it from 7641, having put the digits in descending order of value, you get another permutation of the same digits in ascending order,1467. Is there a four digit number in base 16 which does the same? Another feature (as I note in another comment) is that if you add 6174 to its reverse you get 10890, the same result of adding 1089 to its reverse. The congruent counterpart of 1089 in b16 is 10EF, since you get 10EF0 by doing the same, and also by reverse-and-adding C1D4 and C3B4 - but neither of those two otherwise behave in a very Kaprekar like way. (The three-digiter 7F8 seems to however)
I would like to see this proof.
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."
Great addition...is the source available online?
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.
Take any number that's divisible by 3 and sum the cubes of it's separate digits. Keep doing that and you'll come to 153 and can't go any further since 1 cubed +5 cubed + 3 cubed=153
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.
Try coding instead...
6174 can be an unusual number in Kaprekar’s constant as the old man demonstrated here with Numberphile.
Cool, I wonder what number requires the most steps to get to 6174...
Most step is 7, numbers like 1004, 9985 and some more requires 7 step to become 6174
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This is great. It's fun and very interesting to know. Amazing numbers. 😊👍👍
The number you reach just before you get to Kaprekar's Constant is always some arrangement of the numbers 8, 5, 3, and 2... 24 possible combinations. I think this whole process goes pretty deep...
yes
No 6013
“That’s a wig, isn’t it?”
is there any known relationship between the number you start with and the number of iterations necessary before reaching the constant?
my question too
I might have missed it if someone has already stated this: but if not, here is an interesting fact-- If you start with, say, 8641 and run it until you get to 6174, you will go through the same numbers in the steps if you start with 7530, where each digit is one less than the original number's. Try it. I know the reason, but it is a secret.
@Jim777PS3 thank you... do spread the word for us (or should that be number.. hmmm?)
Hi
Thanks Roger! More number magic for my repertoire.
फरीदाबाद। 15।सही। गलत
Bonjour
@@rajeshsaini7820 bonjour
Thanks to kaprekar
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.
One interesting fact here is that there are at most 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = *54* "meaningfully different" flavors of four digit number, in the following sense.
The rule tells us to compute (A B C D - D C B A), which is equivalent to (999A + 90B - 90C - 999D) = 999(A - D) + 90 (B - C).
So if you give me any four digit number, I'll just compute "A - D" (biggest minus smallest digit) and "B - C" (second biggest minus second smallest digit), and those two quantities together _uniquely_ determine my next number.
Put differently, there are only at most 54 possible numbers at *step 2* of this process, no matter what number you give me at step 1.
Thumbs up if you grab a piece of paper and a pencil to try that.
Paused and laughed at that wig dig for about 2 minutes 😂❤️
short and straight to the point :)
I'll save this video and I'll try write the code for it in my spare time and really confirm that this actually works for all 4 digit numbers where atleast one of them is different. I'll also check what gives the most and least number of iterations and perhaps even find some sort of pattern 🤷♂️. I don't know. I guess I'll just have to wait and see.
Brilliant. It even works if 3 of the 4 digits are the same (I just tried 1112)
It's kind of spooky - I mean, 6174 looks so random, but in starting with any 4 digit number, and so long as even 1 digit is different from the other 3, then you always hit this 'terminus' of this random-looking number.
I hear the Twilight Zone music.
I instantly felt this way.
How does 1112 work
I see that 5173 will get you a Kapekars' constant after the first iteration.
But can you mathematically calculate how many "first iteration Kapekar Constant's" there are?
384
Wow! I love this 1!
I'm studying for an exam in object oriented programming (java), and watched a couple of youtube clips, including this one, in a break. Now i want to try to make a program that can either randomly generate such a 4 digit number or take it as input, and then test how many iterations it needs to become the Kaprekar's Constant. Perhaps even have it test 10.000 random numbers and find how many of them needs how many iterations (statistics), and which one required the most :)
Similarly if we does this with any 3 digit number the will be always be 594
Hahahah grand! Love love love this channel!
How anyone would ever discover this 'constant' is beyond me.
Beyond us
glad to see some more of Dr. Bowley after his retirement. :)
8o al hasil
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?
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
@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
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.
+Austin Ghelli Yes, the second number in the second line should be 0378, not 0387.
wow this is really awesome. 6174 is currently my favourite number now! >:)
Here's another mathematical oddity:
1. Pick ANY number
2. Subtract the number from itself
3. You will ALWAYS get 0
*x-files theme starts playing*
The Manga “We Never Learn” referenced this Constant in Chapter 88 and it brought me here again :)
You will be allright since the subtractions will generate a number containing a number larger than 6 for you. For instance: 5432 - 2345 = 3087 which gives you 8730 to work with.
@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.
did it with 3124, worked after 5 cycles! I can't wait to show my son!
cant wait to show this to my friends
so i tried this with 2 digits 3 digits & 5 digits and 3 digits comes out to be 495 (954-459=495) but 2 and 5 become loops that start with 09 and 74943 respectively (in my mind i liken them to "happy" #s),(9:90-9=81-18=63-36=27,72-27=45,54-45=9);(74943:97443-34479=62964;96642-24669=71973;97731-13779=83952;98532-23589=74943)
same goes for 3 digit numbers will end up being 495
what about 5 digits
This is called a boustrophedonic transform. ( flipping letters and numbers ) Ancient stuff. They used this to create sigils by drawing lines across magic squares, the sigils were formed by the values of the alphanumeric values of the letters in the words.
@SuperLaugh20 you are!
Hi
**ad comes up before vid starts**
*you can skip ad in 56174 seconds*
hey i will tell u some interesting finding...
i will tell you at what step you can exactly get 6174 if you reply
Please tell me
Seems to me that this is probably just a feature of the decimal system. If you did the same with base 3 or base 52, you would probably end up with a different constant.
I’ll try it with binary. Edit: It’s apparently 1110. (Binary for 14)
Edit again for some reason: ok binary doesn’t have one at all because now I tried it starting with 1100 instead of 1000 and got 1100 instead
Edit yet again: Base 3 doesn’t have one. Try starting with 2100.
I've encountered a problem with /6354/...
6354 -> 6543-3456= 3087
3087 -> 8730-0387= 8343
8343 -> 8433-3348 = 5085
5085 -> 8550-0558 = 7992
7992 -> 9972-2799 = 7173
7173 -> 7731-1377 = /6354/
Did I make an error somewhere?
I realize this is ancient history but for posterity's sake, your second subtraction is fault: 8730-0378=8352, not 8343
Austin C. G. Secon row, 0378, not 0387
6354 -> 6543-3456= 3087
3087 -> 8730- 0387-*= 8343 ----------------------------> it must be 0378 NOT 0387
There is a digit you ignored which is zero, 999 is actually 0999 but we ignore the zeros,
so 1112 would be
2111-1112=0999
and arrange that to 9990-0999=8991
8991=> 9981-1899=8082
8082=> 8820-0288=8532
8532=> 8532-2358= the jackpot 6174
Remember you require the numbers you use to be of FOUR digits always..