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
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).
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.]
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?
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.
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...
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.)
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 :)
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
@@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 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
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?)
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.
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
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.)
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
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
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.
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.)!
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.
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)
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.
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)
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.
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.
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
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
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
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.
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
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.
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?
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 :)
@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.
I don't know what A069746 is trying to express. The first one on that list goes 1000-->0999-->8991-->8082-->8532-->6174. Every else, except the monodigits work.
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.
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)
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?
Take any 1 digit (1-9), write 3 of theme an the Last one make one bigger this is the only time this isnt working but it's still enough to disprove 2:28
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."
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.
Yes, you're right. It can be done like that but I still dislike ending up with 3 digit number then placing it in 4 digits since it's numerical value is, in actuality, a 3 digit number. By placing 0 in front of it you increased it's numerical value by 8991(out of nothing) once digits are rearranged. Then again, in order to even do this method of calculation we have to change numerical values of resulting numbers by rearranging their digits, so why not add a digit as well eh? Anyway, good point.
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...
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 :-(
I'd previously known about the procedure for which 6174 is an attractor, but didn't previously know who discovered that attractor 6174. Kaprekar -- might have known it was him. He delights in publishing numerical quirks which rely on how numbers are represented in decimal. OK then, what are the results for other bases and/or number lengths?
I've played around with 5-digit numbers still in decimal and there are a few cycles but all those I've found so far have length that divides 4. (Length 1 if you start with all digits equal. A cycle of length 2 is 59994,53955. )
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.
It does work. First you get 1000 - 0001 = 0999 but you get a zero in the thousands place which gets shifted around to form the subtraction 9990 - 0999 and on you go :)
You must have messed up along the way. I checked 5762 manually and got it to 7 operations: 7652-2567=5085; 8550-0558=7992; 9972-2799=7173; 7731-1377=6354; 6543-3456=3087; 8730-0378=8352; 8532-2358=6174. It might also be so that our perception on what an operation is differ. Whereas I count an occurrence where substraction has been made as an operation(and nothing else), you might have also included the ranking of the numbers.
Turns our were both wrong, you move the digits around, and leave in the '0' that is in the ten-thousands place. So 999 is actually 0999, which gives you: 9990 +0999 So you can keep going.
It's not that those numbers don't work, you just have to remember that you are working with 4 or 3 digit numbers, therefore 9998-8999 does not = 999, it = 0999. 9990-0999 = 8991 9981-1899 = 8082 8820-0288 = 8532 8532-2358 = 6174. 998-899 = 099 990-099 = 891 981-189 = 792 972-279 = 693 963-369 = 594 954-459 = 495 954-456 = 495. I tried nicely spacing these line by line, but youtube keeps stuffing it all into 1 paragraph. If this one does the same I give up.
someone else here in the comments claimed that this is the highest number of possible iterations of 4 digit number to get to the constant. Someone else claimed 8.
It kind of works with 2 digit numbers! If you do this with 2 digit numbers you get a loop: 36-27-45-09 It also works with 1 digit numbers 9-9=0; 8-8=0 :P
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 :)
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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
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).
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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? 😅
Hi
Hi man how are you
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.
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?
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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
"Not everything has to be useful to be appealing and fun"
Too true! :D
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
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
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...
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,
u r 32 now
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
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
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
One of the last non clickbaiting titles on RUclips
Yup
i dunno man, something about 6174 just seems so clickable to me
8750678505))) singal Jodi for wats app
@@unreal-the-ethan +
Q
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)
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.
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
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
@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
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.
@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.
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.
Works with 3 digit also. It will be always 495 at the nth iteration.
Cool also a 9
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
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.
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've chosen 7263.
7632-2367=5265
6552-2556=3996
9963-3699=6264
6642-2466=4176
7641-1467=6174
👍😁
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.
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?
@@steffen5121 3578
Similarly if we does this with any 3 digit number the will be always be 594
Also, the step before the last one is 61974. It's that Kaprekar number with 9 stuck in the middle :)
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
Tried it, seems I can't subtract any more. Thanks school...
The first thought i had when he described the procedure was "i want to make a script for that"
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...
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.
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.
Tried with 5 digits. It does not get to a definite number but repeats the same numbers every 4th iterations after a while.
Another great episode, loving the series so far!
L loom o
⁹⁸⁹⁶⁰⁷²⁶⁰⁴....
Ok
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.
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.
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
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.
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
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
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.
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.
💯💥👈
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.
@Jim777PS3 thank you... do spread the word for us (or should that be number.. hmmm?)
Hi
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
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 :)
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*
@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.
There are other numbers besides those with all digits equal that don't result in 6174 - see OEIS sequence A069746
I don't know what A069746 is trying to express. The first one on that list goes 1000-->0999-->8991-->8082-->8532-->6174. Every else, except the monodigits work.
is there any known relationship between the number you start with and the number of iterations necessary before reaching the constant?
my question too
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.
same goes for 3 digit numbers will end up being 495
what about 5 digits
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)
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
Take any 1 digit (1-9), write 3 of theme an the Last one make one bigger this is the only time this isnt working but it's still enough to disprove 2:28
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.
How anyone would ever discover this 'constant' is beyond me.
Beyond us
@SuperLaugh20 you are!
Hi
Yes, you're right. It can be done like that but I still dislike ending up with 3 digit number then placing it in 4 digits since it's numerical value is, in actuality, a 3 digit number. By placing 0 in front of it you increased it's numerical value by 8991(out of nothing) once digits are rearranged.
Then again, in order to even do this method of calculation we have to change numerical values of resulting numbers by rearranging their digits, so why not add a digit as well eh?
Anyway, good point.
A new channel and more professor Bowley? Oh my, is it Christmas already?
it works for all 3 digit numbers if difference of two of their digits is greater than 1. for example 343 wont work but 353 will :)
What date
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
“Don’t have anything off the top of your head, that’s a wig in it?”
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 :-(
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
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
Ku sms bulik&ibu.di pagi hari.kenudian ku nulis bpk.pada malam hari.prnahananya semua kena.sms.ku.mktn.salam lotry.ok?.
Coba cek hpnya bpk sama ibu.khan ada kemiripan cuman beda.waktunya&jamnya.mktn.salam lotry.ok?.
What starting number needs the MOST iterations to reach 6174?
My favorite professor, PROFESSOR ROGER BOWLEY
6174 can be an unusual number in Kaprekar’s constant as the old man demonstrated here with Numberphile.
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
I'd previously known about the procedure for which 6174 is an attractor, but didn't previously know who discovered that attractor 6174. Kaprekar -- might have known it was him. He delights in publishing numerical quirks which rely on how numbers are represented in decimal.
OK then, what are the results for other bases and/or number lengths?
I've played around with 5-digit numbers still in decimal and there are a few cycles but all those I've found so far have length that divides 4. (Length 1 if you start with all digits equal. A cycle of length 2 is 59994,53955. )
Kaprekar really did have a lot of time on his hands.
**ad comes up before vid starts**
*you can skip ad in 56174 seconds*
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
It works, you must continue until you get a repeating number. 8730 - 0378 = 8352 and 8532 - 2358 = 6174. Think about 9998 number xd.
Thanks Roger! More number magic for my repertoire.
फरीदाबाद। 15।सही। गलत
Bonjour
@@rajeshsaini7820 bonjour
Thanks to kaprekar
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.
What about 2111?
You guys managed to say the same thing 24 times. Congratulations.
2111-1112=0999
9990-0999=8991
9981-1899=8082
8820-0288=8533
8533-3358=5175
7551=1557=5994
9954-4599=5355
5553-3555=1998
9981-1899=8082
8820-0288=8532
8532-2358=6174
MINDBLOWN
line #4 is wrong :
8820-0288=8532 not 8533
then :
8532-2358=6174
@@limbonine1499 YESSSSSSS!!!!!
It does work. First you get 1000 - 0001 = 0999 but you get a zero in the thousands place which gets shifted around to form the subtraction 9990 - 0999 and on you go :)
Thumbs up if you grab a piece of paper and a pencil to try that.
You must have messed up along the way. I checked 5762 manually and got it to 7 operations:
7652-2567=5085; 8550-0558=7992; 9972-2799=7173; 7731-1377=6354; 6543-3456=3087; 8730-0378=8352; 8532-2358=6174.
It might also be so that our perception on what an operation is differ. Whereas I count an occurrence where substraction has been made as an operation(and nothing else), you might have also included the ranking of the numbers.
This is great. It's fun and very interesting to know. Amazing numbers. 😊👍👍
Turns our were both wrong, you move the digits around, and leave in the '0' that is in the ten-thousands place. So 999 is actually 0999, which gives you:
9990
+0999
So you can keep going.
“That’s a wig, isn’t it?”
It's not that those numbers don't work, you just have to remember that you are working with 4 or 3 digit numbers, therefore 9998-8999 does not = 999, it = 0999.
9990-0999 = 8991 9981-1899 = 8082 8820-0288 = 8532 8532-2358 = 6174.
998-899 = 099 990-099 = 891 981-189 = 792 972-279 = 693 963-369 = 594 954-459 = 495 954-456 = 495.
I tried nicely spacing these line by line, but youtube keeps stuffing it all into 1 paragraph. If this one does the same I give up.
"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
Can someone perform this procedure with the number 7248, I tried it but I didn't get 6174.
did it with 3124, worked after 5 cycles! I can't wait to show my son!
But what about the 4 digit no.s that are unique but on arranging in descreasing order, don't even reach 6174, like 1234, 5231, etc. ?
I chose 1236, and I have to do SEVEN iterations to get 6174!
someone else here in the comments claimed that this is the highest number of possible iterations of 4 digit number to get to the constant. Someone else claimed 8.
7 is indeed the highest number of iterations needed for this problem. All of these numbers above require seven iterations: 1010 1021 1031 1040 1042 1050 1052 1053 1060 1062 1086 1095 1096 1097 1152 1162 1163 1172 1196 1240 1250 1251 1253 1260 1261 1263 1264 1271 1273 1297 1350 1352 1354 1361 1362 1364 1365 1372 1374 1398 1420 1453 1462 1463 1465 1473 1475 1520 1521 1530 1532 1543 1554 1563 1564 1574 1576 1590 1620 1621 1631 1632 1642 1643 1653 1654 1664 1665 1675 1680 1690 1691 1721 1732 1743 1754 1765 1776 1790 1792 1860 1893 1950 1960 1961 1970 1972 1983 1994 2041 2043 2050 2051 2053 2054 2061 2063 2087 2096 2097 2098 2140 2150 2151 2153 2160 2161 2163 2164 2171 2173 2197 2263 2273 2274 2283 2340 2350 2351 2360 2361 2362 2364 2371 2372 2374 2375 2382 2384 2410 2430 2450 2461 2463 2465 2472 2473 2475 2476 2483 2485 2510 2530 2531 2540 2564 2573 2574 2576 2584 2586 2610 2630 2631 2632 2641 2643 2654 2665 2674 2675 2685 2687 2690 2731 2732 2742 2743 2753 2754 2764 2765 2775 2776 2780 2786 2790 2791 2832 2843 2854 2865 2870 2876 2887 2890 2960 2970 2971 2980 3042 3051 3052 3054 3062 3064 3097 3098 3150 3152 3154 3161 3162 3164 3165 3172 3174 3198 3240 3250 3251 3260 3261 3262 3264 3271 3272 3274 3275 3282 3284 3374 3384 3385 3394 3420 3440 3450 3451 3460 3461 3462 3471 3472 3473 3475 3482 3483 3485 3486 3493 3495 3510 3520 3521 3540 3541 3550 3561 3572 3574 3576 3583 3584 3586 3587 3594 3596 3620 3621 3640 3641 3642 3651 3675 3684 3685 3687 3695 3697 3721 3741 3742 3743 3752 3754 3765 3776 3785 3786 3790 3796 3798 3842 3843 3853 3854 3864 3865 3875 3876 3880 3886 3887 3890 3891 3897 3943 3954 3965 3970 3976 3980 3981 3987 3990 3998 4010 4021 4032 4043 4052 4053 4063 4065 4090 4098 4120 4153 4162 4163 4165 4173 4175 4210 4230 4250 4261 4263 4265 4272 4273 4275 4276 4283 4285 4320 4340 4350 4351 4360 4361 4362 4371 4372 4373 4375 4382 4383 4385 4386 4393 4395 4430 4485 4495 4496 4520 4530 4531 4550 4551 4560 4561 4562 4571 4572 4573 4582 4583 4584 4586 4593 4594 4596 4597 4621 4630 4631 4632 4650 4651 4652 4661 4672 4683 4685 4687 4694 4695 4697 4698 4731 4732 4751 4752 4753 4762 4786 4795 4796 4798 4832 4852 4853 4854 4863 4865 4876 4887 4890 4896 4897 4953 4954 4964 4965 4975 4976 4980 4986 4987 4990 4991 4997 4998 5010 5020 5021 5031 5032 5042 5043 5053 5054 5064 5080 5090 5091 5120 5121 5130 5132 5143 5154 5163 5164 5174 5176 5190 5210 5230 5231 5240 5264 5273 5274 5276 5284 5286 5310 5320 5321 5340 5341 5350 5361 5372 5374 5376 5383 5384 5386 5387 5394 5396 5420 5430 5431 5450 5451 5460 5461 5462 5471 5472 5473 5482 5483 5484 5486 5493 5494 5496 5497 5530 5540 5541 5596 5631 5640 5641 5642 5660 5661 5662 5671 5672 5673 5682 5683 5684 5693 5694 5695 5697 5732 5741 5742 5743 5761 5762 5763 5772 5783 5794 5796 5798 5842 5843 5862 5863 5864 5873 5897 5910 5943 5963 5964 5965 5974 5976 5987 5990 5998 6010 6021 6032 6043 6054 6065 6081 6090 6091 6092 6120 6121 6131 6132 6142 6143 6153 6154 6164 6165 6175 6180 6190 6191 6210 6230 6231 6232 6241 6243 6254 6265 6274 6275 6285 6287 6290 6320 6321 6340 6341 6342 6351 6375 6384 6385 6387 6395 6397 6421 6430 6431 6432 6450 6451 6452 6461 6472 6483 6485 6487 6494 6495 6497 6498 6531 6540 6541 6542 6560 6561 6562 6571 6572 6573 6582 6583 6584 6593 6594 6595 6597 6641 6650 6651 6652 6742 6751 6752 6753 6771 6772 6773 6782 6783 6784 6793 6794 6795 6810 6843 6852 6853 6854 6872 6873 6874 6883 6894 6910 6920 6953 6954 6973 6974 6975 6984 7082 7091 7092 7093 7121 7132 7143 7154 7165 7176 7190 7192 7231 7232 7242 7243 7253 7254 7264 7265 7275 7276 7280 7286 7290 7291 7321 7341 7342 7343 7352 7354 7365 7376 7385 7386 7390 7396 7398 7431 7432 7451 7452 7453 7462 7486 7495 7496 7498 7532 7541 7542 7543 7561 7562 7563 7572 7583 7594 7596 7598 7642 7651 7652 7653 7671 7672 7673 7682 7683 7684 7693 7694 7695 7752 7761 7762 7763 7820 7853 7862 7863 7864 7882 7883 7884 7893 7894 7895 7910 7920 7921 7930 7954 7963 7964 7965 7983 7984 7985 7994 8050 8061 8072 8083 8092 8093 8094 8160 8193 8232 8243 8254 8265 8270 8276 8287 8290 8342 8343 8353 8354 8364 8365 8375 8376 8380 8386 8387 8390 8391 8397 8432 8452 8453 8454 8463 8465 8476 8487 8490 8496 8497 8542 8543 8562 8563 8564 8573 8597 8610 8643 8652 8653 8654 8672 8673 8674 8683 8694 8720 8753 8762 8763 8764 8782 8783 8784 8793 8794 8795 8830 8863 8872 8873 8874 8920 8930 8931 8940 8964 8973 8974 8975 8993 8994 8995 9040 9050 9051 9060 9061 9062 9071 9072 9073 9082 9083 9084 9093 9094 9095 9150 9160 9161 9170 9172 9183 9194 9260 9270 9271 9280 9343 9354 9365 9370 9376 9380 9381 9387 9390 9398 9453 9454 9464 9465 9475 9476 9480 9486 9487 9490 9491 9497 9498 9510 9543 9563 9564 9565 9574 9576 9587 9590 9598 9610 9620 9653 9654 9673 9674 9675 9684 9710 9720 9721 9730 9754 9763 9764 9765 9783 9784 9785 9794 9820 9830 9831 9840 9864 9873 9874 9875 9893 9894 9895 9930 9940 9941 9950 9974 9983 9984 9985
@@steffen5121 Its pretty easy to write a python program or something to check
@@shawntai "above"
Aizen Nicko Adante lol I meant below
It kind of works with 2 digit numbers!
If you do this with 2 digit numbers you get a loop: 36-27-45-09
It also works with 1 digit numbers 9-9=0; 8-8=0 :P
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 :)