What's the connection between 495 and 6174? Add 495 to its reverse, 594, and you get that other famously cool number 1089. Do the same to 6174 and you get 10890. But then 1089 + 9801 = 10890 too.
Any of these mathematical mysteries will always have 9 as the fulcrum. Add the constants here and (4+9+5=18 or 1+8 = 9) or (6+1+7+4) and the result is 9. Magical 9.
Our no system is decimal , with the largest non-repeated no being 9 we have such a pattern. Similarly for an hexadecimal system u will have 15 aka A as accent point.
Exactly!..... As soon as I saw the video I added up the the digits of both the numbers. Most magical things in this complete universe have a relation with 9. Even Nicola Tesla was amazed of number 9.He said no.9 could explain the universe. For me both 9 and 3 are Magical numbers.Idk why lol.
For 5-digit numbers, there are 3 possible loops: [74943, 62964, 71973, 83952] [63954, 61974, 82962, 75933] [53955, 59994] For 6 digit numers, there are 1 loop and 2 'magic' numbers: [851742, 750843, 840852, 860832, 862632, 642654, 420876] [631764] [549945] The 851742 reminds me of 1/7
It's the number 9. 9 will always equal itself and looks as if it mirrors itself. 9,18,27,36,45 then it flips 54,63,72,81, .. you can add the numbers shown unit you are left with 9.. 1+8=9. First example 6174 6+1+7+4=18 8+1=9. 9 stays 9 always it ends any patterns or cycles. 3 always turns to 6 and the other is 1,2,4,8,7,5.... 5 turns to 0 and is nothing so you start with 1 again. 10 decimal counting system. These are the patterns.
The real magic number is 73. In the 73rd episode of The Big Bang Theory Sheldon explains that: 73 is the 21st prime number, its mirror 37 is the 12th prime number which is also the mirror of 21, the multiplication of 7 and 3 is also 21, and in binary code 73 is a palindrome
something curious I noticed about this: if you do the same sort of thing with two digit numbers, you get nine times the difference between the larger and smaller digit (ie. 64 and 46 become 18, 33 becomes 0, 90 and 9 become 81, etc) and when I tried to do it with five digit numbers, I cycled back to my original set of digits(though I only tested this twice) here's the numbers I got: 41976(start) 82962 75933 63954 61974(this one bugged me considerably given Kaprekar's four digit constant) 82962(end) 82963(start) 74943 62964 71973 83952 74943(end)
For the two digit numbers this happens because, A two number is of the form 10x+y where x,yy then 10x+y > 10y+x Subtracting gives you (10x+y) - (10y+x) = 9(x-y) i.e 9 times the difference of digits where x-y is the difference. It becomes harder to prove the more digits you add because you have to assume which of the digits are greater than which other
As a kid I sometimes played silly games like this. Maybe some mathematicians still have that kind of kid inside ;) which together with some degree of intuition and their expertise can lead further than just playing a silly game?
Note that for 3-digit numbers, the first step necessarily gives 99n, with n being one of 1 through 9, inclusive. So maybe the "homing in" on 495 (which is one of those 9) isn't quite so remarkable. Still, that's pretty interesting. For the 4-digit numbers, you'll always get some multiple of 999 (from the 2 outer digits) plus some multiple of 90 (from the 2 inner digits). 999m + 90n, where m and n are each between 1 and 9, inclusive. That makes 9x9 = 81 possible results each time. [CORRECTION: n can also be 0, because the two inner digits can be equal. This raises the possibilities to 9x10 = 90.] The fact that the process, in both cases, always ends on a value that repeats forever, rather than a cycle of two or more that repeat, is remarkable; the fact that it's always the same final number, is even moreso. For 2-digit numbers, it will just cycle around the first 5 odd multiples of 9, in the order: (81, 63, 27, 45, 09). What does it do for 5-digit and longer numbers? Fred
I tried octal in Python by hand, then noticed the link in the video description, which has a table of constants for bases 2-16 at various digit lengths -- my octal constant matched :)
HOW THE "INTERMEDIATE STEP ANSWERS LOOK" IN A KAPREKAR SEQUENCE OF CALCULATIONS When we go through a Kaprekar Sequence of calculations, we come up with many "intermediate step answers". Each of those intermediate step answers becomes the "starting point" for the next step of calculation, after the digits are "rearranged" as per the rules (largest digit on the left to the smallest digit on the right). Here we are discussing how the digits of the "intermediate step answers" might look (before rearranging them again) and what properties they might have, when 2-digit, 3-digit and 4-digit numbers are used in a Kaprekar Sequence of calculations. (1) For 2-Digit Numbers For 2-digit numbers, all "Intermediate Step Answers" will be "multiples of 9". Example: Start with 91. Then, 91-19 = 72, which is an "intermediate step answer" and that is a multiple of 9. Now subtract 27 from 72 and get 45, which is an "intermediate step answer" and also a multiple of 9. Rearranging 45 as 54 and subtracting 45 from it we get to the Kaprekar Constant "9". You will find that for 2-digit numbers, all "intermediate step answers" in a Kaprekar Sequence will be one of the numbers 9, 18, 27, 36, 45, 54, 63, 72 and 81. NOTE: You can write the number "9" as "09" just to make it appear as a two-digit number. Often you have to do this for 3-digit and 4-digit Kaprekar sequences as well. If a digit is missing, write it such that the first number is "0". (2) For 3-Digit Numbers For 3-digit numbers, in all "Intermediate Step Answers" the "middle digit will be 9", while the "outer two digits will add up to 9". Example: Start with 961. Subtracting 169 we get 792, which is the first "intermediate step answer". Note that the "middle digit is 9" and the two outer digits (which are 7 and 2) add up to 9. Continue by rearranging 792 as 972. Subtracting 279 from 972 we get 693 as an "intermediate step answer". Note that the middle digit is 9 and the outer digits add up to 9. Rearranging 693 as 963 and subtracting 369 we get 594. The center digit is 9 and the outer two add up to 9. Rearranging 594 as 954 and subtracting 459 we get to the Kaprekar Constant of 495. The center digit is 9 and the outer two add up to 9. The number 495 cannot be further reduced because it is the KC for 3-digit numbers. You will find that does not matter what 3-digit number you start with and proceed, in each "intermediate step answer" the center digit will be "9" and the outer digits will add up to "9". (3) 4- Digit Numbers For 4-Digit Numbers, there are two distinct cases. In the first case, we consider a starting number where the 100s digit is greater than the 10s digit. For example, consider the starting number 9871. In this case the 100s digit "8" is greater than the 10s digit "7". This is the first case, which we will call "Sub-Case_1". In the second case, we consider a starting number where the 100s digit is equal to the 10s digit. For example, consider the starting number 9881. In this case, the 100s digit and the 10s digit are both "8". This is the second case, which we will call "Sub-Case_2". We will find that the two sub-cases above will produce different looking "intermediate step answers" and they have to be treated separately. (3) SUB-CASE_1 : 4-Digit Numbers (where the 100s digit is greater than the 10s digit) For 4-digit Numbers where the 100s-digit is "greater than the 10s digit" at the starting point, the next "Intermediate Step Answer" will be such that the "middle two digits will add up to 8" and the "outer two digits will add up to 10". Example: Start with 9871. Note that here the 100s digit is "8" and that is greater than the 10s digit "7". In this case, subtracting 1789 from 9871 we get 8082. Note that the two middle digits add up to "8" (0+8=8) and the two outer digits add up to "10" (8+2=10). Continue by rearranging 8082 as 8820. At this step too, the 100s digit (or 8) is greater than the 10s digit (or 2). Subtracting 0288 (or 288) from 8820 we get 8532. Note that the two middle digits add up to 8 and the two outer digits add up to 10. Now we can subtract 2358 from 8532 and we come to the Kaprekar Constant which is 6174. Here too the two middle digits add up to 8 and the two outer digits add up to 10. You will find that does not matter what 4-digit number you start with, as long as the 100s digit is greater than the 10s digit, in each "intermediate step answer" the middle two digits will add up to "8" and the two outer digits will add up to "10". (3) SUB-CASE_2 : 4-Digit Numbers (where the 100s digit is equal to the 10s digit) For 4-digit Numbers, when the 100s-digit is "equal to the 10s digit" at the starting point, in all Intermediate Steps the "middle two digits" will both be "9" and the "outer two digits" will add up to "9". Example: Start with 9881. Note that here the 100s digit and the 10s digit are the same. In this case, subtracting 1889 from 9881 we get 7992. Note that the two middle digits are both "9" and the outer digits add up to "9" (or 7+2). Continue by rearranging 7992 as 9972. In this case, the 100s digit is greater than the 10s digit and the next intermediate answer will follow "Sub-Case_1" above. Consider another example starting with 9998. Here the 100s and 10s digit are equal. Subtracting 8999 from 9998 we get 999. We should rewrite it as 0999 because that is what it really is. Note that the two middle digits are both "9" and the outer digits add up to "9" (or 0+9=9). You will find that does not matter what 4-digit number you start with, as long as the 100s digit is equal to the 10s digit, in each "intermediate step answer" the middle two digits will be "9" and the two outer digits will add up to "9". END WORD: I have been able to prove "analytically" as to why the above observations are bound to be true in every case. That means I have used only "w, x, y and z" as the individual digits in each case and prove the same, instead of using "specific number examples" as I have done above. My proofs are very simple and I am sure that they can be followed by all high school students. However, it would take some more space and therefore, I am not doing it here. Hopefully people who have come this far liked it. Mathematics is a joyful affair. Is it not?
@@prakhyatpandey5341 Yes it is considered as a digit but number starting with such as 023,024, here "0" is not a significant number and there are 2 significant numbers here whereas if you consider decimals such as 23.0 or 25.06 there are 3 and 4 significant numbers respectively
I can see why you would say that. If this problem interests you. There are some great papers written about this. I recommend reading up on dynamical seuences.
@@randomdude9135 So you are interested. I do not know much about the topic, but papers on the Collatz Conjecture frequently make mention of it. Here is a link. mathworld.wolfram.com/DynamicalSystem.html On that site check out the references at the bottom for further reading.
If we take any 2 digit no say 34 and subtract it with its reverse the ans we get in the first step will always same after 4 iterations with the preceding answers.
I see number 18 as special case of 9 involved here 6+1+7+4=18 4+9+5=18 The sum of digits on each step of subtracting the smaller from the larger add up to 18
Packno1 Hahaha🤣🤣Very funny. This mathematics (Not of this video) might never be used by you. Ok tell me??? How your ekectronic devices work??? Your GPS??? There are so many things, i have no time to make it undersatnd for you
many of the western math axioms and theorems are still very much useless in the real world. instead of criticizing, maybe you'd try to look at the interest it generates when you're at boredom.
I think 9 is the most magical unique number because per example if we have a number like 57 and we add 9 to it (57+9=66) so we have 57 can behave like 3( 5+7=12=1+2=3) and the same thing for 66 (6+6 =12 = 1+2 =3 ) we can rather conclude that 9 behaves like 0 in terms of adding another trick is if we take any number like 653 and we subtract it by it renverse (653 - 356 = 297) we always get a number that can be divided by 9
Karthik Nadig It is because the results were drawn from the subtraction of Largest and smallest number from same digit,which is always a multiple of 9.And any number which is multiple of 9 has a digital sum of 9. Digital sum is the sum of the digits of a number, until you reached one digit number.
There are many more exceptions Eg. 988-889 = 99 Like wise 110-011 221-122 211-112 322-223 332-233 433-334 443-344 And so on... all these numbers gives remainder as 99. Because of thses, numbers like 343, 767, 898, 878 etc. All comes into exception
if you get a two digit number then you must put a 0 at the front. Start with 100 -> 100 - 001 = 099 -> 990 - 099 = 891 -> 981 - 189 = 792 -> 972 - 279 = 693 -> 963 - 369 = 594 - > 954 - 459 = 495. Notice how each iteration the final number of the 'big' number goes up by one and the final number of the 'small' number goes down by one until they eventually meet in the middle (495)
Maximum 3 digit number is 999. You allowed zero in 3 digit numbers, essentially allowing 2 digit numbers. Total one digit numbers are 9, we want to exclude them from our math, so 999 - 9 = 990. 990 ÷ 2 = 495. I.e. 495 is in between 3 digit numbers including zero at either end or middle. When we subtract highest from lowest by rearranging numbers, we thrive towards middle point, which is 495. Something similar logic could be true for 4 digit numbers, I havn't worked out it. Edit : In case of 4 digit numbers, since we allow 3 and 2 digit numbers too, like 2600, 2450, 2000 etc, middle point would be, 9999 + 2 × 999 + 3 × 99 = 12294 12294 ÷ 2 = 6174 So we still thrive towards middle.
It's kind of decimal numbering system's properties. Decimal numbering system is forming from 0 to 9 numbers. And hence 9 got it's magical characteristic property.
Indians had been always done wonderful works either it is invention of surgery, invention of 0, invention of shampoo, invention of sugar, invention of vastu, invention of equation, invention of wheel(mentioned in the Mahabharat), invention of water on moon, invention of plastic sugery, invention of chess, invention of button, invention of economics(Mentioned in book Ashtadhyayi), etc
@@michaeladams2074 Because britishers riined the economy, we are seen as duffer and fools but they don't know that India was one of the richest country before britishers invaded.
I was expecting more discusión about this results, like : is it a property of decimal system or do this numbers exists for all the bases ? Does this work for any. Inner of digits ? (Clearly the numbers will change) etc
No. In fact, this is a bit of wool pulled over the collective's eyes, in as much as 3 and 4 digit repunits (000, 111, 222, ...) all go to 0 in one cycle, so there are two fixed points ("magic numbers") for 3-digit numbers already. For 2 digits there is a "0" fixed point and a cycle of length 5 with the odd multiples of 9 (09-81-63-27-45).For 5 digits there is the "0" fixed point for repunits and 3 cycles of varying length. As far as I know, for all N-digit numbers there is a finite number of cycles depending on N, but 3 and 4 are the only cases where there are only fixed points (cycles of length 1).
For 5 digits numbers it seems you get a loop of 4 numbers that keep repeating themselves, but it seems like they are different for every starting number. Howewer there is a pattern, the 4 loop numbers always have a 9 as a digit and the sum of the other digits is 18.
@MindyourDecisions, did u know that if u take the difference of a number and its twisted version, that its always divisible by 9 for numbers with 2 digits, u need to take the difference of the first digit of those two numbers to actually become the difference divided by 9, for example: 92-29=> 9-2=7, and 7*9= 63 which is 92-29, i figured it out for 3 digital numbers as well; its about the same, but i dont know it for 4 digital numbers
3-digit logic does not work on 101, 998, 566 and similar numbers as well. 4-digit version has lots of exceptions. Like logic will not work if 3 digits will be the same and etc. All in all was very interesting, thanks ^_^
now try doing a proof of why the three digit numbers approach 495 using this algorithm. took me about 1 1/2 pages but finally found that if you start out with N = A*100 + B*10 + C*1 = [A, B, C], N' = [A-C-1, 9, 10+C-A] N'' = [9-A+C, 9, A-C] N''' = [8-A+C, 9, 1+A-C] N'''' = [7-A+C, 9, 2+A-C] ... Plug in A=4, B=2, C=0 (420) into N''' and you will get N''' = [4, 9, 5] Also, you can use this to prove why A=B=C digits won't work because you can define N(prime whatever) as [X-A+C, 9, Y+A-C] Y = 9-X, so N(prime whatever) = [X-A+C, 9, 9-X+A-C] A=C, so N(prime whatever) = [X-2A, 9, 9-X] X-2A = 4 and 9-X = 5 ---> X=4 (4)-2A = 4 2A = 0, which says that the only value where A=B=C works is when A,B,C are zero.
If you want to do this trick with 1 or 2 digit numbers, you'll end up to 0. For 5 digits there is a cycle of 3-4 numbers, which are different for different every input number.
I can visualise what's happening but it'll take too long to describe it. Cool children's party trick tho. Props to the guy who was bored enough to stumble upon it.
594 is one of the numbers before which we get 495 and it loops. I wonder if 549 and 459 are the other possible "loop-entry" numbers. Besides that, numbers with 3 repeating digits like 222 and 777 are 'blackhole' numbers that will never bring you to 495. My intuition is that we can never reach such numbers. I wonder if it is possible to prove this. These are my two thoughts.
The condition is that "all the digits cannot be the same" in the number you start with. That is the ONLY restriction Kaprekar imposed. You can choose 998 but not 999. You can choose any other number - either 3-digit or 4-digit - and it will work. So, you cannot start with 222 or 777. You can start with any 3-digit number but after rearranging, reversing it and then subtracting the latter from the former, you will ALWAYS come up with a number which will have "9" as the "middle digit", while the "other two" will add up to "9". This will hold true of "any intermediate step". There can be no exception to this rule! In your own example you came to 594. Notice that "9" is the center digit and the outer digits add up to "9" (5+4=9). The final KC number 495 also follows this rule, "9" in the middle and the outer two adding up to "9". Therefore, you can never have 549 and 459 as "loop entry numbers" because you will never see them at any intermediate step. In these numbers you have "4" and "5" as the middle digits and the "outer digits" do not add up to "9". Therefore, your examples are impossible at any intermediate step. If you are interested, you can look at my detailed post titled HOW THE "INTERMEDIATE STEP ANSWERS LOOK" IN A KAPREKAR SEQUENCE OF CALCULATIONS, where I have discussed this matter in detail. I have discussed the cases for 2-digit, 3-digit and 4-digit Numbers in that post.
NOTE: I later realized that 2-digit numbers do NOT have a Kaprekar Constant. I have written that as a reply to my own posting under "INTERMEDIATE STEP ANSWERS LOOK" IN A KAPREKAR SEQUENCE OF CALCULATIONS above. If you go there, you can read what I have written.
The trick is found a number. Except those similar digits . Such that arrangement .in the mentioned video doesn't change. 10-01=09 90-9=91 91-19= 72 72-27=45 54-45=09 Now my law. . you start with any 2 digits number you end up with following repeated pattern . .
Didn't work for me. When he said to take a three digit number, I picked my favorite one: 666. That gave me zero at the next step, and stayed there. Only afterwards did it show the note about multiples of 111 not working, a fact which I consider to be fundamentally unfair. ;-) On a more serious note, it would be interesting to see which numbers have the longest path to reach 495/6174 -- Wikipedia says the longest streaks are 6 to reach 495 and 8 to reach 6174 -- and the mathematical proof of why this works. Will have to play around with this some more...
Using 4 digits to get 5174 count the iteration count its always between 0 and 7 very useful music intervals colour 7 colours so check all numbers 0000 to 9999 plot on 100x100 grid top line is 0000 to 0099 next line 0100 to 0199 do all 100 lines asign a colour 0 black 1 red 7 violet look at that patter, now choose key c major compute all 10000 values create chord on interval 0 to 7th wow Now 5 digits do similar but no static number but it cycles at some trigger number 5 , 6 7 digitsvtry these too
doe not work for 5 digits does not converge at least I have found a counter example 👉 I pick the number 63954 then 1)96543-34569=61974 2)97641-14679=82962 3)98622-22689=75933 4)97533-33579=63954 and then it starts all over again lol let's make America great again, I suddenly remember
Weird, if you do 5 digit numbers, it kind of cycles endlessly though a series instead of sticking to one. Still, 9's everywhere though. Also weird, 495*9 = 4455, and 6174*9 = 55566. If you try to reverse the process and do things like 666677, or 344, you get repeating decimals instead of whole numbers. I thought that had something to do with it, but I tried 1111122 (same pattern, repeating digits with two digits one higher on the end, divisible by 9) and it cycles through about 7 or 8 numbers infinitely. Well, screw it, I didn't need sleep tonight anyway. Also, useless fact, 6174/495 is 686/55. Also, math is ultra weird. If you imagine 686 as a pattern of a number on each end, and the middle filled with a number two higher, then 55 is 5's with 0 7s between. 686 is 6's with 1 8 in between. Here's why it's weird. 686/55 is 12.47272727 repeating. If one less is 4's on both ends with "-1" 6 in the middle, just make it 4 (because you have to remove a number). 4/55 is 0.0727272727 repeating. 55/55 is obviously 1, 686/55 is that weird number, but hey...looks familiar. What if we do (686/55)-(4/55)? 682/55 is 12.4. Okay, keep moving up. 7997/55 is 145.4? From here we have two possibilities. 80008 would be next? or is it 8(10)(10)(10)8? If it's the second, we have to assume it's 91108 (as if we're carrying the one, right?) Well... 80008/55 = 1454.690909 repeating (wait, that's almost exactly [7997/55]*10), 91108/55 = 1656.509091... MORE... 911119/55 = 16565.8 (that's awfully similar to the second one above times ten...) The second version would be 1022219 (with carrying), so 1022219/55 = 18585.8... One more... 10222230/55 (ends become 10, last 10 carries one over)? 185858.7273 (Oh...that's...also familiar...), and finally, 11333330/55 (with full carrying) = 206060.5455 So there's this odd pattern where carrying with one is almost exactly 1/10 of the non-carrying version after it, each time raising from 1656... to 1858... to 2060...? At this point it's not even math, it's just nonsense...but I love it. :D
No - you did it wrong. 100-001=099 is correct. Now rearrange it as 990 and then do 990-099=891. Then continue. If you start with a 3-digit number, then the next answer also has to be 3-digit. If you get a 2-digit answer like "99", then make it a 3-digit number by putting a "0" in front (which in this case is 099). You cannot operate with 2-digit numbers at any intermediate step, to get to the Kaprekar Constant of 495 (starting from 3-digit numbers).
But if u write a three digit number as 4 digit number by appendix by 0 at the left then u shd get 6174. So three digit numbers can converge to two constants by K process
You know what's amazing? If you add 1 to a number, it becomes ONE larger.. but if you subtract that SAME, it is reduced by one. This is insane. You can do this all day long. TO ANY NUMBER?! *Mind* BLOWN.
Can you prove what he said if I had heard it correctly. Take any three digit number and do that routine procedure and you end up with 495. ps : Don't take Number Theory as a joke son!
You can do it all the numbers you have chosen. You have followed the wrong steps and that's why it seems you cannot do it. Please see my reply a few posts above.
I tried this for 5 digits and after a few tries I ended up with a few cycles: 63954, 61974, 82962, 75933 83952, 74943, 62964, 71973 For base 2, I got: 2 digits: 1 3 digits: 3 4 digits: 7 or 9 5 digits: too lazy to check
If you are getting 495 in the end, its mathematically proven that death is the end because 4=d, 9=I and 5=e. You 495(die).
Great
what about 6174??
f = 4 a= 1 g= 7 d = 4
Great. Karthik..
@@xxdeathxeaterxx44 so it means I dgaf?
Sorry just found it funny
What's the connection between 495 and 6174? Add 495 to its reverse, 594, and you get that other famously cool number 1089. Do the same to 6174 and you get 10890. But then 1089 + 9801 = 10890 too.
Amazing!
Magic of numbers!
Yeah....the last sum should be.....1089+9810=10899.....and not 1089+9801=10890
Kusum Trivedi Why??? He is saying about its reverse, not about largest number we can get of that.
@@kusumtrivedi1884 wrong
6174 + 4716 = 10890
6+1+7+4 = 4+9+5
@Tokisaki Kurumi this whole video wasn't a proof of anything, it was all just mildly interesting math
This is just another interesting fact
The sum is always 18
@@I_am_Itay yes
@@I_am_Itay
No the sum is 9
1+8= 9
I mean if we push to one digit
That was the point.
V.4.Vanquish L1 This would result same every time because Subtraction of any number with its reverse number is a multiple of 9.
Any of these mathematical mysteries will always have 9 as the fulcrum. Add the constants here and (4+9+5=18 or 1+8 = 9) or (6+1+7+4) and the result is 9. Magical 9.
Our no system is decimal , with the largest non-repeated no being 9 we have such a pattern.
Similarly for an hexadecimal system u will have 15 aka A as accent point.
Yes its true, i figured it out this too
Tesla also had a mystery about 369 number..
@@truebeliever174 Also there is 69 😅😅
Exactly!..... As soon as I saw the video I added up the the digits of both the numbers. Most magical things in this complete universe have a relation with 9. Even Nicola Tesla was amazed of number 9.He said no.9 could explain the universe.
For me both 9 and 3 are Magical numbers.Idk why lol.
For 5-digit numbers, there are 3 possible loops:
[74943, 62964, 71973, 83952]
[63954, 61974, 82962, 75933]
[53955, 59994]
For 6 digit numers, there are 1 loop and 2 'magic' numbers:
[851742, 750843, 840852, 860832, 862632, 642654, 420876]
[631764]
[549945]
The 851742 reminds me of 1/7
Well the 549945 reminds me of 495 and is a palindrome too!
If you shift the second loops in 5 digits so all left, they pair up such that loop 1 minus loop 2 always equals 0990 or 9999.
Clearly this is an amazing result but it would be more interesting to make the proof of this and show the reason of this result..
It's the number 9. 9 will always equal itself and looks as if it mirrors itself. 9,18,27,36,45 then it flips 54,63,72,81, .. you can add the numbers shown unit you are left with 9.. 1+8=9. First example 6174 6+1+7+4=18 8+1=9. 9 stays 9 always it ends any patterns or cycles. 3 always turns to 6 and the other is 1,2,4,8,7,5.... 5 turns to 0 and is nothing so you start with 1 again. 10 decimal counting system. These are the patterns.
The real magic number is 73.
In the 73rd episode of The Big Bang Theory Sheldon explains that:
73 is the 21st prime number,
its mirror 37 is the 12th prime number which is also the mirror of 21, the multiplication of 7 and 3 is also 21, and in binary code 73 is a palindrome
something curious I noticed about this:
if you do the same sort of thing with two digit numbers, you get nine times the difference between the larger and smaller digit (ie. 64 and 46 become 18, 33 becomes 0, 90 and 9 become 81, etc)
and when I tried to do it with five digit numbers, I cycled back to my original set of digits(though I only tested this twice) here's the numbers I got:
41976(start)
82962
75933
63954
61974(this one bugged me considerably given Kaprekar's four digit constant)
82962(end)
82963(start)
74943
62964
71973
83952
74943(end)
For the two digit numbers this happens because,
A two number is of the form 10x+y where x,yy then 10x+y > 10y+x
Subtracting gives you
(10x+y) - (10y+x) = 9(x-y) i.e 9 times the difference of digits where x-y is the difference.
It becomes harder to prove the more digits you add because you have to assume which of the digits are greater than which other
I too did the same and for the same results and I was suspecting 63954
You send up with 9
Which is actually 09
So 90 + 09 = 81
Which results in an infinite loop
I wonder how mathematicians came up with the idea to try this
The explanation of that would make this channel a proper math channel rather than a collection of magic tricks.
As a kid I sometimes played silly games like this. Maybe some mathematicians still have that kind of kid inside ;) which together with some degree of intuition and their expertise can lead further than just playing a silly game?
Simple: when they get bored.
mathematicians are just always number crunching..they play with these and they are great with pattern or sequence series
Like how we scrolled the youtube they scrolled their brain, !
So cool to see one of your older videos for the first time!
Note that for 3-digit numbers, the first step necessarily gives 99n, with n being one of 1 through 9, inclusive.
So maybe the "homing in" on 495 (which is one of those 9) isn't quite so remarkable. Still, that's pretty interesting.
For the 4-digit numbers, you'll always get some multiple of 999 (from the 2 outer digits) plus some multiple of 90 (from the 2 inner digits).
999m + 90n, where m and n are each between 1 and 9, inclusive.
That makes 9x9 = 81 possible results each time.
[CORRECTION: n can also be 0, because the two inner digits can be equal. This raises the possibilities to 9x10 = 90.]
The fact that the process, in both cases, always ends on a value that repeats forever, rather than a cycle of two or more that repeat, is remarkable; the fact that it's always the same final number, is even moreso.
For 2-digit numbers, it will just cycle around the first 5 odd multiples of 9, in the order: (81, 63, 27, 45, 09).
What does it do for 5-digit and longer numbers?
Fred
n can be zero and m ≥ n, there are 54 possible results for 4-digits
Indians here???
Or I am the only one?
Well look at usernames in the comment, most are indians
Here
Yes mai hu😊
Here
@@prachisrivastava6461 true
My question:
How do the numbers change in different bases
5Gonza5 It’s called _Kaprekar’s routine_ and it can be performed for any base. You may look it up on Google.
I tried octal in Python by hand, then noticed the link in the video description, which has a table of constants for bases 2-16 at various digit lengths -- my octal constant matched :)
This is greatly appreciated discovery in India...
I assume the catch is: NO repeat digits?
eric veneto duh and also it is mentioned in the video that it won’t work on all digits same
@Statiscube Not true.
655-565=090
900-9=891
981-189=792
972-279=693
963-369=594
954-459=495
HOW THE "INTERMEDIATE STEP ANSWERS LOOK" IN A KAPREKAR SEQUENCE OF CALCULATIONS
When we go through a Kaprekar Sequence of calculations, we come up with many "intermediate step answers". Each of those intermediate step answers becomes the "starting point" for the next step of calculation, after the digits are "rearranged" as per the rules (largest digit on the left to the smallest digit on the right).
Here we are discussing how the digits of the "intermediate step answers" might look (before rearranging them again) and what properties they might have, when 2-digit, 3-digit and 4-digit numbers are used in a Kaprekar Sequence of calculations.
(1) For 2-Digit Numbers
For 2-digit numbers, all "Intermediate Step Answers" will be "multiples of 9".
Example: Start with 91.
Then, 91-19 = 72, which is an "intermediate step answer" and that is a multiple of 9.
Now subtract 27 from 72 and get 45, which is an "intermediate step answer" and also a multiple of 9.
Rearranging 45 as 54 and subtracting 45 from it we get to the Kaprekar Constant "9".
You will find that for 2-digit numbers, all "intermediate step answers" in a Kaprekar Sequence will be one of the numbers 9, 18, 27, 36, 45, 54, 63, 72 and 81.
NOTE: You can write the number "9" as "09" just to make it appear as a two-digit number. Often you have to do this for 3-digit and 4-digit Kaprekar sequences as well. If a digit is missing, write it such that the first number is "0".
(2) For 3-Digit Numbers
For 3-digit numbers, in all "Intermediate Step Answers" the "middle digit will be 9", while the "outer two digits will add up to 9".
Example: Start with 961.
Subtracting 169 we get 792, which is the first "intermediate step answer". Note that the "middle digit is 9" and the two outer digits (which are 7 and 2) add up to 9.
Continue by rearranging 792 as 972. Subtracting 279 from 972 we get 693 as an "intermediate step answer". Note that the middle digit is 9 and the outer digits add up to 9.
Rearranging 693 as 963 and subtracting 369 we get 594. The center digit is 9 and the outer two add up to 9.
Rearranging 594 as 954 and subtracting 459 we get to the Kaprekar Constant of 495. The center digit is 9 and the outer two add up to 9. The number 495 cannot be further reduced because it is the KC for 3-digit numbers.
You will find that does not matter what 3-digit number you start with and proceed, in each "intermediate step answer" the center digit will be "9" and the outer digits will add up to "9".
(3) 4- Digit Numbers
For 4-Digit Numbers, there are two distinct cases.
In the first case, we consider a starting number where the 100s digit is greater than the 10s digit. For example, consider the starting number 9871. In this case the 100s digit "8" is greater than the 10s digit "7". This is the first case, which we will call "Sub-Case_1".
In the second case, we consider a starting number where the 100s digit is equal to the 10s digit. For example, consider the starting number 9881. In this case, the 100s digit and the 10s digit are both "8". This is the second case, which we will call "Sub-Case_2".
We will find that the two sub-cases above will produce different looking "intermediate step answers" and they have to be treated separately.
(3) SUB-CASE_1 : 4-Digit Numbers (where the 100s digit is greater than the 10s digit)
For 4-digit Numbers where the 100s-digit is "greater than the 10s digit" at the starting point, the next "Intermediate Step Answer" will be such that the "middle two digits will add up to 8" and the "outer two digits will add up to 10".
Example: Start with 9871. Note that here the 100s digit is "8" and that is greater than the 10s digit "7".
In this case, subtracting 1789 from 9871 we get 8082. Note that the two middle digits add up to "8" (0+8=8) and the two outer digits add up to "10" (8+2=10).
Continue by rearranging 8082 as 8820. At this step too, the 100s digit (or 8) is greater than the 10s digit (or 2). Subtracting 0288 (or 288) from 8820 we get 8532. Note that the two middle digits add up to 8 and the two outer digits add up to 10.
Now we can subtract 2358 from 8532 and we come to the Kaprekar Constant which is 6174. Here too the two middle digits add up to 8 and the two outer digits add up to 10.
You will find that does not matter what 4-digit number you start with, as long as the 100s digit is greater than the 10s digit, in each "intermediate step answer" the middle two digits will add up to "8" and the two outer digits will add up to "10".
(3) SUB-CASE_2 : 4-Digit Numbers (where the 100s digit is equal to the 10s digit)
For 4-digit Numbers, when the 100s-digit is "equal to the 10s digit" at the starting point, in all Intermediate Steps the "middle two digits" will both be "9" and the "outer two digits" will add up to "9".
Example: Start with 9881. Note that here the 100s digit and the 10s digit are the same.
In this case, subtracting 1889 from 9881 we get 7992. Note that the two middle digits are both "9" and the outer digits add up to "9" (or 7+2).
Continue by rearranging 7992 as 9972. In this case, the 100s digit is greater than the 10s digit and the next intermediate answer will follow "Sub-Case_1" above.
Consider another example starting with 9998.
Here the 100s and 10s digit are equal. Subtracting 8999 from 9998 we get 999. We should rewrite it as 0999 because that is what it really is. Note that the two middle digits are both "9" and the outer digits add up to "9" (or 0+9=9).
You will find that does not matter what 4-digit number you start with, as long as the 100s digit is equal to the 10s digit, in each "intermediate step answer" the middle two digits will be "9" and the two outer digits will add up to "9".
END WORD:
I have been able to prove "analytically" as to why the above observations are bound to be true in every case. That means I have used only "w, x, y and z" as the individual digits in each case and prove the same, instead of using "specific number examples" as I have done above. My proofs are very simple and I am sure that they can be followed by all high school students. However, it would take some more space and therefore, I am not doing it here.
Hopefully people who have come this far liked it. Mathematics is a joyful affair. Is it not?
A
This happens with 45 as well. I too discovered this long ago 😊
🇮🇳🇮🇳Proud to be an Indian 🇮🇳🇮🇳
I am too
Proud to be a Bengali.
Proud to be an Earthian
@@PhysicsMath Bharat Mata ki Jay
@@alhadbhagwat6142
Yesss. Proud to be son of mataa Bharat..
Doesn't it always work if you reverse the digits and subtract the smaller from the larger?
Really interesting! Mathematics is God.
When you don't notice the "594" in every 3 digit number
3:00 420... i see what you did there
What?
@@notAYUSHGAMING 024 is not a 3 digit no.
@@justalazyguy.0_0 It is. By the modern society of math, 0 at the start is considered a digit!
@@prakhyatpandey5341 Yes it is considered as a digit but number starting with such as 023,024, here "0" is not a significant number and there are 2 significant numbers here whereas if you consider decimals such as 23.0 or 25.06 there are 3 and 4 significant numbers respectively
Yes... i think of that too but actually, that 2nd subtraction problem, is taken from the result of 1st subtraction
(6*9)+(6+9)=69
It's true
No
It's true
*FACTS*
True for all integral a:
a*9 + a+9 = 10*a+9
*For all Real a
Only India can teach whole world via mathematics
72-27=45
54-45=09
90-09=81
81-18=63
63-36=27
2 digit loop!
Mateusz Dziewierz hmm,
10-01=09
90-09=81
81-18=63
63-36=27
WOAHH
27 HAS TO BE THE 2-DIGIT VERSION OF THESE NUMBERS
@@daffa_fm4583 Not quite. 27 doesn't loop to itself. What I would imagine is that a 2 digit number inevitably ends up somewhere in this loop.
reminds me of 3x+1 (Collatz) problem
I can see why you would say that. If this problem interests you. There are some great papers written about this. I recommend reading up on dynamical seuences.
@@alexandertownsend3291 link?
@@randomdude9135 So you are interested. I do not know much about the topic, but papers on the Collatz Conjecture frequently make mention of it. Here is a link.
mathworld.wolfram.com/DynamicalSystem.html
On that site check out the references at the bottom for further reading.
By the way earlier I said dynamical sequences, but I meant to say dynamical systems.
Sorry about the mix up. My mistake I just noticed it. I hope this helps.
If we take any 2 digit no say 34 and subtract it with its reverse the ans we get in the first step will always same after 4 iterations with the preceding answers.
Presh talwalker: *Any 3 digit number*
Me: *Holding 999* God forgive my sin which i m going to commit
... I... How did I... not think of that?
I overthought!
He did correct himself to say almost any, and put the exception in both summaries.
@@Utred2012 I dunno man
MEME MAN that is right. He clearly says in the video... to exclude numbers like 111 , 222,333,444,555,666,777,888 and 999
Why in every magical math facts this number "9" is always includes
Amazing mathematics
🇮🇳
I see number 18 as special case of 9 involved here
6+1+7+4=18
4+9+5=18
The sum of digits on each step of subtracting the smaller from the larger add up to 18
@River Wood yeah, but for me, as a numerologist, I see a more specific info with the 18. 18 appears to be like a gateway.
These numbers are always divisible by 9 and for the two-digits number, too.
Try a two digit number it will go through the variation of 9>81>63>27>45>9
Same as two digit number is related to magical number 9✌️
@ 3:50 Haven't heard this in math class because this has absolutely no useful application whatsoever.
almost all math has no useful application whatsoever
Packno1 except it does.
Don't just search for everything's usefulness. Sometimes you must also search for interesting patterns and beauty.
Packno1 Hahaha🤣🤣Very funny. This mathematics (Not of this video) might never be used by you. Ok tell me??? How your ekectronic devices work??? Your GPS??? There are so many things, i have no time to make it undersatnd for you
many of the western math axioms and theorems are still very much useless in the real world. instead of criticizing, maybe you'd try to look at the interest it generates when you're at boredom.
I think 9 is the most magical unique number
because per example if we have a number like 57 and we add 9 to it (57+9=66) so we have 57 can behave like 3( 5+7=12=1+2=3) and the same thing for 66 (6+6 =12 = 1+2 =3 )
we can rather conclude that 9 behaves like 0 in terms of adding
another trick is if we take any number like 653 and we subtract it by it renverse (653 - 356 = 297) we always get a number that can be divided by 9
What's curious is that five digit numbers always seem to create a repeating loop, and never settle on a single solution.
6+1+7+4 is same as 4+9+5 which is 18 or simply 1+8=9. Interesting! Wonder what application can come out of this.
Karthik Nadig It is because the results were drawn from the subtraction of Largest and smallest number from same digit,which is always a multiple of 9.And any number which is multiple of 9 has a digital sum of 9.
Digital sum is the sum of the digits of a number, until you reached one digit number.
@@RAJSINGH-of9iy ya I got that.....I'm wondering what application can be realised out of these two numbers. Number 9 and 11 are my favorites.
what an observation!
It seems like 100 and 101 are also exceptions for 3 digit numbers.
And 1000 is an exception too for 4 digits
There are many more exceptions
Eg. 988-889 = 99
Like wise
110-011
221-122
211-112
322-223
332-233
433-334
443-344
And so on... all these numbers gives remainder as 99.
Because of thses, numbers like 343, 767, 898, 878 etc. All comes into exception
none of these are actually exceptions, you just have to write 99 as 099.
if you get a two digit number then you must put a 0 at the front. Start with 100 -> 100 - 001 = 099 -> 990 - 099 = 891 -> 981 - 189 = 792 -> 972 - 279 = 693 -> 963 - 369 = 594 - > 954 - 459 = 495. Notice how each iteration the final number of the 'big' number goes up by one and the final number of the 'small' number goes down by one until they eventually meet in the middle (495)
Maximum 3 digit number is 999.
You allowed zero in 3 digit numbers, essentially allowing 2 digit numbers.
Total one digit numbers are 9, we want to exclude them from our math, so
999 - 9 = 990.
990 ÷ 2 = 495.
I.e. 495 is in between 3 digit numbers including zero at either end or middle.
When we subtract highest from lowest by rearranging numbers, we thrive towards middle point, which is 495.
Something similar logic could be true for 4 digit numbers, I havn't worked out it.
Edit :
In case of 4 digit numbers, since we allow 3 and 2 digit numbers too, like 2600, 2450, 2000 etc, middle point would be,
9999 + 2 × 999 + 3 × 99 = 12294
12294 ÷ 2 = 6174
So we still thrive towards middle.
Hello presh tall walker ,, where are you from ? [ Country]
With the 6174 magic number you get to it with a maximum of seven calculations
If anyone is wondering for 5 digit numbers, it will hit 74943, 62964, 71973, 83952 in a loop.
Sir ,
Does the rule also apply for 5 digit number ,
and beyond ? Does this have any physical
significance ? e.g . In Statistics or Random
Sampling .
These does not hold also for :- 112,122,223,233,334,344,445,455,556,566,667,677,778,788,889,899 & 1112,2223,3334,4445,5556,6667,7778,8889.
It's kind of decimal numbering system's properties. Decimal numbering system is forming from 0 to 9 numbers. And hence 9 got it's magical characteristic property.
INDIANS ARE always smart
It's very easy and intuitive to prove why it is 495. I have proved it. And so it is special but not magical number.
Proof please please!
Indians had been always done wonderful works either it is invention of surgery, invention of 0, invention of shampoo, invention of sugar, invention of vastu, invention of equation, invention of wheel(mentioned in the Mahabharat), invention of water on moon, invention of plastic sugery, invention of chess, invention of button, invention of economics(Mentioned in book Ashtadhyayi), etc
Definitely underrated in American culture :)
@@michaeladams2074 Because britishers riined the economy, we are seen as duffer and fools but they don't know that India was one of the richest country before britishers invaded.
I was expecting more discusión about this results, like : is it a property of decimal system or do this numbers exists for all the bases ? Does this work for any. Inner of digits ? (Clearly the numbers will change) etc
Is there a magic number for every N-digit variation?
No. In fact, this is a bit of wool pulled over the collective's eyes, in as much as 3 and 4 digit repunits (000, 111, 222, ...) all go to 0 in one cycle, so there are two fixed points ("magic numbers") for 3-digit numbers already. For 2 digits there is a "0" fixed point and a cycle of length 5 with the odd multiples of 9 (09-81-63-27-45).For 5 digits there is the "0" fixed point for repunits and 3 cycles of varying length.
As far as I know, for all N-digit numbers there is a finite number of cycles depending on N, but 3 and 4 are the only cases where there are only fixed points (cycles of length 1).
For 5 digits numbers it seems you get a loop of 4 numbers that keep repeating themselves, but it seems like they are different for every starting number. Howewer there is a pattern, the 4 loop numbers always have a 9 as a digit and the sum of the other digits is 18.
his voice was so different here compare to the present haha
What about 111, 222, 333 and so on?
Stabilize at 000.
wow amazing
That's frickin cool
@MindyourDecisions, did u know that if u take the difference of a number and its twisted version, that its always divisible by 9 for numbers with 2 digits, u need to take the difference of the first digit of those two numbers to actually become the difference divided by 9, for example: 92-29=> 9-2=7, and 7*9= 63 which is 92-29, i figured it out for 3 digital numbers as well; its about the same, but i dont know it for 4 digital numbers
Nice video
3-digit logic does not work on 101, 998, 566 and similar numbers as well. 4-digit version has lots of exceptions. Like logic will not work if 3 digits will be the same and etc. All in all was very interesting, thanks ^_^
It works
101: 101-011=090
090:900-009=891
891:981-189=792
792:972-279=693
693:963-369=594
594:954-459=495
@@MuraliKrishna-rq1ie If you consider that there is a number 090, it will work. P.S. 101 is not the largest number that possible to make from 101.
Mind blown
now try doing a proof of why the three digit numbers approach 495 using this algorithm. took me about 1 1/2 pages but finally found that if you start out with N = A*100 + B*10 + C*1 = [A, B, C],
N' = [A-C-1, 9, 10+C-A]
N'' = [9-A+C, 9, A-C]
N''' = [8-A+C, 9, 1+A-C]
N'''' = [7-A+C, 9, 2+A-C]
...
Plug in A=4, B=2, C=0 (420) into N''' and you will get N''' = [4, 9, 5]
Also, you can use this to prove why A=B=C digits won't work because you can define N(prime whatever) as [X-A+C, 9, Y+A-C]
Y = 9-X, so N(prime whatever) = [X-A+C, 9, 9-X+A-C]
A=C, so N(prime whatever) = [X-2A, 9, 9-X]
X-2A = 4 and 9-X = 5 ---> X=4
(4)-2A = 4
2A = 0, which says that the only value where A=B=C works is when A,B,C are zero.
You can do the same thing with 1 digit numbers:
It will always end up at 0.
If you want to do this trick with 1 or 2 digit numbers, you'll end up to 0.
For 5 digits there is a cycle of 3-4 numbers, which are different for different every input number.
What about 111 , 222 , 333 , 444 ,555 , 666 , 777 , 888 , 999
5 digit magic #?
I can visualise what's happening but it'll take too long to describe it. Cool children's party trick tho. Props to the guy who was bored enough to stumble upon it.
594 is one of the numbers before which we get 495 and it loops. I wonder if 549 and 459 are the other possible "loop-entry" numbers.
Besides that, numbers with 3 repeating digits like 222 and 777 are 'blackhole' numbers that will never bring you to 495. My intuition is that we can never reach such numbers. I wonder if it is possible to prove this.
These are my two thoughts.
The condition is that "all the digits cannot be the same" in the number you start with. That is the ONLY restriction Kaprekar imposed. You can choose 998 but not 999. You can choose any other number - either 3-digit or 4-digit - and it will work. So, you cannot start with 222 or 777.
You can start with any 3-digit number but after rearranging, reversing it and then subtracting the latter from the former, you will ALWAYS come up with a number which will have "9" as the "middle digit", while the "other two" will add up to "9". This will hold true of "any intermediate step". There can be no exception to this rule!
In your own example you came to 594. Notice that "9" is the center digit and the outer digits add up to "9" (5+4=9). The final KC number 495 also follows this rule, "9" in the middle and the outer two adding up to "9". Therefore, you can never have 549 and 459 as "loop entry numbers" because you will never see them at any intermediate step. In these numbers you have "4" and "5" as the middle digits and the "outer digits" do not add up to "9". Therefore, your examples are impossible at any intermediate step.
If you are interested, you can look at my detailed post titled HOW THE "INTERMEDIATE STEP ANSWERS LOOK" IN A KAPREKAR SEQUENCE OF CALCULATIONS, where I have discussed this matter in detail. I have discussed the cases for 2-digit, 3-digit and 4-digit Numbers in that post.
NOTE: I later realized that 2-digit numbers do NOT have a Kaprekar Constant. I have written that as a reply to my own posting under "INTERMEDIATE STEP ANSWERS LOOK" IN A KAPREKAR SEQUENCE OF CALCULATIONS above. If you go there, you can read what I have written.
Wow, very cool!
What happens if you used the number 100? Would the result be 495?
Nope...we ain't get
Yes we do. But we need to write 1 as 001 and 99 as 099.
ua-cam.com/video/FrL8gqs2B8w/v-deo.html
The trick is found a number. Except those similar digits .
Such that arrangement .in the mentioned video doesn't change.
10-01=09
90-9=91
91-19= 72
72-27=45
54-45=09
Now my law. . you start with any 2 digits number you end up with following repeated pattern . .
Nice video. Interesting, but why? NEED DEMOSTRATION
*Good night from Tajikistan*
*a ☆0*
*ab ☆9*
*abc ☆495*
*abcd ☆6174*
*...*
This is what i was searchig for
We obtain 99*(biggest digit - smallest digit)
Only 9 possibles and those converge to 495
Except 000...
Didn't work for me. When he said to take a three digit number, I picked my favorite one: 666. That gave me zero at the next step, and stayed there. Only afterwards did it show the note about multiples of 111 not working, a fact which I consider to be fundamentally unfair. ;-)
On a more serious note, it would be interesting to see which numbers have the longest path to reach 495/6174 -- Wikipedia says the longest streaks are 6 to reach 495 and 8 to reach 6174 -- and the mathematical proof of why this works. Will have to play around with this some more...
Someone in the comment above yours has a proof in the reply section of the theory behind why does it always end up on 495.
Nordic Exile didn’t you see in the video it said except for numbers with the same digit.
How about a follow-up video with the formulas demonstrating why this is so?
Thanks!
Using 4 digits to get 5174 count the iteration count its always between 0 and 7 very useful music intervals colour 7 colours so check all numbers 0000 to 9999 plot on 100x100 grid top line is 0000 to 0099 next line 0100 to 0199 do all 100 lines asign a colour 0 black 1 red 7 violet look at that patter, now choose key c major compute all 10000 values create chord on interval 0 to 7th wow
Now 5 digits do similar but no static number but it cycles at some trigger number 5 , 6 7 digitsvtry these too
Typo meant 6174 not 5174 sorry
4 = d
9 = i
5 = e
This video shows that everyone dies in the end.
doe not work for 5 digits does not converge at least I have found a counter example
👉 I pick the number 63954 then
1)96543-34569=61974
2)97641-14679=82962
3)98622-22689=75933
4)97533-33579=63954
and then it starts all over again lol let's make America great again, I suddenly remember
For two digits is 9 a Kaprekar number ? For four digits ?
If you apply this to a 2 digit number you will end up getting 9
Weird, if you do 5 digit numbers, it kind of cycles endlessly though a series instead of sticking to one. Still, 9's everywhere though.
Also weird, 495*9 = 4455, and 6174*9 = 55566. If you try to reverse the process and do things like 666677, or 344, you get repeating decimals instead of whole numbers. I thought that had something to do with it, but I tried 1111122 (same pattern, repeating digits with two digits one higher on the end, divisible by 9) and it cycles through about 7 or 8 numbers infinitely.
Well, screw it, I didn't need sleep tonight anyway.
Also, useless fact, 6174/495 is 686/55. Also, math is ultra weird. If you imagine 686 as a pattern of a number on each end, and the middle filled with a number two higher, then 55 is 5's with 0 7s between. 686 is 6's with 1 8 in between.
Here's why it's weird. 686/55 is 12.47272727 repeating.
If one less is 4's on both ends with "-1" 6 in the middle, just make it 4 (because you have to remove a number). 4/55 is 0.0727272727 repeating. 55/55 is obviously 1, 686/55 is that weird number, but hey...looks familiar. What if we do (686/55)-(4/55)? 682/55 is 12.4. Okay, keep moving up. 7997/55 is 145.4?
From here we have two possibilities. 80008 would be next? or is it 8(10)(10)(10)8? If it's the second, we have to assume it's 91108 (as if we're carrying the one, right?) Well...
80008/55 = 1454.690909 repeating (wait, that's almost exactly [7997/55]*10), 91108/55 = 1656.509091...
MORE...
911119/55 = 16565.8 (that's awfully similar to the second one above times ten...) The second version would be 1022219 (with carrying), so 1022219/55 = 18585.8...
One more...
10222230/55 (ends become 10, last 10 carries one over)? 185858.7273 (Oh...that's...also familiar...), and finally, 11333330/55 (with full carrying) = 206060.5455
So there's this odd pattern where carrying with one is almost exactly 1/10 of the non-carrying version after it, each time raising from 1656... to 1858... to 2060...?
At this point it's not even math, it's just nonsense...but I love it. :D
The first 3 digit number I tried didn’t follow the sequence. It actually got to the 495 backstop after 2 cycles rather than 3. 417
The second 3-digit number I tried didn’t work at all. 222
Is there any constant for numbers with more digits?
Nice very interesting...I learned something new today.
What about 100
100-1=99
99-99=0
🤔🤔🤔
No - you did it wrong. 100-001=099 is correct. Now rearrange it as 990 and then do 990-099=891. Then continue. If you start with a 3-digit number, then the next answer also has to be 3-digit. If you get a 2-digit answer like "99", then make it a 3-digit number by putting a "0" in front (which in this case is 099). You cannot operate with 2-digit numbers at any intermediate step, to get to the Kaprekar Constant of 495 (starting from 3-digit numbers).
Thanks sarba😊
But if u write a three digit number as 4 digit number by appendix by 0 at the left then u shd get 6174. So three digit numbers can converge to two constants by K process
it is not working with 101 also
@MindYourDecisions Presh... Ever heard of the Collatz Conjecture?
You know what's amazing? If you add 1 to a number, it becomes ONE larger.. but if you subtract that SAME, it is reduced by one. This is insane. You can do this all day long. TO ANY NUMBER?! *Mind* BLOWN.
You are dumber than a pig
Can you prove what he said if I had heard it correctly. Take any three digit number and do that routine procedure and you end up with 495.
ps : Don't take Number Theory as a joke son!
The real magic is that when you multiply any number by 1, the number doesn't change, even if it is 0.
100 - 001 = 99
99 - 99 = 0
0 - 0 = 0
oops, i disproved it. you can also do this with 1000 and get 0 in the end
You can do it all the numbers you have chosen. You have followed the wrong steps and that's why it seems you cannot do it. Please see my reply a few posts above.
This is called an *algorithm*
I tried this for 5 digits and after a few tries I ended up with a few cycles:
63954, 61974, 82962, 75933
83952, 74943, 62964, 71973
For base 2, I got:
2 digits: 1
3 digits: 3
4 digits: 7 or 9
5 digits: too lazy to check
Luke Balgan There is another cycle for 5 digit number, 59994, 53955
@@ymblcza2557 I just didnt check all of them
What are the applications of kaprekar constant
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495 and 6174 makes 9 in addition of digits.
what about the number 221,112 and so on
How about those which are not 3-digit or 4-digit?
What about 999