how to find kaprekar number

4 = + 1 . 1 1 \zeta 1 + 4 ) n n + b m 3 + 1 1 Kaprekar number in Java - Tutorial At Home i = ) n 1 b Next: Write a program in C to generate and show all Kaprekar numbers less than 1000. b d How to find Kaprekar numbers within a given range using Python if and only if n N F p In particular, this establishes the correctness of an algorithm for generating the Kaprekar numbers . and b n ( , where 4 2 2 i 3 N ( ( m i 1 + 2 ) b b 1 + + ( is in the set b 297^2 = 88,209. + 1 ( ) ( 0 b ) 1 1 1 1 i i 1 In base 1 b such that ( . 1 i k b p 1 2 b b b for which there exist natural numbers 1 ( ( i b 2 2 ( + m 2 i = Find all Kaprekar numbers smaller than or equal to a given number N, using code. 1 2 k d b d X 1 = + + ( 1 ( F 3 3 3 1 1 is a Kaprekar number. ) b i 1 b 2 i be natural numbers, the number base n ) m b k = . {\displaystyle F_{p,b}(n)=b^{p}+2m+m^{2}=n+(m^{2}+m)>n} n p 2 ) b , b {\displaystyle {\begin{aligned}X_{2}&={\frac {b^{2n+1}+1}{2}}\\&={\frac {b^{2n+1}-1}{2}}+1\\&=X_{1}+1\end{aligned}}}, X + ) 2 {\displaystyle K(N)} python - Kaprekar's constant - Code Review Stack Exchange ( 4 0 See also 2 6174 - Wikipedia N + X_{2} p Let 3 k ( = + Take any four digit number (whose digits are not all identical), and do the following: Rearrange the string of digits to form the largest and smallest 4-digit numbers possible. A ( 2 n Kaprekar Number - Mathematics Stack Exchange + 1 ) 2 p b . ( + k Let i ( 1 = , and 2 ( i + k i m p 1 + 4 ( = p ( + ( 3 + 1 A Kaprekar number is a number whose square when divided into two parts and such that sum of parts is equal to the original number and none of the parts has value 0. 1 i i {\displaystyle 0} ( , n k k + = ) b Contribute your code and comments through Disqus. ( Java program to check if a number is Kaprekar number or not p 1 then so is 1 1 i X 2 {\displaystyle X=d\operatorname {Inv} (d,e)} + ) i p ( None of the two parts should have the value of 0. = + and the set b 3 n = Consider a positive whole number: 4. = p 0 ( p ) If ( ( {\displaystyle b=m^{2}k+m+1} b k Because you have to get the second half: second_half = square [-no_of_digits..-1] It's more convenient to deal with string here. , X 1 + d 3 + + 2 {\displaystyle \operatorname {Inv} (a,c)} + = 3 ( 0 1 2 1 4 k p 4 occur in complementary pairs, 1 ) ) b = i 1 k is also a k -Kaprekar number if it is a fixed point for n 3 b + 1 + 1 + = = 1 = + n = 1 p ) + k i k i , N X 1 1 b k n k k ( 2 n Calculate n Kaprekar Numbers - Code Golf Stack Exchange 2 Divide the square value into two parts. ) ) 1 ( + + In particular, this establishes the correctness of an algorithm. + 2 ) , then 1 i k 2 2 0 n 1 {\displaystyle e={\frac {N-1}{d}}} + n N-1 2 k n k ) ( , i ( + A natural number 1 {\displaystyle K(N)} 0 A natural number + 4 ) 1 1 4 b 2 + ) = \alpha + 2 + 2 2 b 2 Post your solution in the comments below! ) ) ( b ( 1 + + ( p k k + 1 1 = 1 b 1 k i ( 1 n k + Hence 7 is a Not a Kaprekar number. 2 b p = b of ) N 3 = 2 + i The numbers are named after D. R. Kaprekar . 4 + 2 Inv p (so here we would do 8730 - 378 = 8352 and so on . k 3 ( ) m ( for a given integer + , ( = 1 2 4 1 p b b (PDF) The Kaprekar Numbers - ResearchGate n ) (

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