## Kaprekar Numbers

### September 21, 2010

Wolfram’s MathWorld describes Kaprekar numbers like this:

Consider an

n-digit numberk. Square it and add the rightndigits to the leftnorn-1 digits. If the resultant sum isk, thenkis called a Kaprekar number. For example, 9 is a Kaprekar number since 9^{2}= 81 and 8 + 1 = 9 and 297 is a Kaprekar number since 297^{2}= 88209 and 88 + 209 = 297.

Your task is to write a function that identifies Kaprekar numbers and to determine the Kaprekar numbers less than a thousand. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Pages: 1 2

#lang racket

(require (planet soegaard/math/math))

(for/fold ([result ‘()])

([x (in-range 1000)])

(let ([d (digits (* x x))])

(let-values ([(left right) (split-at d (quotient (length d) 2))])

(if (equal? x (+ (digits->number left)

(digits->number right)))

(cons x result)

result))))

This is a good problem for python.

def kaprekar(n):

l = str(n * n)

length = len(l)

return n == int(l[0:length/2]) + int(l[length/2:])

for i in range(4,10000000):

if kaprekar(i):

print i

A simple ruby version …

A haskell solution

which gives

*Main> kaprekarLessThan 1000

[1,9,45,55,99,297,703,999]

And yes, yes I did misspell Kaprekar.

Answered in Python, with some pre-computating values and working only with digits (as opposed to strings) for speed:

Oops! Here we go:

ruby-1.9.2-p0 > require "kaprekar"

=> true

ruby-1.9.2-p0 > kaprekar? 9

=> true

ruby-1.9.2-p0 > kaprekar? 297

=> true

ruby-1.9.2-p0 > kaprekar? 50

=> false

ruby-1.9.2-p0 > kaprekar_to 1000

=> [1, 9, 45, 55, 99, 297, 703, 999]

;; First time I ever use the prelude…

(define (kaprekar? n)

(let* ((n2 (ipow n 2))

(size (+ 1 (ilog 10 n2)))

(size-l (if (even? size) (/ size 2) (/ (- size 1) 2)))

(size-r (if (even? size) size-l (+ 1 size-l))))

(let* ((right (modulo n2 (ipow 10 size-r)))

(left (/ (- n2 right) (ipow 10 size-r))))

(= n (+ left right)))))

(define (kaprekar n)

(filter kaprekar? (range 0 n)))

;; err, that was the wrong version.

(define (kaprekar? n)

(let* ((n2 (ipow n 2))

(size (+ 1 (ilog 10 n2)))

(size-r (/ (if (even? size) size (+ 1 size)) 2)))

(let* ((mask (ipow 10 size-r))

(right (modulo n2 mask))

(left (/ (- n2 right) mask)))

(= n (+ left right)))))

Common Lisp solution. No strings manipulation, but arithmetic (log, expt, ceiling, floor).

Same as model solution in scheme.

A soultion in Haskell using Data.Digits

Anyone have an idea, how to implement it without reversing the list twice?

Hackish code will come back to it when I have time.

Mine in F #

C Implementation

This is in python:

[…] Below is my solution to the Programming Praxis problem from Sept. 21, 2010. The problem was to find all Kaprekar numbers less than 1000. I wrote my solution in C++ to refresh my skills a little. For more info, see this link. […]

[…] A few days ago I attempted to put my Ruby skills to use, given a programming puzzle centred on Kaprekar numbers. […]

Finally, I have understood kaprekar number. Thanks guys

C++:

using namespace std;

bool isKep(int num)

{

int n = num;

int digit = 1;

int right = 0;

int squere = n*n;

while (n)

{

right += digit*(squere%10);

n /= 10;

squere /= 10;

digit *= 10;

}

if (squere+right == num) {return true;}

else {return false;}

}

int main()

{

for (int i=1;i<1000;i++){

if (isKep(i)) {cout << i << " ";}

}

return 0;

}

September 21st, 2010.c:Output:On an Apple Power Mac G4 (AGP Graphics) (450MHz processor, 1GB memory) to run the solution took approximately one second on both Mac OS 9.2.2 (International English) (the solution interpreted using Leonardo 3.4.1) and Mac OS X 10.4.11 (the solution compiled using Xcode 2.2.1).

(I’m just trying to solve the problems posed by this ‘site whilst I try to get a job; I’m well aware that my solutions are far from the best – but, in my defence, I don’t have any traditional qualifications in computer science :/ )