We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Implementation
  4. Modified Kaprekar Numbers
  5. Discussions

Modified Kaprekar Numbers

Sort by

recency

|

1037 Discussions

|

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can what the explanation here : https://youtu.be/MttrQCHGu3w

    int getSum(long a){
    long sq = a * a;
    int digit = (int)log10(a) + 1;
    int result = sq % (int)pow(10, digit);
    int rest = (int)log10(sq) + 1 - digit;
    if(rest > 0) result += stoi(to_string(sq).substr(0, rest));
    return result;
}

void kaprekarNumbers(int p, int q) {
    bool valid_range = false;
    for(int i = p; i <= q; i++){
        int s = getSum(i);
        if(s == i){
           cout << i << " ";
           valid_range = true; 
        }  
    }
    if(!valid_range) cout << "INVALID RANGE";
}

  • g_azabache1
     + 0 comments

    Hi, this is my solution in C# 🐿️

        public static void kaprekarNumbers(int p, int q)
    {
        List<int> numbers = new List<int>();
        for (int i = p; i <= q; i++)
        {
            var digits = i.ToString().ToCharArray().Count();
            var square = Math.Pow(i, 2);
            var square_digits = square.ToString().ToCharArray().Count();
            int left = 0;
            int right = 0;
            if (square_digits > 1)
            {
                int digitsForLeft = square_digits - digits;
                left = Convert.ToInt32(square.ToString().Substring(0, digitsForLeft));
                right = Convert.ToInt32(square.ToString().Substring(digitsForLeft));
            }
            else
            {
                right = (int)square;
            }
            if (left + right == i)
                numbers.Add(i);
        }
        Console.WriteLine(numbers.Count > 0 ? string.Join(" ", numbers) : "INVALID RANGE");
    }

  • abhayad268207
     + 0 comments

    Solution of Kaprekar Numbers in Python :

    def kaprekarNumbers(p, q):
    # Write your code here
    li=[]
    for i in range(p , q+1):
        if(i ==1): 
            li.append(i)
                
        elif(i == 2 or i==3 ):
            continue

        elif(i>3):
            num= i*i
            digitsize = len(str(i))
            strnum=str(num)
            leftsize =  len(strnum)-digitsize
            left = int(strnum[:leftsize])
            right= int(strnum[leftsize:])
            add = left + right
            if( add == i ):
                li.append(i)
                                     
    if(len(li) >0):
        for i in range(0,len(li)):
            print(li[i] ,end=" ")
    else:
        print("INVALID RANGE")

  • vutrantrung14821
     + 0 comments
    g=[]
    for i in range(p,q+1):
      a=str(i**2)
      b=a[:len(a)//2]
      c=a[len(a)//2:]
      if c=='':
        c=0
      elif b=='':
        b=0
      if int(b)+int(c)==i:g.append(i)
    print('INVALID RANGE')if g==[]else print(*g)

  • alex_k5
     + 0 comments

    def kaprekarNumbers(p, q): arr = [] if p == 1: arr.append(1) start = p if p >= 9 else 9 for i in range(start, q+1): left = str(i**2)[:-len(str(i))] right = str(i**2)[-len(str(i)):] if int(left) + int(right) == i: arr.append(i) print(*arr) if arr else print("INVALID RANGE")