Modify N by adding its smallest positive divisor exactly K times

Given two positive integers N and K, the task is to find the value of N after incrementing the value of N in each operation by its smallest divisor exceeding N ( exceeding 1 ), exactly K times.

Examples:

Input: N = 5, K = 2 
Output: 12 
Explanation:
Smallest divisor of N (= 5) is 5. Therefore, N = 5 + 5 = 10. 
Smallest divisor of N (= 10) is 2. Therefore, N = 5 + 2 = 12. 
Therefore, the required output is 12.

Input: N = 6, K = 4 
Output: 14

Naive Approach: The simplest approach to solve this problem to iterate over the range [1, K] using variable i and in each operation, find the smallest divisors greater than 1 of N and increment the value of N by the smallest divisor greater than 1 of N. Finally, print the value of N.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the smallest
// divisor of N greater than 1
int smallestDivisorGr1(int N)
{
    for (int i = 2; i <= sqrt(N);
         i++) {

        // If i is a divisor
        // of N
        if (N % i == 0) {
            return i;
        }
    }

    // If N is a prime number
    return N;
}

// Function to find the value of N by
// performing the operations K times
int findValOfNWithOperat(int N, int K)
{

    // Iterate over the range [1, K]
    for (int i = 1; i <= K; i++) {

        // Update N
        N += smallestDivisorGr1(N);
    }

    return N;
}

// Driver Code
int main()
{
    int N = 6, K = 4;

    cout << findValOfNWithOperat(N, K);
    return 0;
}

Java

// Java program to implement
// the above approach

class GFG{

// Function to find the smallest
// divisor of N greater than 1
static int smallestDivisorGr1(int N)
{
    for (int i = 2; i <= Math.sqrt(N);
         i++) {

        // If i is a divisor
        // of N
        if (N % i == 0) {
            return i;
        }
    }

    // If N is a prime number
    return N;
}

// Function to find the value of N by
// performing the operations K times
static int findValOfNWithOperat(int N, int K)
{

    // Iterate over the range [1, K]
    for (int i = 1; i <= K; i++)
    {

        // Update N
        N += smallestDivisorGr1(N);
    }

    return N;
}

// Driver Code
public static void main(String[] args)
{
    int N = 6, K = 4;

    System.out.print(findValOfNWithOperat(N, K));
}
}

// This code is contributed by shikhasingrajput

Python3

# Python 3 program to implement
# the above approach
import math

# Function to find the smallest
# divisor of N greater than 1
def smallestDivisorGr1(N):
    for i in range(2, int(math.sqrt(N))+1):

        # If i is a divisor
        # of N
        if (N % i == 0):
            return i

    # If N is a prime number
    return N

# Function to find the value of N by
# performing the operations K times
def findValOfNWithOperat(N, K):

    # Iterate over the range [1, K]
    for i in range(1, K + 1):

        # Update N
        N += smallestDivisorGr1(N)

    return N

# Driver Code
if __name__ == "__main__":

    N = 6
    K = 4

    print(findValOfNWithOperat(N, K))

    # This code is contributed by ukasp.

C#

// C# program to implement
// the above approach
using System;
public class GFG
{

  // Function to find the smallest
  // divisor of N greater than 1
  static int smallestDivisorGr1(int N)
  {
    for (int i = 2; i <= Math.Sqrt(N);
         i++) {

      // If i is a divisor
      // of N
      if (N % i == 0) {
        return i;
      }
    }

    // If N is a prime number
    return N;
  }

  // Function to find the value of N by
  // performing the operations K times
  static int findValOfNWithOperat(int N, int K)
  {

    // Iterate over the range [1, K]
    for (int i = 1; i <= K; i++)
    {

      // Update N
      N += smallestDivisorGr1(N);
    }

    return N;
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int N = 6, K = 4;

    Console.Write(findValOfNWithOperat(N, K));
  }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// JavaScript program to implement
// the above approach

// Function to find the smallest
// divisor of N greater than 1
function smallestDivisorGr1(N)
{
    for (let i = 2; i <= Math.sqrt(N);
        i++) {

        // If i is a divisor
        // of N
        if (N % i == 0)
        {
            return i;
        }
    }

    // If N is a prime number
    return N;
}

// Function to find the value of N by
// performing the operations K times
function findValOfNWithOperat(N, K)
{

    // Iterate over the range [1, K]
    for (let i = 1; i <= K; i++)
    {

        // Update N
        N += smallestDivisorGr1(N);
    }
    return N;
}

// Driver Code
let N = 6, K = 4;
document.write(findValOfNWithOperat(N, K));

// This code is contributed by Surbhi Tyagi.
</script>

Output:

14

Time Complexity: O(K * √N)
Auxiliary Space: O(1)

Efficient Approach: Follow the steps below to solve the problem:

• If N is an even number then update the value of N to (N + K * 2).
• Otherwise, find the smallest positive divisor greater than 1 of N say, smDiv and update the value N to (N + smDiv + (K – 1) * 2)
• Finally, print the value of N.

Below is the implementation of the above approach:

C++14

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the smallest
// divisor of N greater than 1
int smallestDivisorGr1(int N)
{
    for (int i = 2; i <= sqrt(N);
         i++) {

        // If i is a divisor
        // of N
        if (N % i == 0) {
            return i;
        }
    }

    // If N is a prime number
    return N;
}

// Function to find the value of N by
// performing the operations K times
int findValOfNWithOperat(int N, int K)
{
    // If N is an even number
    if (N % 2 == 0) {

        // Update N
        N += K * 2;
    }

    // If N is an odd number
    else {

        // Update N
        N += smallestDivisorGr1(N)
             + (K - 1) * 2;
    }

    return N;
}

// Driver Code
int main()
{
    int N = 6, K = 4;

    cout << findValOfNWithOperat(N, K);
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{

// Function to find the smallest
// divisor of N greater than 1
static int smallestDivisorGr1(int N)
{
    for(int i = 2; i <= Math.sqrt(N); i++)
    {

        // If i is a divisor
        // of N
        if (N % i == 0)
        {
            return i;
        }
    }

    // If N is a prime number
    return N;
}

// Function to find the value of N by
// performing the operations K times
static int findValOfNWithOperat(int N, int K)
{

    // If N is an even number
    if (N % 2 == 0)
    {

        // Update N
        N += K * 2;
    }

    // If N is an odd number
    else
    {

        // Update N
        N += smallestDivisorGr1(N) + (K - 1) * 2;
    }
    return N;
}

// Driver Code
public static void main(String[] args)
{
    int N = 6, K = 4;

    System.out.print(findValOfNWithOperat(N, K));
}
}

// This code is contributed by target_2

Python3

# Python program to implement
# the above approach

# Function to find the smallest
# divisor of N greater than 1
def smallestDivisorGr1(N):
    for i in range (sqrt(N)):
         i += 1

        # If i is a divisor
        # of N
    if(N % i == 0):
            return i

    # If N is a prime number
    return N

# Function to find the value of N by
# performing the operations K times
def findValOfNWithOperat(N, K):

  # If N is an even number
  if (N % 2 == 0):

    # Update N
    N += K * 2

  # If N is an odd number
  else:
    # Update N
    N += smallestDivisorGr1(N) + (K - 1) * 2

  return N

# Driver Code
N = 6
K = 4
print(findValOfNWithOperat(N, K))

# This code is contributed by shivanisinghss2110

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to find the smallest
// divisor of N greater than 1
static int smallestDivisorGr1(int N)
{
    for(int i = 2; i <= Math.Sqrt(N); i++)
    {

        // If i is a divisor
        // of N
        if (N % i == 0)
        {
            return i;
        }
    }

    // If N is a prime number
    return N;
}

// Function to find the value of N by
// performing the operations K times
static int findValOfNWithOperat(int N, int K)
{

    // If N is an even number
    if (N % 2 == 0)
    {

        // Update N
        N += K * 2;
    }

    // If N is an odd number
    else
    {

        // Update N
        N += smallestDivisorGr1(N) + (K - 1) * 2;
    }
    return N;
}

// Driver code
static public void Main()
{
    int N = 6, K = 4;

    Console.Write(findValOfNWithOperat(N, K));
}
}

// This code is contributed by Khushboogoyal499

Javascript

<script>

// Function to find the smallest
// divisor of N greater than 1
function smallestDivisorGr1( N)
{
    for (var i = 2; i <= Math.sqrt(N);
        i++) {

        // If i is a divisor
        // of N
        if (N % i == 0) {
            return i;
        }
    }

    // If N is a prime number
    return N;
}

// Function to find the value of N by
// performing the operations K times
function findValOfNWithOperat(N, K)
{
    // If N is an even number
    if (N % 2 == 0) {

        // Update N
        N += K * 2;
    }

    // If N is an odd number
    else {

        // Update N
        N += smallestDivisorGr1(N)
            + (K - 1) * 2;
    }

    return N;
}

// Driver Code
var N = 6, K = 4;

    document.write(findValOfNWithOperat(N, K));



</script>

Output:

14

Time Complexity: O(√N)
Auxiliary Space: O(1)