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Sum of first K numbers which are not divisible by N
  • Last Updated : 07 May, 2021

Given two numbers N and K, the task is to find the sum of first K numbers which are not divisible by N.
Examples: 
 

Input: N = 5, K = 10 
Output: 63 
Explanation: Sum of { 1, 2, 3, 4, 6, 7, 8, 9, 11, 12 } is 63.
Input: N = 3, k = 13 
Output: 127 
Explanation: Sum of { 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19 } is 127. 
 

 

Approach: In order to solve this problem, we ned to follow the following steps: 
 

  • Calculate the last multiple of N by (K / (N – 1)) * N
  • Calculate K%(N – 1). If the remainder is 0, (K / (N – 1)) * N – 1 gives the last value which is not divisible by N. Otherwise, we need to add the remainder to (K / (N – 1)) * N.
  • Calculate the sum up to the last value which is not divisible by N obtained in the above step.
  • Calculate the sum of multiples of N and subtract from the sum calculated in the above step to get the desired result.

Below code is the implementation of the above approach:
 



C++




// C++ Program to calculate
// the sum of first K
// numbers not divisible by N
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum
int findSum(int n, int k)
{
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
 
    int rem = k % (n - 1);
 
    // Find the K-th non-multiple of N
    if (rem == 0) {
        val = val - 1;
    }
    else {
        val = val + rem;
    }
 
    // Calculate the  sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
 
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples
        = (x
           * (x + 1) * n)
          / 2;
    sum -= sum_of_multiples;
 
    return sum;
}
 
// Driver code
int main()
{
    int n = 7, k = 13;
    cout << findSum(n, k)
         << endl;
}

Java




// Java program to calculate
// the sum of first K numbers
// not divisible by N
import java.util.*;
 
class GFG {
 
// Function to find the sum
static int findSum(int n, int k)
{
     
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
 
    int rem = k % (n - 1);
 
    // Find the K-th non-multiple of N
    if (rem == 0)
    {
        val = val - 1;
    }
    else
    {
        val = val + rem;
    }
 
    // Calculate the sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
 
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples = (x * (x + 1) * n) / 2;
    sum -= sum_of_multiples;
 
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 7, k = 13;
 
    System.out.println(findSum(n, k));
}
}
 
// This code is contributed by offbeat

Python3




# Python3 Program to calculate
# the sum of first K
# numbers not divisible by N
 
# Function to find the sum
def findSum(n, k):
 
    # Find the last multiple of N
    val = (k // (n - 1)) * n;
 
    rem = k % (n - 1);
 
    # Find the K-th non-multiple of N
    if (rem == 0):
        val = val - 1;
     
    else:
        val = val + rem;
     
    # Calculate the sum of
    # all elements from 1 to val
    sum = (val * (val + 1)) // 2;
 
    # Calculate the sum of
    # all multiples of N
    # between 1 to val
    x = k // (n - 1);
    sum_of_multiples = (x * (x + 1) * n) // 2;
    sum -= sum_of_multiples;
 
    return sum;
 
# Driver code
n = 7; k = 13;
print(findSum(n, k))
 
# This code is contributed by Code_Mech

C#




// C# program to calculate
// the sum of first K numbers
// not divisible by N
using System;
 
class GFG{
 
// Function to find the sum
static int findSum(int n, int k)
{
     
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
 
    int rem = k % (n - 1);
 
    // Find the K-th non-multiple of N
    if (rem == 0)
    {
        val = val - 1;
    }
    else
    {
        val = val + rem;
    }
 
    // Calculate the sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
 
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples = (x * (x + 1) * n) / 2;
    sum -= sum_of_multiples;
 
    return sum;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 7, k = 13;
 
    Console.WriteLine(findSum(n, k));
}
}
 
// This code is contributed by 29AjayKumar
Output: 
99

 

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

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