Find sum of modulo K of first N natural number Last Updated : 12 Sep, 2022 Comments Improve Suggest changes Like Article Like Report Given two integer N ans K, the task is to find sum of modulo K of first N natural numbers i.e 1%K + 2%K + ..... + N%K. Examples : Input : N = 10 and K = 2. Output : 5 Sum = 1%2 + 2%2 + 3%2 + 4%2 + 5%2 + 6%2 + 7%2 + 8%2 + 9%2 + 10%2 = 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 = 5.Recommended PracticeReverse CodingTry It! Method 1: Iterate a variable i from 1 to N, evaluate and add i%K. Below is the implementation of this approach: C++ // C++ program to find sum of // modulo K of first N natural numbers. #include <bits/stdc++.h> using namespace std; // Return sum of modulo K of // first N natural numbers. int findSum(int N, int K) { int ans = 0; // Iterate from 1 to N && // evaluating and adding i % K. for (int i = 1; i <= N; i++) ans += (i % K); return ans; } // Driver Program int main() { int N = 10, K = 2; cout << findSum(N, K) << endl; return 0; } Java // Java program to find sum of modulo // K of first N natural numbers. import java.io.*; class GFG { // Return sum of modulo K of // first N natural numbers. static int findSum(int N, int K) { int ans = 0; // Iterate from 1 to N && evaluating // and adding i % K. for (int i = 1; i <= N; i++) ans += (i % K); return ans; } // Driver program static public void main(String[] args) { int N = 10, K = 2; System.out.println(findSum(N, K)); } } // This code is contributed by vt_m. Python3 # Python3 program to find sum # of modulo K of first N # natural numbers. # Return sum of modulo K of # first N natural numbers. def findSum(N, K): ans = 0; # Iterate from 1 to N && # evaluating and adding i % K. for i in range(1, N + 1): ans += (i % K); return ans; # Driver Code N = 10; K = 2; print(findSum(N, K)); # This code is contributed by mits C# // C# program to find sum of modulo // K of first N natural numbers. using System; class GFG { // Return sum of modulo K of // first N natural numbers. static int findSum(int N, int K) { int ans = 0; // Iterate from 1 to N && evaluating // and adding i % K. for (int i = 1; i <= N; i++) ans += (i % K); return ans; } // Driver program static public void Main() { int N = 10, K = 2; Console.WriteLine(findSum(N, K)); } } // This code is contributed by vt_m. PHP <?php // PHP program to find sum // of modulo K of first N // natural numbers. // Return sum of modulo K of // first N natural numbers. function findSum($N, $K) { $ans = 0; // Iterate from 1 to N && // evaluating and adding i % K. for ($i = 1; $i <= $N; $i++) $ans += ($i % $K); return $ans; } // Driver Code $N = 10; $K = 2; echo findSum($N, $K), "\n"; // This code is contributed by ajit ?> JavaScript <script> // JavaScript program to find sum // of modulo K of first N natural // numbers. // Return sum of modulo K of // first N natural numbers. function findSum(N, K) { let ans = 0; // Iterate from 1 to N && evaluating // and adding i % K. for(let i = 1; i <= N; i++) ans += (i % K); return ans; } // Driver Code let N = 10, K = 2; document.write(findSum(N, K)); // This code is contributed by code_hunt </script> Output : 5 Time Complexity : O(N). Auxiliary Space: O(1) Method 2 : Two cases arise in this method.Case 1: When N < K, for each number i, N >= i >= 1, will give i as result when operate with modulo K. So, the required sum will be the sum of the first N natural number, N*(N+1)/2.Case 2: When N >= K, then integers from 1 to K in natural number sequence will produce, 1, 2, 3, ....., K - 1, 0 as result when operate with modulo K. Similarly, from K + 1 to 2K, it will produce same result. So, the idea is to count how many numbers of times this sequence appears and multiply it with the sum of first K - 1 natural numbers. Below is the implementation of this approach: C++ // C++ program to find sum of modulo // K of first N natural numbers. #include <bits/stdc++.h> using namespace std; // Return sum of modulo K of // first N natural numbers. int findSum(int N, int K) { int ans = 0; // Counting the number of times 1, 2, .., // K-1, 0 sequence occurs. int y = N / K; // Finding the number of elements left which // are incomplete of sequence Leads to Case 1 type. int x = N % K; // adding multiplication of number of // times 1, 2, .., K-1, 0 sequence occurs // and sum of first k natural number and sequence // from case 1. ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2; return ans; } // Driver program int main() { int N = 10, K = 2; cout << findSum(N, K) << endl; return 0; } Java // Java program to find sum of modulo // K of first N natural numbers. import java.io.*; class GFG { // Return sum of modulo K of // first N natural numbers. static int findSum(int N, int K) { int ans = 0; // Counting the number of times 1, 2, .., // K-1, 0 sequence occurs. int y = N / K; // Finding the number of elements left which // are incomplete of sequence Leads to Case 1 type. int x = N % K; // adding multiplication of number of times // 1, 2, .., K-1, 0 sequence occurs and sum // of first k natural number and sequence // from case 1. ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2; return ans; } // Driver program static public void main(String[] args) { int N = 10, K = 2; System.out.println(findSum(N, K)); } } // This Code is contributed by vt_m. Python3 # Python3 program to find sum of modulo # K of first N natural numbers. # Return sum of modulo K of # first N natural numbers. def findSum(N, K): ans = 0; # Counting the number of times # 1, 2, .., K-1, 0 sequence occurs. y = N / K; # Finding the number of elements # left which are incomplete of # sequence Leads to Case 1 type. x = N % K; # adding multiplication of number # of times 1, 2, .., K-1, 0 # sequence occurs and sum of # first k natural number and # sequence from case 1. ans = ((K * (K - 1) / 2) * y + (x * (x + 1)) / 2); return int(ans); # Driver Code N = 10; K = 2; print(findSum(N, K)); # This code is contributed by mits C# // C# program to find sum of modulo // K of first N natural numbers. using System; class GFG { // Return sum of modulo K of // first N natural numbers. static int findSum(int N, int K) { int ans = 0; // Counting the number of times 1, 2, .., // K-1, 0 sequence occurs. int y = N / K; // Finding the number of elements left which // are incomplete of sequence Leads to Case 1 type. int x = N % K; // adding multiplication of number of times // 1, 2, .., K-1, 0 sequence occurs and sum // of first k natural number and sequence // from case 1. ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2; return ans; } // Driver program static public void Main() { int N = 10, K = 2; Console.WriteLine(findSum(N, K)); } } // This code is contributed by vt_m. PHP <?php // PHP program to find sum of modulo // K of first N natural numbers. // Return sum of modulo K of // first N natural numbers. function findSum($N, $K) { $ans = 0; // Counting the number of times // 1, 2, .., K-1, 0 sequence occurs. $y = $N / $K; // Finding the number of elements // left which are incomplete of // sequence Leads to Case 1 type. $x = $N % $K; // adding multiplication of number // of times 1, 2, .., K-1, 0 // sequence occurs and sum of // first k natural number and // sequence from case 1. $ans = ($K * ($K - 1) / 2) * $y + ($x * ($x + 1)) / 2; return $ans; } // Driver program $N = 10; $K = 2; echo findSum($N, $K) ; // This code is contributed by anuj_67. ?> JavaScript <script> // Javascript program to find sum of modulo // K of first N natural numbers. // Return sum of modulo K of // first N natural numbers. function findSum(N, K) { let ans = 0; // Counting the number of times // 1, 2, .., K-1, 0 sequence occurs. let y = N / K; // Finding the number of elements // left which are incomplete of // sequence Leads to Case 1 type. let x = N % K; // adding multiplication of number // of times 1, 2, .., K-1, 0 // sequence occurs and sum of // first k natural number and // sequence from case 1. ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2; return ans; } // Driver code let N = 10; let K = 2; document.write(findSum(N, K)); // This code is contributed by _saurabh_jaiswal </script> Output : 5 Time Complexity : O(1). Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article How to compute mod of a big number? A Anuj Chauhan Improve Article Tags : Misc Mathematical DSA Modular Arithmetic Practice Tags : MathematicalMiscModular Arithmetic Similar Reads Modular Arithmetic Modular arithmetic is a system of arithmetic for numbers where numbers "wrap around" after reaching a certain value, called the modulus. It mainly uses remainders to get the value after wrap around. It is often referred to as "clock arithmetic. 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For example for p = 5, the new sequence would be 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4 … The minimal zero of the 5 min read Modulo of a large Binary StringGiven a large binary string str and an integer K, the task is to find the value of str % K.Examples: Input: str = "1101", K = 45 Output: 13 decimal(1101) % 45 = 13 % 45 = 13 Input: str = "11010101", K = 112 Output: 101 decimal(11010101) % 112 = 213 % 112 = 101 Approach: It is known that (str % K) wh 5 min read Modular multiplicative inverse from 1 to nGive a positive integer n, find modular multiplicative inverse of all integer from 1 to n with respect to a big prime number, say, 'prime'. The modular multiplicative inverse of a is an integer 'x' such that. a x ? 1 (mod prime) Examples: Input : n = 10, prime = 17 Output : 1 9 6 13 7 3 5 15 2 12 Ex 9 min read Modular exponentiation (Recursive)Given three numbers a, b and c, we need to find (ab) % cNow why do “% c†after exponentiation, because ab will be really large even for relatively small values of a, b and that is a problem because the data type of the language that we try to code the problem, will most probably not let us store suc 6 min read Chinese Remainder TheoremIntroduction to Chinese Remainder TheoremWe are given two arrays num[0..k-1] and rem[0..k-1]. In num[0..k-1], every pair is coprime (gcd for every pair is 1). 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This generator polynomial represents key 15+ min read Using Chinese Remainder Theorem to Combine Modular equationsGiven N modular equations: A ? x1mod(m1) . . A ? xnmod(mn) Find x in the equation A ? xmod(m1*m2*m3..*mn) where mi is prime, or a power of a prime, and i takes values from 1 to n. The input is given as two arrays, the first being an array containing values of each xi, and the second array containing 12 min read Like