Program to find remainder without using modulo or % operator Last Updated : 11 Jul, 2022 Summarize Suggest changes Share Like Article Like Report Given two numbers 'num' and 'divisor', find remainder when 'num' is divided by 'divisor'. The use of modulo or % operator is not allowed.Examples : Input: num = 100, divisor = 7 Output: 2 Input: num = 30, divisor = 9 Output: 3 Method 1 : C++ // C++ program to find remainder without using // modulo operator #include <iostream> using namespace std; // This function returns remainder of num/divisor // without using % (modulo) operator int getRemainder(int num, int divisor) { return (num - divisor * (num / divisor)); } // Driver program to test above functions int main() { // cout << 100 %0; cout << getRemainder(100, 7); return 0; } Java // Java program to find remainder without // using modulo operator import java.io.*; class GFG { // This function returns remainder of // num/divisor without using % (modulo) // operator static int getRemainder(int num, int divisor) { return (num - divisor * (num / divisor)); } // Driver program to test above functions public static void main(String[] args) { // print 100 % 0; System.out.println(getRemainder(100, 7)); } } // This code is contributed by Sam007. Python3 # Python program to find remainder # without using modulo operator # This function returns remainder of # num / divisor without using % (modulo) # operator def getRemainder(num, divisor): return (num - divisor * (num // divisor)) # Driver program to test above functions num = 100 divisor = 7 print(getRemainder(num, divisor)) # This code is contributed by Danish Raza C# // C# program to find remainder without using // modulo operator using System; class GFG { // This function returns remainder of // num/divisor without using % // (modulo) operator static int getRemainder(int num, int divisor) { return (num - divisor * (num / divisor)); } // Driver program to test above functions public static void Main() { // print 100 % 0; Console.Write(getRemainder(100, 7)); } } // This code is contributed by Sam007. PHP <?php // PHP program to find remainder // without using modulo operator // This function returns remainder // of num/divisor without using // % (modulo) operator function getRemainder($num, $divisor) { $t = ($num - $divisor * (int)($num / $divisor)); return $t; } // Driver Code echo getRemainder(100, 7); // This code is contributed by ajit ?> JavaScript <script> // Javascript program to find remainder // without using modulo operator // This function returns remainder // of num/divisor without using // % (modulo) operator function getRemainder(num, divisor) { let t = (num - divisor * parseInt(num / divisor)); return t; } // Driver Code document.write(getRemainder(100, 7)); // This code is contributed by _saurabh_jaiswal </script> Output : 2 Time Complexity: O(1) Auxiliary Space: O(1)This method is contributed by Bishal Kumar Dubey Method 2 The idea is simple, we run a loop to find the largest multiple of 'divisor' that is smaller than or equal to 'num'. Once we find such a multiple, we subtract the multiple from 'num' to find the divisor.Following is the implementation of above idea. Thanks to eleventyone for suggesting this solution in a comment. C++ // C++ program to find remainder without using modulo operator #include <iostream> using namespace std; // This function returns remainder of num/divisor without // using % (modulo) operator int getRemainder(int num, int divisor) { // Handle divisor equals to 0 case if (divisor == 0) { cout << "Error: divisor can't be zero \n"; return -1; } // Handle negative values if (divisor < 0) divisor = -divisor; if (num < 0) num = -num; // Find the largest product of 'divisor' that is smaller // than or equal to 'num' int i = 1; int product = 0; while (product <= num) { product = divisor * i; i++; } // return remainder return num - (product - divisor); } // Driver program to test above functions int main() { // cout << 100 %0; cout << getRemainder(100, 7); return 0; } Java // Java program to find remainder without // using modulo operator import java.io.*; class GFG { // This function returns remainder // of num/divisor without using % // (modulo) operator static int getRemainder(int num, int divisor) { // Handle divisor equals to 0 case if (divisor == 0) { System.out.println("Error: divisor " + "can't be zero \n"); return -1; } // Handle negative values if (divisor < 0) divisor = -divisor; if (num < 0) num = -num; // Find the largest product of 'divisor' // that is smaller than or equal to 'num' int i = 1; int product = 0; while (product <= num) { product = divisor * i; i++; } // return remainder return num - (product - divisor); } // Driver program to test above functions public static void main(String[] args) { // print 100 % 0; System.out.println(getRemainder(100, 7)); } } // This code is contributed by Sam007. Python3 # Python program to find remainder without # using modulo operator. This function # returns remainder of num / divisor without # using % (modulo) operator def getRemainder(num, divisor): # Handle divisor equals to 0 case if (divisor == 0): return False # Handle negative values if (divisor < 0): divisor = -divisor if (num < 0): num = -num # Find the largest product of 'divisor' # that is smaller than or equal to 'num' i = 1 product = 0 while (product <= num): product = divisor * i i += 1 # return remainder return num - (product - divisor) # Driver program to test above functions num = 100 divisor = 7 print(getRemainder(num, divisor)) # This code is contributed by Danish Raza C# // C# program to find remainder without // using modulo operator using System; class GFG { // This function returns remainder // of num/divisor without using % // (modulo) operator static int getRemainder(int num, int divisor) { // Handle divisor equals to 0 case if (divisor == 0) { Console.WriteLine("Error: divisor " + "can't be zero \n"); return -1; } // Handle negative values if (divisor < 0) divisor = -divisor; if (num < 0) num = -num; // Find the largest product of 'divisor' // that is smaller than or equal to 'num' int i = 1; int product = 0; while (product <= num) { product = divisor * i; i++; } // return remainder return num - (product - divisor); } // Driver program to test above functions public static void Main() { // print 100 %0; Console.Write(getRemainder(100, 7)); } } // This code is contributed by Sam007. PHP <?php // php program to find remainder without // using modulo operator // This function returns remainder of // num/divisor without using % (modulo) // operator function getRemainder($num, $divisor) { // Handle divisor equals to 0 case if ($divisor == 0) { echo "Error: divisor can't be zero \n"; return -1; } // Handle negative values if ($divisor < 0) $divisor = -$divisor; if ($num < 0) $num = -$num; // Find the largest product of 'divisor' // that is smaller than or equal to 'num' $i = 1; $product = 0; while ($product <= $num) { $product = $divisor * $i; $i++; } // return remainder return $num - ($product - $divisor); } // Driver program to test above functions echo getRemainder(100, 7); // This code is contributed by ajit. ?> JavaScript // Javascript program to find remainder without // using modulo operator // This function returns remainder of // num/divisor without using % (modulo) // operator function getRemainder(num, divisor) { // Handle divisor equals to 0 case if (divisor == 0) { document.write("Error: divisor can't be zero <br>"); return -1; } // Handle negative values if (divisor < 0) divisor = -divisor; if (num < 0) num = -num; // Find the largest product of 'divisor' // that is smaller than or equal to 'num' let i = 1; let product = 0; while (product <= num) { product = divisor * i; i++; } // return remainder return num - (product - divisor); } // Driver program to test above functions document.write(getRemainder(100, 7)); // This code is contributed by _saurabh_jaiswal Output : 2 Time Complexity: O(n) Auxiliary Space: O(1) Method 3 Keep subtracting the denominator from numerator until the numerator is less than the denominator. C++ // C++ implementation of the approach #include <iostream> using namespace std; // Function to return num % divisor // without using % (modulo) operator int getRemainder(int num, int divisor) { // While divisor is smaller // than n, keep subtracting // it from num while (num >= divisor) num -= divisor; return num; } // Driver code int main() { int num = 100, divisor = 7; cout << getRemainder(num, divisor); return 0; } Java // A Java implementation of the approach import java.util.*; class GFG { // Function to return num % divisor // without using % (modulo) operator static int getRemainder(int num, int divisor) { // While divisor is smaller // than n, keep subtracting // it from num while (num >= divisor) num -= divisor; return num; } // Driver code public static void main(String[] args) { int num = 100, divisor = 7; System.out.println(getRemainder(num, divisor)); } } // This code is contributed by Princi Singh Python3 # Python3 implementation of the approach # Function to return num % divisor # without using % (modulo) operator def getRemainder(num, divisor): # While divisor is smaller # than n, keep subtracting # it from num while (num >= divisor): num -= divisor; return num; # Driver code if __name__ == '__main__': num = 100; divisor = 7; print(getRemainder(num, divisor)); # This code is contributed by Princi Singh C# // C# implementation of the approach using System; class GFG { // Function to return num % divisor // without using % (modulo) operator static int getRemainder(int num, int divisor) { // While divisor is smaller // than n, keep subtracting // it from num while (num >= divisor) num -= divisor; return num; } // Driver code public static void Main(String[] args) { int num = 100, divisor = 7; Console.WriteLine(getRemainder(num, divisor)); } } // This code is contributed by PrinciRaj1992 JavaScript // Javascript implementation of the approach // Function to return num % divisor // without using % (modulo) operator function getRemainder(num, divisor) { // While divisor is smaller // than n, keep subtracting // it from num while (num >= divisor) num -= divisor; return num; } // Driver code let num = 100, divisor = 7; document.write(getRemainder(num, divisor)); // This code is contributed by _saurabh_jaiswal Output : 2 Time Complexity: O(n) Auxiliary Space: O(1) Advertise with us Next Article Modular multiplicative inverse K kartik Similar Reads Number Theory for DSA & Competitive Programming What is Number Theory?Number theory is a branch of pure mathematics that deals with the properties and relationships of numbers, particularly integers. 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The strings will only contain lowercase English alphabets.While reporting the results, use 1-based indexing (i.e., the first character of the text is at po 12 min read Measure one litre using two vessels and infinite water supplyThere are two vessels of capacities 'a' and 'b' respectively. We have infinite water supply. Give an efficient algorithm to make exactly 1 litre of water in one of the vessels. You can throw all the water from any vessel any point of time. Assume that 'a' and 'b' are Coprimes.Following are the steps 15 min read Program to find last digit of n'th Fibonacci NumberGiven a number 'n', write a function that prints the last digit of n'th ('n' can also be a large number) Fibonacci number. Examples : Input : n = 0 Output : 0 Input: n = 2 Output : 1 Input : n = 7 Output : 3 Recommended PracticeThe Nth FibonnaciTry It! Method 1 : (Naive Method) Simple approach is to 13 min read GCD of two numbers when one of them can be very largeGiven two numbers 'a' and 'b' such that (0 <= a <= 10^12 and b <= b < 10^250). Find the GCD of two given numbers.Examples : Input: a = 978 b = 89798763754892653453379597352537489494736 Output: 6 Input: a = 1221 b = 1234567891011121314151617181920212223242526272829 Output: 3 Solution : In 9 min read Find Last Digit of a^b for Large NumbersYou are given two integer numbers, the base a (number of digits d, such that 1 <= d <= 1000) and the index b (0 <= b <= 922*10^15). You have to find the last digit of a^b.Examples: Input : 3 10Output : 9Input : 6 2Output : 6Input : 150 53Output : 0 After taking few examples, we can notic 9 min read Remainder with 7 for large numbersGiven a large number as a string, find the remainder of number when divided by 7. Examples : Input : num = 1234 Output : 2 Input : num = 1232 Output : 0 Input : num = 12345 Output : 4Recommended PracticeRemainder with 7Try It! Simple Approach is to convert a string into number and perform the mod op 8 min read Find (a^b)%m where 'a' is very largeGiven three numbers a, b and m where 1<=b,m<=10^6 and 'a' may be very large and contains upto 10^6 digits. The task is to find (a^b)%m. Examples: Input : a = 3, b = 2, m = 4 Output : 1 Explanation : (3^2)%4 = 9%4 = 1 Input : a = 987584345091051645734583954832576, b = 3, m = 11 Output: 10Recomm 15+ min read Find sum of modulo K of first N natural numberGiven 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 PracticeReve 9 min read Count sub-arrays whose product is divisible by kGiven an integer K and an array arr[], the task is to count all the sub-arrays whose product is divisible by K.Examples: Input: arr[] = {6, 2, 8}, K = 4 Output: 4 Required sub-arrays are {6, 2}, {6, 2, 8}, {2, 8}and {8}.Input: arr[] = {9, 1, 14}, K = 6 Output: 1 Naive approach: Run nested loops and 15+ min read Partition a number into two divisible partsGiven a number (as string) and two integers a and b, divide the string in two non-empty parts such that the first part is divisible by a and the second part is divisible by b. If the string can not be divided into two non-empty parts, output "NO", else print "YES" with the two parts. Examples: Input 15+ min read Find power of power under mod of a primeGiven four numbers A, B, C and M, where M is prime number. Our task is to compute A raised to power (B raised to power C) modulo M. Example: Input : A = 2, B = 4, C = 3, M = 23Output : 643 = 64 so,2^64(mod 23) = 6 A Naive Approach is to calculate res = BC and then calculate Ares % M by modular expon 7 min read Rearrange an array in maximum minimum form in O(1) extra spaceGiven a sorted array of positive integers, rearrange the array alternately i.e first element should be the maximum value, second minimum value, third-second max, fourth-second min and so on. Examples:Input: arr[] = {1, 2, 3, 4, 5, 6, 7} Output: arr[] = {7, 1, 6, 2, 5, 3, 4}Explanation: First 7 is th 8 min read Subset with no pair sum divisible by KGiven an array of integer numbers, we need to find maximum size of a subset such that sum of each pair of this subset is not divisible by K. Examples : Input : arr[] = [3, 7, 2, 9, 1] K = 3 Output : 3 Maximum size subset whose each pair sum is not divisible by K is [3, 7, 1] because, 3+7 = 10, 3+1 = 7 min read Number of substrings divisible by 6 in a string of integersGiven a string consisting of integers 0 to 9. The task is to count the number of substrings which when convert into integer are divisible by 6. Substring does not contain leading zeroes. Examples: Input : s = "606". Output : 5 Substrings "6", "0", "6", "60", "606" are divisible by 6. Input : s = "48 9 min read Miscellaneous Practice ProblemsHow to compute mod of a big number?Given a big number 'num' represented as string and an integer x, find value of "num % a" or "num mod a". Output is expected as an integer. Examples : Input: num = "12316767678678", a = 10 Output: num (mod a) ? 8 The idea is to process all digits one by one and use the property that xy (mod a) ? ((x 4 min read BigInteger Class in JavaBigInteger class is used for the mathematical operation which involves very big integer calculations that are outside the limit of all available primitive data types. In this way, BigInteger class is very handy to use because of its large method library and it is also used a lot in competitive progr 6 min read Modulo 10^9+7 (1000000007)In most programming competitions, we are required to answer the result in 10^9+7 modulo. The reason behind this is, if problem constraints are large integers, only efficient algorithms can solve them in an allowed limited time.What is modulo operation: The remainder obtained after the division opera 10 min read How to avoid overflow in modular multiplication?Consider below simple method to multiply two numbers. C++ #include <iostream> using namespace std; #define ll long long // Function to multiply two numbers modulo mod ll multiply(ll a, ll b, ll mod) { // Multiply two numbers and then take modulo to prevent // overflow return ((a % mod) * (b % 6 min read RSA Algorithm in CryptographyRSA(Rivest-Shamir-Adleman) Algorithm is an asymmetric or public-key cryptography algorithm which means it works on two different keys: Public Key and Private Key. The Public Key is used for encryption and is known to everyone, while the Private Key is used for decryption and must be kept secret by t 13 min read Article Tags : Mathematical DSA Chalk Studio Modular Arithmetic Practice Tags : MathematicalModular Arithmetic Like