Given a large number as string, find remainder of number when divided by 7.

Examples:

Input : num = 1234 Output : 2 Input : num = 1232 Output : 0 Input : num = 12345 Output : 4

Simple Approach is to convert string into number and perform mod operation. But this approach will not work for long strings.

The exist an approach that works for large numbers also. Below are the steps.

Let the given number be “num”

- We use a series 1, 3, 2, -1, -3, -2 to find the remainder (Intuition behind the series is discussed below).
- Initialize the result as 0.
- Traverse num from end and above series from beginning. For every traversed digit, multiply it with next digit of series, and add the multiplication to result.
- Keep repeating step 3 while there are more digits to process. If there are more than 6(number of terms in series) digits, then start traversing the series from beginning.
- After each step, we keep doing result = result % 7 to make sure that result remains less than 7.

**Illustration : **

let us take above Example where number is 12345. We reverse the number from end and series from the beginning and keep adding multiplication to the result. (12345 % 7) = (5*1 + 4*3 + 3*2 + 2*(-1) + 1*(-3)) % 7 = (5 + 12 + 6 - 2 - 3)%7 = (18) % 7 = 4 hence 4 will be the remainder when we divide the number 12345 by 7.

**How does this series formula work?**

Below is the intuition behind the series

1 % 7 = 1 10 % 7 = 3 100 % 7 = 2 1000 % 7 = 6 = (-1) % 7 10000 % 7 = 4 = (-3) % 7 100000 % 7 = 3 = (-2) % 7 The series repeats itself for larger powers 1000000 % 7 = 1 10000000 % 7 = 3 .............. ............... The above property of modular division with 7 and associative properties of modular arithmetic are the basis of the approach user here.

**Implementation:**

## C++

// C++ program to find remainder of a large // number when divided by 7. #include<iostream> using namespace std; /* Function which return Remainder after dividing the number by 7 */ int remainderWith7(string num) { // This series is used to find remainder with 7 int series[] = {1, 3, 2, -1, -3, -2}; // Index of next element in series int series_index = 0; int result = 0; // Initialize result // Traverse num from end for (int i=num.size()-1; i>=0; i--) { /* Find current digit of nun */ int digit = num[i] - '0'; // Add next term to result result += digit * series[series_index]; // Move to next term in series series_index = (series_index + 1) % 6; // Make sure that result never goes beyond 7. result %= 7; } // Make sure that remainder is positive if (result < 0) result = (result + 7) % 7; return result; } /* Driver program */ int main() { string str = "12345"; cout << "Remainder with 7 is " << remainderWith7(str); return 0; }

## Java

// Java program to find remainder of a large // number when divided by 7. class GFG { // Function which return Remainder // after dividingthe number by 7 static int remainderWith7(String num) { // This series is used to // find remainder with 7 int series[] = {1, 3, 2, -1, -3, -2}; // Index of next element in series int series_index = 0; // Initialize result int result = 0; // Traverse num from end for (int i = num.length() - 1; i >= 0; i--) { /* Find current digit of nun */ int digit = num.charAt(i) - '0'; // Add next term to result result += digit * series[series_index]; // Move to next term in series series_index = (series_index + 1) % 6; // Make sure that result never goes beyond 7. result %= 7; } // Make sure that remainder is positive if (result < 0) result = (result + 7) % 7; return result; } // Driver code public static void main (String[] args) { String str = "1232"; System.out.print("Remainder with 7 is " +remainderWith7(str)); } } // This code is contributed by Anant Agarwal.

Output:

4

This article is contributed by ** Kuldeep Singh **. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.