# To check whether a large number is divisible by 7

• Difficulty Level : Medium
• Last Updated : 03 Aug, 2022

You are given an n-digit large number, you have to check whether it is divisible by 7.
A (r+1)-digit integer n whose digital form is (ar ar-1 ar-2….a2 a1 a0) is divisible by 7 if and only if the alternate series of numbers (a2 a1 a0) – (a5 a4 a3) + (a8 a7 a6) – … is divisible by 7.

The triplets of digits within parenthesis represent a 3-digit number in digital form.

The given number n can be written as a sum of powers of 1000 as follows.
n= (a2 a1 a0) + (a5 a4 a3)*1000 + (a8 a7 a6)*(1000*1000) +….
As 1000 = (-1)(mod 7), 1000 as per congruence relation.
For a positive integer n, two numbers a and b are said to be congruent modulo n, if their difference
(a – b) is an integer multiple of n (that is, if there is an integer k such that a – b = kn). This congruence relation is typically considered when a and b are integers, and is denoted Hence we can write:
n = { (a2a1a0) + (a5a4a3)* (-1) + (a8a7a6)* (-1)*(-1)+…..}(mod 7),
Thus n is divisible by 7 if and if only if the series is divisible by 7.

Examples:

Input : 8955795758
Output : Divisible by 7
Explanation:
We express the number in terms of triplets
of digits as follows.
(008)(955)(795)(758)
Now, 758- 795 + 955 - 8 = 910, which is
divisible by 7

Input : 100000000000
Output : Not Divisible by 7
Explanation:
We express the number in terms of triplets
of digits as follows.
(100)(000)(000)(000)
Now, 000- 000 + 000 - 100 = -100, which is
not divisible by 7

Note that the number of digits in n may not be multiple of 3. In that case, we pass zero(s) on the left side of the remaining digits(s) after taking out all the triplets (from the right side of n) to form the last triplet.
A simple and efficient method is to take input in form of a string (make its length in form of 3*m by adding 0 to left of the number if required) and then you have to add the digits in blocks of three from the right to left until it becomes a 3 digit number to form an alternate series and check whether the series is divisible by 7 or not.

Here the program implementation to check the divisibility of 7 is done.

Recommended Practice

## C++

 // C++ code to check divisibility of a// given large number by 7#includeusing namespace std; int isdivisible7(string num){    int n = num.length(), gSum=0;    if (n == 0)        return 1;     // Append required 0s at the beginning.    if (n % 3 == 1) {        num="00" + num;        n += 2;    }    else if (n % 3 == 2) {        num= "0" + num;        n++;    }     // add digits in group of three in gSum    int i, GSum = 0, p = 1;    for (i = n - 1; i >= 0; i--) {         // group saves 3-digit group        int group = 0;        group += num[i--] - '0';        group += (num[i--] - '0') * 10;        group += (num[i] - '0') * 100;         gSum = gSum + group * p;         // generate alternate series of plus        // and minus        p *= (-1);    }     return (gSum % 7 == 0);} // Driver codeint main(){    // Driver method    string num= "8955795758";    if (isdivisible7(num))        cout << "Divisible by 7";    else        cout << "Not Divisible by 7";    return 0;}

## C

 // C code to check divisibility of a// given large number by 7#include #include int isdivisible7(char num[]){    int n = strlen(num), gSum=0;    char final[n+3];    if (n == 0 && num == '\n')        return 1;     // Append required 0s at the beginning.    if (n % 3 == 1) {        final='0';        final='0';        strcat(final,num);        n += 2;    }    else if (n % 3 == 2) {        final='0';        strcat(final,num);        n++;    }     // add digits in group of three in gSum    int i, GSum = 0, p = 1;    for (i = n - 1; i >= 0; i--) {         // group saves 3-digit group        int group = 0;        group += final[i--] - '0';        group += (final[i--] - '0') * 10;        group += (final[i] - '0') * 100;         gSum = gSum + group * p;         // generate alternate series of plus        // and minus        p *= (-1);    }     return (gSum % 7 == 0);} // Driver codeint main(){    // Driver method    char num[] = "8955795758";    if (isdivisible7(num))        printf("Divisible by 7");    else        printf("Not Divisible by 7");    return 0;}

## Java

 // Java code to check divisibility of a given large number by 7 class Test {    // Method to check divisibility    static boolean isDivisible7(String num)    {        int n = num.length();        if (n == 0 && num.charAt(0) == '0')            return true;         // Append required 0s at the beginning.        if (n % 3 == 1)            num = "00" + num;        if (n % 3 == 2)            num = "0" + num;        n = num.length();         // add digits in group of three in gSum        int gSum = 0, p = 1;        for (int i = n - 1; i >= 0; i--) {             // group saves 3-digit group            int group = 0;            group += num.charAt(i--) - '0';            group += (num.charAt(i--) - '0') * 10;            group += (num.charAt(i) - '0') * 100;            gSum = gSum + group * p;            // generate alternate series of plus and minus            p = p * -1;        }         // calculate result till 3 digit sum        return (gSum % 7 == 0);    }     // Driver method    public static void main(String args[])    {        String num = "8955795758";         System.out.println(isDivisible7(num) ? "Divisible by 7" : "Not Divisible  by 7");    }}

## Python3

 # Python 3 code to check divisibility# of a given large number by 7 def isdivisible7(num):    n = len(num)    if (n == 0 and num == '\n'):        return 1     # Append required 0s at the beginning.    if (n % 3 == 1) :        num = "00" + str(num)        n += 2         elif (n % 3 == 2) :        num = "0" + str(num)        n += 1     # add digits in group of three in gSum    GSum = 0    p = 1    i = n-1    while i>=0 :         # group saves 3-digit group        group = 0        group += ord(num[i]) - ord('0')        i -= 1        group += (ord(num[i]) - ord('0')) * 10        i -= 1        group += (ord(num[i]) - ord('0')) * 100         GSum = GSum + group * p         # generate alternate series of        # plus and minus        p *= (-1)        i -= 1     return (GSum % 7 == 0) # Driver codeif __name__ == "__main__":         num = "8955795758"    if (isdivisible7(num)):        print("Divisible by 7")    else :        print("Not Divisible by 7") # This code is contributed by ChitraNayal

## C#

 // C# code to check divisibility of a// given large number by 7using System; class GFG {     // Method to check divisibility    static bool isDivisible7(String num)    {        int n = num.Length;        if (n == 0 && num == '0')            return true;         // Append required 0s at the beginning.        if (n % 3 == 1)            num = "00" + num;         if (n % 3 == 2)            num = "0" + num;         n = num.Length;         // add digits in group of three in gSum        int gSum = 0, p = 1;        for (int i = n - 1; i >= 0; i--) {             // group saves 3-digit group            int group = 0;            group += num[i--] - '0';            group += (num[i--] - '0') * 10;            group += (num[i] - '0') * 100;            gSum = gSum + group * p;             // generate alternate series            // of plus and minus            p = p * -1;        }         // calculate result till 3 digit sum        return (gSum % 7 == 0);    }     // Driver code    static public void Main()    {        String num = "8955795758";         // Function calling        Console.WriteLine(isDivisible7(num) ? "Divisible by 7" : "Not Divisible by 7");    }} // This code is contributed by Ajit.

## PHP

 = 0; $i--) {  // group saves 3-digit group $group = 0;        $group += $num[$i--] - '0'; $group += ($num[$i--] - '0') * 10;        $group += ($num[$i] - '0') * 100; $gSum = $gSum + $group * $p;   // generate alternate series // of plus and minus $p = $p * -1; }  // calculate result till 3 digit sum return ($gSum % 7 == 0);} // Driver Code$num = "8955795758"; echo (isDivisible7($num) ?        "Divisible by 7" :        "Not Divisible by 7"); // This code is contributed by Ryuga?>

## Javascript

 

Output

Divisible by 7

Method 2: Checking given number is divisible by 7 or not by using modulo division operator “%”.

## C++

 #include using namespace std;int main(){    //input    long long int n=100000000000;              // finding given number is divisible by 7 or not    if (n%7==0)    {        cout << "Yes";    }    else    {        cout << "No";    }       return 0;}

## Java

 // Java code// To check whether the given number is divisible by 7 or not import java.io.*;import java.util.*;  class GFG{     public static void main(String[] args)  {    //input    long n=100000000000L;    // finding given number is divisible by 7 or not         if ((n)%7==0)    {        System.out.println("Yes");    }    else    {        System.out.println("No");    }       }}//this code is contributed by aditya942003patil

## Python3

 # Python code# To check whether the given number is divisible by 7 or not #inputn=100000000000# the above input can also be given as n=input() -> taking input from user# finding given number is divisible by 7 or notif int(n)%7==0:  print("Yes")else:  print("No")     # this code is contributed by gangarajula laxmi

## C#

 // c# code to check whether the given// number is diivisible by 7 or notusing System;  public class GFG {          public static void Main()    {        //input        long n=100000000000;                 // the above input can also be given as n=input() -> taking input from user        // finding given number is divisible by 7 or not        if (n%7==0)        {            Console.Write("Yes");        }        else        {            Console.Write("No");        }    }}// This code is contributed by aditya942003patil

## Javascript

 

## PHP

  taking input from user    // finding given number is divisible by 7 or not    if (\$n%7==0)    {        echo "Yes";    }    else    {        echo "No";    }  // This code is contributed by laxmigangarajula03?>

Output

No

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