Open In App

# To check whether a large number is divisible by 7

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 : 8955795758Output : 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 7Input : 100000000000Output : 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[0] == '\n')        return 1;     // Append required 0s at the beginning.    if (n % 3 == 1) {        final[0]='0';        final[1]='0';        strcat(final,num);        n += 2;    }    else if (n % 3 == 2) {        final[0]='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[0] == '\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] == '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.

## Javascript

 

## 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?>

Output

Divisible by 7



Time Complexity: O(n), where n is the number of digits in num.
Auxiliary Space: O(1)

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 17 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 17 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 71 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 711 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;                 // 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

 

Output

No



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

Method 3: Checking given String(containing digits 0-9) is divisible by 7 or not by checking each character of the string and taking its modulo % .

Firstly we take first character of the String and take its modulo with 7 .

Then we add the remainder with the next character and take modulo of that and so.. on , until we reach the end of the string .

If at last we have remainder as 0 we get string divisible by 7 otherwise we get it as not divisible by 7 .

## C++

 // C++ code for the above approach#include using namespace std; bool isDivisibleBySeven(string s){    int remainder = 0;    for (int i = 0; i < s.length(); i++) {        int digit = remainder * 10 + (s[i] - '0');        remainder = digit % 7;    }     return remainder == 0;}int main(){    string s = "8955795758";    bool check = isDivisibleBySeven(s);    cout << ((check) ? "String is divisible by 7"                     : "String is not divisible by 7");    return 0;} // This code is contributed by karandeep1234

## Java

 /*package whatever //do not write package name here */ import java.io.*;import java.util.Scanner; class GFG {    public boolean isDivisibleBySeven(String s)    {        int remainder = 0;        for (int i = 0; i < s.length(); i++) {            int digit                = remainder * 10 + (s.charAt(i) - '0');            remainder = digit % 7;        }         return remainder == 0;    }    public static void main(String[] args)    {        GFG obj = new GFG();        String s = "8955795758";        boolean check = obj.isDivisibleBySeven(s);        System.out.print(            (check) ? "String is divisible by 7"                    : "String is not divisible by 7");    }}

## Python3

 # Python code for the above approachdef isDivisibleBySeven(s):    remainder = 0    for i in range(len(s)):        digit = remainder * 10 + int(s[i]) - 0        remainder = digit % 7     return remainder == 0  s = "8955795758"check = isDivisibleBySeven(s)print("String is divisible by 7" if check else "String is not divisible by 7") # This code is contributed by lokesh

## C#

 // C# code for the above approach using System; public class GFG {     public bool isDivisibleBySeven(String s)    {        int remainder = 0;        for (int i = 0; i < s.Length; i++) {            int digit = remainder * 10 + (s[i] - '0');            remainder = digit % 7;        }         return remainder == 0;    }     static public void Main()    {         GFG obj = new GFG();        string s = "8955795758";        bool check = obj.isDivisibleBySeven(s);        Console.Write((check)                          ? "String is divisible by 7"                          : "String is not divisible by 7");    }} // This code is contributed by lokesh.

## Javascript

 // JavaScript code for the above approach function isDivisibleBySeven(s) {    let remainder = 0;    for (let i = 0; i < s.length; i++) {        let digit = remainder * 10 + (s.charCodeAt(i) - '0'.charCodeAt(0));        remainder = digit % 7;    }    return remainder === 0;} let s = "8955795758";let check = isDivisibleBySeven(s);console.log(check ? "String is divisible by 7" : "String is not divisible by 7"); // This code is contributed by lokeshmvs21.

Output

String is divisible by 7



Time Complexity: O(n) , n being length of the input string .
Auxiliary Space: O(1)

### Method 4:

We can use the following property to check whether a number is divisible by 7 or not:

• If we subtract twice the last digit of the number from the remaining prefix, the resulting number will be divisible by 7 if and only if the original number is divisible by 7.

We can repeatedly apply this property until we are left with a 2-digit number, and then check whether it is divisible by 7 or not.

Steps:

• Initialize the given number as n.
• While n is greater than or equal to 100, do the following:
a. Extract the last digit of n.
b. Update n as n // 10 – 2 * last_digit.
• Check whether the remaining 2-digit number is divisible by 7 or not.
• If the remaining 2-digit number is divisible by 7, return “Divisible by 7”, otherwise return “Not divisible by 7”.

## C++

 #include using namespace std; string is_divisible_by_7(long long n){    while (n >= 100) {        int last_digit = n % 10;        n = n / 10 - 2 * last_digit;    }    return (n % 7 == 0) ? "Divisible by 7"                        : "Not divisible by 7";} // example usageint main(){    long long number = 8955795758;    cout << is_divisible_by_7(number) << endl;    return 0;}

## Java

 import java.util.*; public class Main {    public static String isDivisibleBy7(long n)    {        while (n >= 100) {            int lastDigit = (int)(n % 10);            n = n / 10 - 2 * lastDigit;        }        return (n % 7 == 0) ? "Divisible by 7"                            : "Not divisible by 7";    }     public static void main(String[] args)    {        long number = 8955795758L;        System.out.println(isDivisibleBy7(number));    }}

## Python3

 def is_divisible_by_7(n):    while n >= 100:        last_digit = n % 10        n = n // 10 - 2 * last_digit    return "Divisible by 7" if n % 7 == 0 else "Not divisible by 7"  # example usagenumber = 8955795758print(is_divisible_by_7(number))

## C#

 using System; class GFG{    static string IsDivisibleBy7(long n)    {        while (n >= 100)        {            int lastDigit = (int)(n % 10);            n = n / 10 - 2 * lastDigit;        }        return (n % 7 == 0) ? "Divisible by 7" : "Not divisible by 7";    }     static void Main(string[] args)    {        long number = 8955795758;        Console.WriteLine(IsDivisibleBy7(number));    }}

## Javascript

 // JavaScript function to check if a number is divisible by 7function isDivisibleBy7(n) {    // While the number is greater than or equal to 100    while (n >= 100) {        // Get the last digit of the number        var lastDigit = n % 10;                 // Update the number by removing the last digit and subtracting twice the last digit from it        n = Math.floor(n / 10) - 2 * lastDigit;    }         // Check if the resulting number is divisible by 7    return (n % 7 === 0) ? "Divisible by 7" : "Not divisible by 7";} // Test the function with the given numbervar number = 8955795758;console.log(isDivisibleBy7(number));

Output

Divisible by 7



Time complexity: O(log10 n)

Auxiliary space: O(1)