Check if a large number is divisible by 9 or not

Given a large number as a string s, determine if it is divisible by 9.
Note: The number might be so large that it can't be stored in standard data types like long long.

Examples: 

Input : s = "69354"
Output: Yes
Explanation: 69354 is divisible by 9.

Input: s = "234567876799333"
Output: No
Explanation: 234567876799333 is not divisible by 9.

Input: s = "3635883959606670431112222"
Output: No
Explanation: 3635883959606670431112222 is not divisible by 9.

We can solve this by using this property that a number is divisible by 9 if sum of its digits is divisible by 9.

Let us take an example, s = "69354"
Sum of digits = 6 + 9 + 3 + 5 + 4 = 27
Since 27 is divisible by 9.So Answer will be Yes.

How does this work?  

Consider the number 1332. We can break it down as follows:

69354 = 6×10000 + 9×1000 + 3×100 + 5×10 + 4

Key Observation:

When dividing powers of 10 by 9, the remainder is always 1. This means that:

Remainder of 10ⁱ (for any i) when divided by 9 is 1.

This property allows us to simplify the number for divisibility by 9.

Proof:

Now, let’s find the remainder when dividing 69354 = 6×10000 + 9×1000 + 3×100 + 5×10 + 4 by 9.

• The remainder when dividing 6×10000, 9×1000, 3×100, 5×10 and 4 by 9 is the same as the remainder when dividing:

6×1 + 9×1 + 3×1 + 5×1 +4

• This simplifies to:

6 + 9 + 3 + 5 + 4 = 27

Since 27 is divisible by 9, the number 69354 is also divisible by 9.

#include<iostream>
using namespace std;

// Function to find that number divisible by 9 or not
int check(string s)
{
    // Compute sum of digits
    int n = s.length();
    int sum = 0;
    for (int i=0; i<n; i++)
        sum += (s[i]-'0');

    // Check if sum of digits is divisible by 9.
    return (sum % 9 == 0);
}

int main()
{
    string s = "69354";
    check(s)?  cout << "Yes" : cout << "No ";
    return 0;
}

import java.util.*;

// Function to find that number divisible by 9 or not
public class GfG{
    static boolean check(String s) {
        // Compute sum of digits
        int sum = 0;
        for (int i = 0; i < s.length(); i++)
            sum += (s.charAt(i) - '0');

        // Check if sum of digits is divisible by 9.
        return (sum % 9 == 0);
    }

    public static void main(String[] args) {
        String s = "69354";
        System.out.println(check(s) ? "Yes" : "No");
    }
}

# Function to find that number divisible by 9 or not
def check(s):
    # Compute sum of digits
    sum_digits = sum(int(digit) for digit in s)
    
    # Check if sum of digits is divisible by 9.
    return sum_digits % 9 == 0

s = "69354"
print("Yes" if check(s) else "No")

using System;

class GfG{
    static bool Check(string s) {
        // Compute sum of digits
        int sum = 0;
        foreach (char c in s)
            sum += (c - '0');

        // Check if sum of digits is divisible by 9.
        return sum % 9 == 0;
    }

    static void Main() {
        string s = "69354";
        Console.WriteLine(Check(s) ? "Yes" : "No");
    }
}

function check(s) {
    // Compute sum of digits
    let sum = 0;
    for (let i = 0; i < s.length; i++)
        sum += (s.charAt(i) - '0');

    // Check if sum of digits is divisible by 9.
    return sum % 9 === 0;
}

let s = "69354";
console.log(check(s) ? "Yes" : "No");

Yes

Time Complexity: O(n), n is length of string.
Auxiliary Space: O(1)