Given a number, the task is to find if the number is divisible by 9 or not. The input number may be large and it may not be possible to store even if we use long long int.
Examples:
Input : n = 69354 Output : Yes Input : n = 234567876799333 Output : No Input : n = 3635883959606670431112222 Output : No
Since input number may be very large, we cannot use n % 9 to check if a number is divisible by 9 or not, especially in languages like C/C++. The idea is based on following fact.
A number is divisible by 9 if sum of its digits is divisible by 9.
Illustration:
For example n = 9432
Sum of digits = 9 + 4 + 3 + 2
= 18
Since sum is divisible by 9,
answer is Yes.
How does this work?
Let us consider 1332, we can write it as 1332 = 1*1000 + 3*100 + 3*10 + 2 The proof is based on below observation: Remainder of 10i divided by 9 is 1 So powers of 10 only results in remainder 1 when divided by 9. Remainder of "1*1000 + 3*100 + 3*10 + 2" divided by 9 can be written as : 1*1 + 3*1 + 3*1 + 2 = 9 The above expression is basically sum of all digits. Since 9 is divisible by 9, answer is yes.
Below is the implementation of above idea.
C++
// C++ program to find if a number is divisible by // 9 or not #include<bits/stdc++.h> using namespace std; // Function to find that number divisible by 9 or not int check(string str) { // Compute sum of digits int n = str.length(); int digitSum = 0; for (int i=0; i<n; i++) digitSum += (str[i]-'0'); // Check if sum of digits is divisible by 9. return (digitSum % 9 == 0); } // Driver code int main() { string str = "99333"; check(str)? cout << "Yes" : cout << "No "; return 0; } |
Java
// Java program to find if a number is // divisible by 9 or not class IsDivisible { // Function to find that number // is divisible by 9 or not static boolean check(String str) { // Compute sum of digits int n = str.length(); int digitSum = 0; for (int i=0; i<n; i++) digitSum += (str.charAt(i)-'0'); // Check if sum of digits is divisible by 9. return (digitSum % 9 == 0); } // main function public static void main (String[] args) { String str = "99333"; if(check(str)) System.out.println("Yes"); else System.out.println("No"); } } |
Python3
# Python 3 program to # find if a number is # divisible by # 9 or not # Function to find that # number divisible by 9 # or not def check(st) : # Compute sum of digits n = len(st) digitSum = 0 for i in range(0,n) : digitSum = digitSum + (int)(st[i]) # Check if sum of digits # is divisible by 9. return (digitSum % 9 == 0) # Driver code st = "99333" if(check(st)) : print("Yes") else : print("No") # This code is contributed by Nikita Tiwari. |
C#
// C# program to find if a number is // divisible by 9 or not. using System; class GFG { // Function to find that number // is divisible by 9 or not static bool check(String str) { // Compute sum of digits int n = str.Length; int digitSum = 0; for (int i = 0; i < n; i++) digitSum += (str[i] - '0'); // Check if sum of digits is // divisible by 9. return (digitSum % 9 == 0); } // main function public static void Main () { String str = "99333"; if(check(str)) Console.Write("Yes"); else Console.Write("No"); } } // This code is Contributed by // nitin mittal. |
PHP
<?php // PHP program to find if a number // is divisible by 9 or not // Function to find that // number divisible by 9 or not function check($str) { // Compute sum of digits $n = strlen($str); $digitSum = 0; for ($i = 0; $i < $n; $i++) $digitSum += ($str[$i] - '0'); // Check if sum of digits // is divisible by 9. return ($digitSum % 9 == 0); } // Driver code $str = "99333"; $x = check($str) ? "Yes" : "No "; echo($x); // This code is contributed by Ajit. ?> |
Output:
Yes
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