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 the above idea.
C++
#include<bits/stdc++.h>
using namespace std;
int check(string str)
{
int n = str.length();
int digitSum = 0;
for ( int i=0; i<n; i++)
digitSum += (str[i]- '0' );
return (digitSum % 9 == 0);
}
int main()
{
string str = "99333" ;
check(str)? cout << "Yes" : cout << "No " ;
return 0;
}
|
Java
import java.io.*;
class IsDivisible
{
static boolean check(String str)
{
int n = str.length();
int digitSum = 0 ;
for ( int i= 0 ; i<n; i++)
digitSum += (str.charAt(i)- '0' );
return (digitSum % 9 == 0 );
}
public static void main (String[] args)
{
String str = "99333" ;
if (check(str))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
def check(st) :
n = len (st)
digitSum = 0
for i in range ( 0 ,n) :
digitSum = digitSum + ( int )(st[i])
return (digitSum % 9 = = 0 )
st = "99333"
if (check(st)) :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG {
static bool check(String str)
{
int n = str.Length;
int digitSum = 0;
for ( int i = 0; i < n; i++)
digitSum += (str[i] - '0' );
return (digitSum % 9 == 0);
}
public static void Main ()
{
String str = "99333" ;
if (check(str))
Console.Write( "Yes" );
else
Console.Write( "No" );
}
}
|
PHP
<?php
function check( $str )
{
$n = strlen ( $str );
$digitSum = 0;
for ( $i = 0; $i < $n ; $i ++)
$digitSum += ( $str [ $i ] - '0' );
return ( $digitSum % 9 == 0);
}
$str = "99333" ;
$x = check( $str ) ? "Yes" : "No " ;
echo ( $x );
?>
|
Javascript
<script>
function check(str)
{
let n = str.length;
let digitSum = 0;
for (let i = 0; i < n; i++)
digitSum += (str[i] - '0' );
return (digitSum % 9 == 0);
}
let str = "99333" ;
let x = check(str) ? "Yes" : "No " ;
document.write(x);
</script>
|
Time Complexity: O(logN), as we are traversing the digits which will effectively costs logN time.
Auxiliary Space: O(1), as we are not using any extra space.
Method 2: Checking given number is divisible by 9 or not by using the modulo division operator “%”.
C++
#include <iostream>
using namespace std;
int main() {
long long int n = 3635883959606670431112222;
if (n % 9 == 0) {
cout<< "Yes" ;
}
else {
cout<< "No" ;
}
return 0;
}
|
Java
import java.io.*;
import java.math.BigInteger;
class GFG {
public static void main (String[] args)
{
BigInteger n, b1,b2;
n = new BigInteger( "3635883959606670431112222" );
b1 = new BigInteger( "9" );
BigInteger result = n.mod(b1);
b2 = new BigInteger( "0" );
int comparevalue = result.compareTo(b2);
if (comparevalue== 0 ) {
System.out.println( "Yes" );
}
else {
System.out.println( "No" );
}
}
}
|
Python3
n = 3635883959606670431112222
if int (n) % 9 = = 0 :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
public class GFG {
static public void Main()
{
double n = 36358839596066;
if (n % 9 == 0) {
Console.Write( "Yes" );
}
else {
Console.Write( "No" );
}
}
}
|
Javascript
<script>
var n=3635883959606670431112222
if (n%9==0)
document.write( "Yes" )
else
document.write( "No" )
</script>
|
PHP
<?php
$num = 3635883959606670431112222;
if ( $num % 9 == 0)
echo " divisible" ;
else
echo "not divisible" ;
?>
|
Time complexity: O(1) it is performing constant operations
Auxiliary space: O(1)
This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@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.
Please Login to comment...