# Check if a large number is divisible by 3 or not

Given a number, the task is that we divide number by 3. The input number may be large and it may not be possible to store even if we use long long int.

Examples:

```Input  : n = 769452
Output : Yes

Input  : n = 123456758933312
Output : No

Input  : n = 3635883959606670431112222
Output : Yes```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Since input number may be very large, we cannot use n % 3 to check if a number is divisible by 3 or not, especially in languages like C/C++. The idea is based on following fact.

A number is divisible by 3 if sum of its digits is divisible by 3.

Illustration:

```For example n = 1332
Sum of digits = 1 + 3 + 3 + 2
= 9
Since sum is divisible by 3,
```

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 3 is 1
So powers of 10 only result in value 1.

Remainder of "1*1000 + 3*100 + 3*10 + 2"
divided by 3 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 3, answer is yes.
```

Below is the implementation of above fact :

## C++

```// C++ program to find if a number is divisible by
// 3 or not
#include<bits/stdc++.h>
using namespace std;

// Function to find that number divisble by 3 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 3.
return (digitSum % 3 == 0);
}

// Driver code
int main()
{
string str = "1332";
check(str)?  cout << "Yes" : cout << "No ";
return 0;
}
```

## Java

```// Java program to find if a number is
// divisible by 3 or not
class IsDivisible
{
// Function to find that number
// divisble by 3 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 3.
return (digitSum % 3 == 0);
}

// main function
public static void main (String[] args)
{
String str = "1332";
if(check(str))
System.out.println("Yes");
else
System.out.println("No");
}
}
```

## Python

```# Python program to find if a number is
# divisible by 3 or not

# Function to find that number
# divisble by 3 or not
def check(num) :

# Compute sum of digits
digitSum = 0
while num > 0 :
rem = num % 10
digitSum = digitSum + rem
num = num / 10

# Check if sum of digits is
# divisible by 3.
return (digitSum % 3 == 0)

# main function
num = 1332
if(check(num)) :
print "Yes"
else :
print "No"

# This code is contributed by Nikita Tiwari.
```

Output:

`Yes`

This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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