# Decimal representation of given binary string is divisible by 5 or not

The problem is to check whether the decimal representation of the given binary number is divisible by 5 or not. Take care, the number could be very large and may not fit even in long long int. The approach should be such that there are zero or minimum number of multiplication and division operations. No leading 0’s are there in the input.

Examples:

```Input : 1010
Output : YES
(1010)2 = (10)10,
and 10 is divisible by 5.

Input : 10000101001
Output : YES
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach: The following steps are:

1. Convert the binary number to base 4.
2. Numbers in base 4 contains only 0, 1, 2, 3 as their digits.
3. 5 in base 4 is equivalent to 11.
4. Now apply the rule of divisibility by 11 where you add all the digits at odd places and add all the digits at even places and then subtract one from the other. If the result is divisible by 11(which remember is 5), then the binary number is divisible by 5.

How to covert binary number to base 4 representation?

1. Check whether the length of binary string is even or odd.
2. If odd, the add ‘0’ in the beginning of the string.
3. Now, traverse the string from left to right.
4. One by extract substrings of size 2 and add their equivalent decimal to the resultant string.

## C++

```// C++ implementation to check whether decimal representation
// of given binary number is divisible by 5 or not
#include <bits/stdc++.h>

using namespace std;

// function to return equivalent base 4 number
// of the given binary number
int equivalentBase4(string bin)
{
if (bin.compare("00") == 0)
return 0;
if (bin.compare("01") == 0)
return 1;
if (bin.compare("10") == 0)
return 2;
return 3;
}

// function to check whether the given binary
// number is divisible by 5 or not
string isDivisibleBy5(string bin)
{
int l = bin.size();

if (l % 2 != 0)
// add '0' in the beginning to make
// length an even number
bin = '0' + bin;

// to store sum of digits at odd and
// even places respectively
int odd_sum, even_sum = 0;

// variable check for odd place and
// even place digit
int isOddDigit = 1;
for (int i = 0; i<bin.size(); i+= 2)
{
// if digit of base 4 is at odd place, then
if (isOddDigit)
odd_sum += equivalentBase4(bin.substr(i, 2));
// else digit of base 4 is at even place,
else
even_sum += equivalentBase4(bin.substr(i, 2));

isOddDigit ^= 1;
}

// if this diff is divisible by 11(which is 5 in decimal)
// then, the binary number is divisible by 5
if (abs(odd_sum - even_sum) % 5 == 0)
return "Yes";

// else not divisible by 5
return "No";

}

// Driver program to test above
int main()
{
string bin = "10000101001";
cout << isDivisibleBy5(bin);
return 0;
}
```

## Java

```//Java implementation to check whether decimal representation
//of given binary number is divisible by 5 or not

class GFG
{
// Method to return equivalent base 4 number
// of the given binary number
static int equivalentBase4(String bin)
{
if (bin.compareTo("00") == 0)
return 0;
if (bin.compareTo("01") == 0)
return 1;
if (bin.compareTo("10") == 0)
return 2;
return 3;
}

// Method to check whether the given binary
// number is divisible by 5 or not
static String isDivisibleBy5(String bin)
{
int l = bin.length();

if (l % 2 != 0)
// add '0' in the beginning to make
// length an even number
bin = '0' + bin;

// to store sum of digits at odd and
// even places respectively
int odd_sum=0, even_sum = 0;

// variable check for odd place and
// even place digit
int isOddDigit = 1;
for (int i = 0; i<bin.length(); i+= 2)
{
// if digit of base 4 is at odd place, then
if (isOddDigit != 0)
odd_sum += equivalentBase4(bin.substring(i, i+2));
// else digit of base 4 is at even place,
else
even_sum += equivalentBase4(bin.substring(i, i+2));

isOddDigit ^= 1;
}

// if this diff is divisible by 11(which is 5 in decimal)
// then, the binary number is divisible by 5
if (Math.abs(odd_sum - even_sum) % 5 == 0)
return "Yes";

// else not divisible by 5
return "No";

}

public static void main (String[] args)
{
String bin = "10000101001";
System.out.println(isDivisibleBy5(bin));
}
}
```

Output:

```YES
```

Time Complexity: O(n), where n is the number of digits in the binary number.

This article is contributed by Ayush Jauhari. 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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