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

• Difficulty Level : Hard
• Last Updated : 12 Jul, 2022

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 Practice

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 convert 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.

Implementation:

## C++

 `// C++ implementation to check whether decimal representation``// of given binary number is divisible by 5 or not``#include ` `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

## 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

## Python 3

 `# Python3 implementation to check whether``# decimal representation of given binary``# number is divisible by 5 or not` `# function to return equivalent base 4``# number of the given binary number``def` `equivalentBase4(``bin``):``    ``if``(``bin` `=``=` `"00"``):``        ``return` `0``    ``if``(``bin` `=``=` `"01"``):``        ``return` `1``    ``if``(``bin` `=``=` `"10"``):``        ``return` `2``    ``if``(``bin` `=``=` `"11"``):``        ``return` `3``    ` `# function to check whether the given``# binary number is divisible by 5 or not    ``def` `isDivisibleBy5(``bin``):``    ``l ``=` `len``(``bin``)``    ``if``((l ``%` `2``) ``=``=` `1``):``        ` `    ``# 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``    ``odd_sum ``=` `0``    ``even_sum ``=` `0``    ``isOddDigit ``=` `1``    ``for` `i ``in` `range``(``0``, ``len``(``bin``), ``2``):``        ` `        ``# if digit of base 4 is at odd place,``        ``# then add it to odd_sum``        ``if``(isOddDigit):``            ``odd_sum ``+``=` `equivalentBase4(``bin``[i:i ``+` `2``])``            ` `        ``# else digit of base 4 is at``        ``# even place, add it to even_sum    ``        ``else``:``            ``even_sum ``+``=` `equivalentBase4(``bin``[i:i ``+` `2``])``            ` `        ``isOddDigit ``=` `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``:``        ``return` `"No"` `# Driver Code``if` `__name__``=``=``"__main__"``:``    ``bin` `=` `"10000101001"``    ``print``(isDivisibleBy5(``bin``))` `# This code is contributed``# by Sairahul Jella`

## C#

 `// C# implementation to check whether``// decimal representation of given``// binary number is divisible by 5 or not``using` `System;` `class` `GFG``{``// Method to return equivalent base``// 4 number of the given binary number``public` `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``public` `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 add it to odd_sum``        ``if` `(isOddDigit != 0)``        ``{``            ``odd_sum += equivalentBase4(``                             ``bin.Substring(i, 2));``        ``}``        ` `        ``// else digit of base 4 is at even ``        ``// place, add it to even_sum``        ``else``        ``{``            ``even_sum += equivalentBase4(``                              ``bin.Substring(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"``;` `}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``string` `bin = ``"10000101001"``;``    ``Console.WriteLine(isDivisibleBy5(bin));``}``}` `// This code is contributed by Shrikant13`

## Javascript

 ``

Output

`No`

Time Complexity: O(n), where n is the number of digits in the binary number.
Auxiliary Space: O(1)

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