# Decimal representation of given binary string is divisible by 10 or not

The problem is to check whether the decimal representation of the given binary number is divisible by 10 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 : 101000
Output : Yes
(101000)2 = (40)10
and 40 is divisible by 10.

Input : 11000111001110
Output : Yes
```

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

Approach: First of all we need to know that last digit of pow(2, i) = 2, 4, 8, 6 if i % 4 is equal to 1, 2, 3, 0 respectively, where i is greater than equal to 1. So, in the binary representation we need to know the position of digit ‘1’ from the right, so as to know the perfect power of 2 with which it is going to be multiplied. This will help us to obtain the last digit of the required perfect power’s of 2. We can add these digits and then check whether the last digit of the sum is 0 or not which implies that the number is divisible by 10 or not. Note that if the last digit in the binary representation is ‘1’ then it represents an odd number, and thus not divisible by 10.

## C++

 `// C++ implementation to check whether decimal ` `// representation of given binary number is  ` `// divisible by 10 or not ` `#include ` `using` `namespace` `std; ` ` `  `// function to check whether decimal representation  ` `// of given binary number is divisible by 10 or not ` `bool` `isDivisibleBy10(string bin) ` `{ ` `    ``int` `n = bin.size(); ` `     `  `    ``// if last digit is '1', then ` `    ``// number is not divisible by 10 ` `    ``if` `(bin[n-1] == ``'1'``) ` `        ``return` `false``; ` `     `  `    ``// to accumulate the sum of last digits ` `    ``// in perfect powers of 2 ` `    ``int` `sum = 0; ` `     `  `    ``// traverse from the 2nd last up to 1st digit ` `    ``// in 'bin' ` `    ``for` `(``int` `i=n-2; i>=0; i--)     ` `    ``{ ` `        ``// if digit in '1' ` `        ``if` `(bin[i] == ``'1'``) ` `        ``{ ` `            ``// calculate digit's position from ` `            ``// the right ` `            ``int` `posFromRight = n - i - 1; ` `             `  `            ``// according to the digit's position,  ` `            ``// obtain the last digit of the applicable  ` `            ``// perfect power of 2 ` `            ``if` `(posFromRight % 4 == 1) ` `                ``sum = sum + 2; ` `            ``else` `if` `(posFromRight % 4 == 2) ` `                ``sum = sum + 4; ` `            ``else` `if` `(posFromRight % 4 == 3) ` `                ``sum = sum + 8; ` `            ``else` `if` `(posFromRight % 4 == 0) ` `                ``sum = sum + 6;             ` `        ``} ` `    ``} ` `     `  `    ``// if last digit is 0, then ` `    ``// divisible by 10 ` `    ``if` `(sum % 10 == 0) ` `        ``return` `true``; ` `     `  `    ``// not divisible by 10     ` `    ``return` `false``;     ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``string bin = ``"11000111001110"``; ` `     `  `    ``if` `(isDivisibleBy10(bin)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``;     ` `         `  `    ``return` `0; ` `}  `

## Java

 `// Java implementation to check whether decimal ` `// representation of given binary number is  ` `// divisible by 10 or not ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``// function to check whether decimal  ` `    ``// representation of given binary number ` `    ``// is divisible by 10 or not ` `    ``static` `boolean` `isDivisibleBy10(String bin) ` `    ``{ ` `        ``int` `n = bin.length(); ` `          `  `        ``// if last digit is '1', then ` `        ``// number is not divisible by 10 ` `        ``if` `(bin.charAt(n - ``1``) == ``'1'``) ` `            ``return` `false``; ` `          `  `        ``// to accumulate the sum of last  ` `        ``// digits in perfect powers of 2 ` `        ``int` `sum = ``0``; ` `          `  `        ``// traverse from the 2nd last up to ` `        ``// 1st digit in 'bin' ` `        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--)     ` `        ``{ ` `            ``// if digit in '1' ` `            ``if` `(bin.charAt(i) == ``'1'``) ` `            ``{ ` `                ``// calculate digit's position ` `                ``// from the right ` `                ``int` `posFromRight = n - i - ``1``; ` `                  `  `                ``// according to the digit's  ` `                ``// position, obtain the last  ` `                ``// digit of the applicable  ` `                ``// perfect power of 2 ` `                ``if` `(posFromRight % ``4` `== ``1``) ` `                    ``sum = sum + ``2``; ` `                ``else` `if` `(posFromRight % ``4` `== ``2``) ` `                    ``sum = sum + ``4``; ` `                ``else` `if` `(posFromRight % ``4` `== ``3``) ` `                    ``sum = sum + ``8``; ` `                ``else` `if` `(posFromRight % ``4` `== ``0``) ` `                    ``sum = sum + ``6``;             ` `            ``} ` `        ``} ` `          `  `        ``// if last digit is 0, then ` `        ``// divisible by 10 ` `        ``if` `(sum % ``10` `== ``0``) ` `            ``return` `true``; ` `          `  `        ``// not divisible by 10     ` `        ``return` `false``;     ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``String bin = ``"11000111001110"``; ` `          `  `        ``if` `(isDivisibleBy10(bin)) ` `            ``System.out.print(``"Yes"``); ` `        ``else` `            ``System.out.print(``"No"``);    ` `         `  `        ``} ` `    ``} ` ` `  `// This code is contributed by Arnav Kr. Mandal.     `

## Python

 `# Python implementation to check whether ` `# decimal representation of given binary ` `# number is divisible by 10 or not ` ` `  ` `  `# function to check whether decimal  ` `# representation of given binary number ` `# is divisible by 10 or not ` `def` `isDivisibleBy10(``bin``) : ` `    ``n ``=` `len``(``bin``) ` `     `  `    ``#if last digit is '1', then ` `    ``# number is not divisible by 10 ` `    ``if` `(``bin``[n ``-` `1``] ``=``=` `'1'``) : ` `        ``return` `False` `         `  `    ``# to accumulate the sum of last  ` `    ``# digits in perfect powers of 2 ` `    ``sum` `=` `0` `     `  `    ``#traverse from the 2nd last up to ` `    ``# 1st digit in 'bin' ` `     `  `    ``i ``=` `n ``-` `2` `    ``while` `i >``=` `0` `: ` `         `  `        ``# if digit in '1' ` `        ``if` `(``bin``[i] ``=``=` `'1'``) : ` `            ``# calculate digit's position ` `            ``# from the right ` `            ``posFromRight ``=` `n ``-` `i ``-` `1` `             `  `            ``#according to the digit's  ` `            ``# position, obtain the last  ` `            ``# digit of the applicable  ` `            ``# perfect power of 2 ` `            ``if` `(posFromRight ``%` `4` `=``=` `1``) : ` `                ``sum` `=` `sum` `+` `2` `            ``elif` `(posFromRight ``%` `4` `=``=` `2``) : ` `                ``sum` `=` `sum` `+` `4` `            ``elif` `(posFromRight ``%` `4` `=``=` `3``) : ` `                ``sum` `=` `sum` `+` `8` `            ``elif` `(posFromRight ``%` `4` `=``=` `0``) : ` `                ``sum` `=` `sum` `+` `6` `             `  `        ``i ``=` `i ``-` `1` `         `  `    ``# if last digit is 0, then ` `    ``# divisible by 10 ` `    ``if` `(``sum` `%` `10` `=``=` `0``) : ` `        ``return` `True` `         `  `    ``# not divisible by 10  ` `    ``return` `False` `         `  ` `  `# Driver program to test above function ` ` `  `bin` `=` `"11000111001110"` `if` `(isDivisibleBy10(``bin``)``=``=` `True``) : ` `    ``print``(``"Yes"``) ` `else` `: ` `    ``print``(``"No"``) ` ` `  `# This code is contributed by Nikita Tiwari.  `

## C#

 `// C# implementation to check whether decimal ` `// representation of given binary number is ` `// divisible by 10 or not ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// function to check whether decimal ` `    ``// representation of given binary number ` `    ``// is divisible by 10 or not ` `    ``static` `bool` `isDivisibleBy10(String bin) ` `    ``{ ` `        ``int` `n = bin.Length; ` ` `  `        ``// if last digit is '1', then ` `        ``// number is not divisible by 10 ` `        ``if` `(bin[n - 1] == ``'1'``) ` `            ``return` `false``; ` ` `  `        ``// to accumulate the sum of last ` `        ``// digits in perfect powers of 2 ` `        ``int` `sum = 0; ` ` `  `        ``// traverse from the 2nd last up to ` `        ``// 1st digit in 'bin' ` `        ``for` `(``int` `i = n - 2; i >= 0; i--) { ` `             `  `            ``// if digit in '1' ` `            ``if` `(bin[i] == ``'1'``) { ` `                 `  `                ``// calculate digit's position ` `                ``// from the right ` `                ``int` `posFromRight = n - i - 1; ` ` `  `                ``// according to the digit's ` `                ``// position, obtain the last ` `                ``// digit of the applicable ` `                ``// perfect power of 2 ` `                ``if` `(posFromRight % 4 == 1) ` `                    ``sum = sum + 2; ` `                ``else` `if` `(posFromRight % 4 == 2) ` `                    ``sum = sum + 4; ` `                ``else` `if` `(posFromRight % 4 == 3) ` `                    ``sum = sum + 8; ` `                ``else` `if` `(posFromRight % 4 == 0) ` `                    ``sum = sum + 6; ` `            ``} ` `        ``} ` ` `  `        ``// if last digit is 0, then ` `        ``// divisible by 10 ` `        ``if` `(sum % 10 == 0) ` `            ``return` `true``; ` ` `  `        ``// not divisible by 10 ` `        ``return` `false``; ` `    ``} ` ` `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``String bin = ``"11000111001110"``; ` ` `  `        ``if` `(isDivisibleBy10(bin)) ` `            ``Console.Write(``"Yes"``); ` `        ``else` `            ``Console.Write(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 `= 0; ``\$i``--)  ` `    ``{ ` `        ``// if digit in '1' ` `        ``if` `(``\$bin``[``\$i``] == ``'1'``) ` `        ``{ ` `            ``// calculate digit's  ` `            ``// position from the right ` `            ``\$posFromRight` `= ``\$n` `- ``\$i` `- 1; ` `             `  `            ``// according to the digit's  ` `            ``// position, obtain the last  ` `            ``// digit of the applicable  ` `            ``// perfect power of 2 ` `            ``if` `(``\$posFromRight` `% 4 == 1) ` `                ``\$sum` `= ``\$sum` `+ 2; ` `            ``else` `if` `(``\$posFromRight` `% 4 == 2) ` `                ``\$sum` `= ``\$sum` `+ 4; ` `            ``else` `if` `(``\$posFromRight` `% 4 == 3) ` `                ``\$sum` `= ``\$sum` `+ 8; ` `            ``else` `if` `(``\$posFromRight` `% 4 == 0) ` `                ``\$sum` `= ``\$sum` `+ 6;          ` `        ``} ` `    ``} ` `     `  `    ``// if last digit is 0, then ` `    ``// divisible by 10 ` `    ``if` `(``\$sum` `% 10 == 0) ` `        ``return` `true; ` `     `  `    ``// not divisible by 10  ` `    ``return` `false;  ` `} ` ` `  `// Driver Code ` `\$bin` `= ``"11000111001110"``; ` `if``(isDivisibleBy10(``\$bin``)) ` `    ``echo` `"Yes"``; ` `else` `    ``echo` `"No"``; ` ` `  `// This code is contributed by mits.  ` `?> `

Output:

```Yes
```

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

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Improved By : Mithun Kumar

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