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# 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 0â€™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```

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` `            ``else` `if` `(posFromRight ``%` `4` `=``=` `2``) :` `                ``sum` `=` `sum` `+` `4` `            ``else` `if` `(posFromRight ``%` `4` `=``=` `3``) :` `                ``sum` `=` `sum` `+` `8` `            ``else` `if` `(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. ` `?>`

## Javascript

 `    `

Output:

`Yes`

Time Complexity: O(N)
Auxiliary Space: O(1)

Method: Convert the given binary string in to decimal using int function then check if it is divisible by 10 or not using modulo division.

## C++

 `#include ` `using` `namespace` `std;`   `int` `main()` `{`   `    ``char` `s[] = ``"1010"``;` `  `  `    ``// converting binary string in to` `    ``// decimal number using stoi function` `    ``int` `n = stoi(s, 0, 2);`   `    ``if` `(n % 10 == 0) {` `        ``cout << ``"Yes"``;` `    ``}` `    ``else` `{` `        ``cout << ``"No"``;` `    ``}` `    ``return` `0;` `}`   ` ``// this code is contributed by Gangarajula Laxmi`

## Python3

 `# Python code to check` `# decimal representation of` `# a given binary string is` `# divisible by 10 or not`   `str1 ``=` `"101000"` `# converting binary string in to` `# decimal number using int function` `decnum ``=` `int``(str1, ``2``)` `# checking if number is divisible by 10` `# or not if divisible print yes else no` `if` `decnum ``%` `10` `=``=` `0``:` `    ``print``(``"Yes"``)` `else``:` `    ``print``(``"No"``)`   `    ``# this code is contributed by gangarajula laxmi`

## Java

 `// java code to check` `// decimal representation of` `// a given binary string is` `// divisible by 10 or not`   `import` `java.io.*;`   `class` `GFG {` `    ``public` `static` `void` `main (String[] args) {` `      ``String s=``"1010"``; ` `      ``//converting binary string in to` `//decimal number using Convert.ToInt function` `    ``int` `n=Integer.parseInt(s,``2``);` `    `  `     ``if` `(n%``10``==``0``)` `        ``{` `           ``System.out.println(``"Yes"``);` `        ``}` `        ``else` `        ``{` `            ``System.out.println(``"No"``);` `        ``}` `      `  `    ``}` `}`

## Javascript

 ``

## C#

 `// C# code to check` `// decimal representation of` `// a given binary string is` `// divisible by 10 or not` `using` `System;`   `public` `class` `GFG{`   `    ``static` `public` `void` `Main (){` `        ``string` `s=``"1010"``; ` `      ``//converting binary string in to` `//decimal number using Convert.ToInt function` `    ``int` `n=Convert.ToInt32(s,2);` `    `  `     ``if` `(n%10==0)` `        ``{` `            ``Console.Write(``"Yes"``);` `        ``}` `        ``else` `        ``{` `            ``Console.Write(``"No"``);` `        ``}` `    ``}` `}`

Output

`Yes`

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

Auxiliary Space: O(1)

Method 3: Using the concept of modular arithmetic
We can also solve this problem without converting the binary string to decimal by using the concept of modular arithmetic. The idea is to iterate through the binary string from right to left, and keep track of the remainder after dividing the current number by 10. If the final remainder is 0, then the binary string is divisible by 10.

## Python3

 `def` `is_divisible_by_10(binary_str):` `    ``remainder ``=` `0` `    ``for` `digit ``in` `binary_str:` `        ``remainder ``=` `(remainder ``*` `2` `+` `int``(digit)) ``%` `10` `    ``return` `remainder ``=``=` `0`   `# Example usage:` `binary_str ``=` `'101010'` `print``(is_divisible_by_10(binary_str))`

## Javascript

 `// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL` `function` `is_divisible_by_10(binary_str) {` `  ``let remainder = 0;` `  ``for` `(let i = 0; i < binary_str.length; i++) {` `    ``remainder = (remainder * 2 + parseInt(binary_str[i])) % 10;` `  ``}` `  ``return` `remainder == 0;` `}`   `// Example usage:` `let binary_str = ``'101010'``;` `console.log(is_divisible_by_10(binary_str));`

## Java

 `// Define a public class named GFG` `public` `class` `GFG {`   `    ``public` `static` `boolean` `    ``is_divisible_by_10(String binary_str)` `    ``{` `        ``int` `remainder = ``0``;` `        ``// Iterate over the binary digits of the input` `        ``// string` `        ``for` `(``int` `i = ``0``; i < binary_str.length(); i++) {` `            ``// Convert the current binary digit to an` `            ``// integer` `            ``char` `digit = binary_str.charAt(i);` `            ``int` `num = Character.getNumericValue(digit);` `            ``// Update the remainder by shifting the bits to` `            ``// the left by 1 position and adding the current` `            ``// binary digit` `            ``remainder = (remainder * ``2` `+ num) % ``10``;` `        ``}` `        ``// Return true if the remainder is 0 (i.e., the` `        ``// binary number is divisible by 10), otherwise` `        ``// return false` `        ``return` `remainder == ``0``;` `    ``}`   `    ``// Define the main method` `    ``// The main method calls the is_divisible_by_10 method` `    ``// with an example binary string and prints the result` `    ``// to the console` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String binary_str = ``"101010"``;` `        ``System.out.println(is_divisible_by_10(binary_str));` `    ``}` `}`

## C++

 `#include ` `#include ` `using` `namespace` `std;`   `bool` `is_divisible_by_10(string binary_str) {` `    ``int` `remainder = 0;` `    ``for` `(``int` `i = 0; i < binary_str.length(); i++) {` `        ``char` `digit = binary_str[i];` `        ``int` `num = digit - ``'0'``;` `        ``remainder = (remainder * 2 + num) % 10;` `    ``}` `    ``return` `remainder == 0;` `}`   `int` `main() {` `    ``string binary_str = ``"101010"``;` `    ``cout << is_divisible_by_10(binary_str) << endl;` `    ``return` `0;` `}`

## C#

 `using` `System;`   `class` `Program` `{` `    ``static` `bool` `IsDivisibleBy10(``string` `binaryStr)` `    ``{` `        ``int` `remainder = 0;` `        ``foreach` `(``char` `digit ``in` `binaryStr)` `        ``{` `            ``remainder = (remainder * 2 + ``int``.Parse(digit.ToString())) % 10;` `        ``}` `        ``return` `remainder == 0;` `    ``}`   `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``string` `binaryStr = ``"101010"``;` `        ``Console.WriteLine(IsDivisibleBy10(binaryStr));` `    ``}` `}`

Output

`False`

Time complexity: O(n)
Auxiliary Space: O(1)

This article is contributed by Ayush Jauhari. 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.