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

• Difficulty Level : Medium
• Last Updated : 08 Sep, 2022

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 notbool 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 aboveint 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 notimport 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 notdef 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 notusing 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 functiondecnum = int(str1, 2)# checking if number is divisible by 10# or not if divisible print yes else noif 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");        }           }}



## C#

 // C# code to check// decimal representation of// a given binary string is// divisible by 10 or notusing 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)
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