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

25 Jun, 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
Approach: The following steps are:
Convert the binary number to base 4.Numbers in base 4 contains only 0, 1, 2, 3 as their digits.5 in base 4 is equivalent to 11.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.
Convert the binary number to base 4.
Numbers in base 4 contains only 0, 1, 2, 3 as their digits.
5 in base 4 is equivalent to 11.
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?
Check whether the length of binary string is even or odd.If odd, the add ‘0’ in the beginning of the string.Now, traverse the string from left to right.One by extract substrings of size 2 and add their equivalent decimal to the resultant string.
Check whether the length of binary string is even or odd.
If odd, the add ‘0’ in the beginning of the string.
Now, traverse the string from left to right.
One by extract substrings of size 2 and add their equivalent decimal to the resultant string.
C++
Java
Python 3
C#
Javascript
// 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 numberint 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 notstring 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        // add it to odd_sum        if (isOddDigit)            odd_sum += equivalentBase4(bin.substr(i, 2));        // else digit of base 4 is at even place,        // add it to even_sum        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 aboveint main(){    string bin = "10000101001";    cout << isDivisibleBy5(bin);    return 0;}
//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            // add it to odd_sum            if (isOddDigit != 0)                odd_sum += equivalentBase4(bin.substring(i, i+2));            // else digit of base 4 is at even place,            // add it to even_sum            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));    }}
# 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 numberdef 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 Codeif __name__=="__main__":    bin = "10000101001"    print(isDivisibleBy5(bin)) # This code is contributed# by Sairahul Jella
// C# implementation to check whether// decimal representation of given// binary number is divisible by 5 or notusing System; class GFG{// Method to return equivalent base// 4 number of the given binary numberpublic 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 notpublic 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 Codepublic static void Main(string[] args){    string bin = "10000101001";    Console.WriteLine(isDivisibleBy5(bin));}} // This code is contributed by Shrikant13
<script> // Javascript 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 numberfunction 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   function isDivisibleBy5(bin){    let l = bin.length    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    let odd_sum = 0    let even_sum = 0    let isOddDigit = 1    for(let i=0;i<bin.length;i+=2){                 // if digit of base 4 is at odd place,        // then add it to odd_sum        if(isOddDigit)            odd_sum += equivalentBase4(bin.substring(i,i + 2))                     // else digit of base 4 is at        // even place, add it to even_sum           else            even_sum += equivalentBase4(bin.substring(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(Math.abs(odd_sum - even_sum) % 5 == 0)        return "Yes"    else        return "No"} // Driver Code let    bin = "10000101001"document.writeln(isDivisibleBy5(bin)) // This code is contributed by shinjanpatra </script>
Output:
YES
Time Complexity: O(n), where n is the number of digits in the binary number. Auxiliary Space: O(1)
References: https://stackoverflow.com/questions/18473730/algorithm-in-hardware-to-find-out-if-number-is-divisible-by-five This article is contributed by Aarti_Rathi and 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. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
shrikanth13
Sairahul Jella
nidhi_biet
shinjanpatra
sachinvinod1904
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