Multiples of 4 (An Interesting Method)

Given a number n, the task is to check whether this number is a multiple of 4 or not without using +, -, * ,/ and % operators.

Examples :

Input: n = 4  Output - Yes
n = 20 Output - Yes
n = 19 Output - No

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Using XOR)
An interesting fact for n > 1 is, we do XOR of all numbers from 1 to n and if the result is equal to n, then n is a multiple of 4 else not.

C++

 // An interesting XOR based method to check if // a number is multiple of 4. #include using namespace std;    // Returns true if n is a multiple of 4. bool isMultipleOf4(int n) {     if (n == 1)        return false;        // Find XOR of all numbers from 1 to n     int XOR = 0;     for (int i = 1; i <= n; i++)         XOR = XOR ^ i;        // If XOR is equal n, then return true     return (XOR == n); }    // Driver code to print multiples of 4 int main() {     // Printing multiples of 4 using above method     for (int n=0; n<=42; n++)        if (isMultipleOf4(n))          cout << n << " ";     return 0; }

Java

 // An interesting XOR based method to check if // a number is multiple of 4.    class Test {     // Returns true if n is a multiple of 4.     static boolean isMultipleOf4(int n)     {         if (n == 1)            return false;                 // Find XOR of all numbers from 1 to n         int XOR = 0;         for (int i = 1; i <= n; i++)             XOR = XOR ^ i;                 // If XOR is equal n, then return true         return (XOR == n);     }            // Driver method     public static void main(String[] args)      {         // Printing multiples of 4 using above method         for (int n=0; n<=42; n++)            System.out.print(isMultipleOf4(n) ? n : " ");     } }

Python 3

 # An interesting XOR based # method to check if a  # number is multiple of 4.    # Returns true if n is a # multiple of 4. def isMultipleOf4(n):        if (n == 1):         return False        # Find XOR of all numbers     # from 1 to n     XOR = 0     for i in range(1, n + 1):         XOR = XOR ^ i        # If XOR is equal n, then     # return true     return (XOR == n)    # Driver code to print  # multiples of 4 Printing # multiples of 4 using # above method for n in range(0, 43):     if (isMultipleOf4(n)):         print(n, end = " ")    # This code is contributed # by Smitha

C#

 // An interesting XOR based method // to check if a number is multiple // of 4. using System; class GFG {            // Returns true if n is a      // multiple of 4.     static bool isMultipleOf4(int n)     {         if (n == 1)         return false;                // Find XOR of all numbers          // from 1 to n         int XOR = 0;         for (int i = 1; i <= n; i++)             XOR = XOR ^ i;                // If XOR is equal n, then          // return true         return (XOR == n);     }            // Driver method     public static void Main()      {                    // Printing multiples of 4          // using above method         for (int n = 0; n <= 42; n++)         {             if (isMultipleOf4(n))                 Console.Write(n+" ");         }     } }    // This code is contributed by Smitha.

PHP



Output :

0 4 8 12 16 20 24 28 32 36 40

How does this work?
When we do XOR of numbers, we get 0 as XOR value just before a multiple of 4. This keeps repeating before every multiple of 4.

Number Binary-Repr  XOR-from-1-to-n
1         1           
2        10           
3        11           

Method 2 (Using Bitwise Shift Operators)
The idea is to remove last two bits using >>, then multiply with 4 using <<. If final result is same as n, then last two bits were 0, hence number was a multiple of four.

C++

 // An interesting XOR based method to check if // a number is multiple of 4. #include using namespace std;    // Returns true if n is a multiple of 4. bool isMultipleOf4(long long n) {     if (n==0)         return true;        return (((n>>2)<<2) == n); }    // Driver code to print multiples of 4 int main() {     // Printing multiples of 4 using above method     for (int n=0; n<=42; n++)         if (isMultipleOf4(n))             cout << n << " ";     return 0; }

Java

 // An interesting XOR based method to check if // a number is multiple of 4.    class Test {     // Returns true if n is a multiple of 4.     static boolean isMultipleOf4(long n)     {         if (n==0)             return true;                 return (((n>>2)<<2) == n);     }            // Driver method     public static void main(String[] args)      {         // Printing multiples of 4 using above method         for (int n=0; n<=42; n++)            System.out.print(isMultipleOf4(n) ? n : " ");     } }

C#

 // An interesting XOR based method to // check if a number is multiple of 4. using System;    class GFG {            // Returns true if n is a multiple     // of 4.     static bool isMultipleOf4(int n)     {         if (n == 0)             return true;                return (((n >> 2) << 2) == n);     }            // Driver code to print multiples     // of 4     static void Main()      {                    // Printing multiples of 4 using         // above method         for (int n = 0; n <= 42; n++)             if (isMultipleOf4(n))                 Console.Write(n + " ");     } }    // This code is contributed by Anuj_67

PHP

 > 2) << 2) == \$n); }    // Driver Code    // Printing multiples of 4 // using above method for (\$n = 0; \$n <= 42; \$n++)     if (isMultipleOf4(\$n))         echo \$n , " ";            // This code is contributed by anuj_67. ?>

Output :

0 4 8 12 16 20 24 28 32 36 40

As we can see that the main idea to find multiplicity of 4 is to check the least two significant bits of the given number. We know that for any even number, the least significant bit is always ZERO (i.e. 0). Similarly, for any number which is multiple of 4 will have least two significant bits as ZERO. And with the same logic, for any number to be multiple of 8, least three significant bits will be ZERO. That’s why we can use AND operator (&) as well with other operand as 0x3 to find multiplicity of 4.

This article is contributed by Sahil Chhabra(KILLER). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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