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**1. **A simple method is to take log of the given number on base 4, and if we get an integer then number is power of 4.

**2. **Another solution is to keep dividing the number by 4, i.e, do n = n/4 iteratively. In any iteration, if n%4 becomes non-zero and n is not 1 then n is not a power of 4, otherwise n is a power of 4.

#include<stdio.h> #define bool int /* Function to check if x is power of 4*/ bool isPowerOfFour(int n) { if(n == 0) return 0; while(n != 1) { if(n%4 != 0) return 0; n = n/4; } return 1; } /*Driver program to test above function*/ int main() { int test_no = 64; if(isPowerOfFour(test_no)) printf("%d is a power of 4", test_no); else printf("%d is not a power of 4", test_no); getchar(); }

**3. **A number n is a power of 4 if following conditions are met.

a) There is only one bit set in the binary representation of n (or n is a power of 2)

b) The count of zero bits before the (only) set bit is even.

For example: 16 (10000) is power of 4 because there is only one bit set and count of 0s before the set bit is 4 which is even.

Thanks to Geek4u for suggesting the approach and providing the code.

#include<stdio.h> #define bool int bool isPowerOfFour(unsigned int n) { int count = 0; /*Check if there is only one bit set in n*/ if ( n && !(n&(n-1)) ) { /* count 0 bits before set bit */ while(n > 1) { n >>= 1; count += 1; } /*If count is even then return true else false*/ return (count%2 == 0)? 1 :0; } /* If there are more than 1 bit set then n is not a power of 4*/ return 0; } /*Driver program to test above function*/ int main() { int test_no = 64; if(isPowerOfFour(test_no)) printf("%d is a power of 4", test_no); else printf("%d is not a power of 4", test_no); getchar(); }

Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.