FizzBuzz with recursion and no multiplication or division

fizzbuzz-recursive-bit.c

#include <stdio.h>
#include <stdbool.h>

int addDigitsHex(int num);
void recursiveFizzBuzz(int start, int end);

int main (int argc, char const *argv[]) {
	recursiveFizzBuzz(1, 100);
	return 0;
}

int addDigitsHex(int num) {
	// Add up each hex digit, resulting in a single hex digit sum
	// Ex: 111 => 3
	// Ex: FC3 => 1E => F
	int workingNum = num;
	int accum = 0;
	while(workingNum > 0){
		int temp = workingNum & 0xF;
		accum += temp;
		workingNum >>= 4;
	}
	if(accum > 0xF){
		return addDigitsHex(accum);
	}else{
		return accum;
	}
}

void recursiveFizzBuzz(int start, int limit) {
	if(start <= limit){
		switch(addDigitsHex(start)){
			//if divisible by 5, If you add up all of the hex digits, it will equal 5, A, or F
			case 0x5:
			case 0xA:
				printf("Buzz\n");
				break;
			//if divisible by 3, If you add up all of the hex digits, it will equal 3, 6, 9, C or F
			case 0x3:
			case 0x6:
			case 0x9:
				printf("Fizz\n");
				break;
			// F is the only intersection between the possibilities of divisible by 3 and 5
			case 0xF:
				printf("FizzBuzz\n");
				break;
			default:
				printf("%d\n", start);
				break;
		}
		recursiveFizzBuzz(++start, limit);
	}
}