Problem : Write a function to recursively print out an integer in any base from base 2 to base 9.

Problem : Write a recursive function int count_digit(int n, int digit); to count the number of digits in a number n (n > 0) that are equal to a specified digit. For example, if the digit we're searching for were 2 and the number we're searching were 220, the answer would be 2.

Problem : For some reason, the computer you're working on doesn't allow you to use the modulo operator % to compute the remainder of a division. Your friend proposes the following function to do it:

int remainder(int num, int den)
{
	if (num < den) return num;
	else return(remainder(num - den, den));
}

Problem : The following function iteratively computes $x^n$:

int exponentiate_i(int x, int n)
{
	int i, result = 1;
	for(i=0; i<n; i++) result *= x;
	return result;
}

Problem : Use the knowledge that $x^n == (x^2)^(n/2)$ when $n$ is even to write a more efficient solution to the above problem.

Problem : The classic fibonacci problem, where the next term in the sequence is the sum of the previous two terms, is often called fib2. One could also imagine a sequence fibN, where $N$ is the number of previous terms to sum up. Write this function recursively.

Problem : What operation does the following function implement when p is 0, 1, and 2?

int mystery(n, m, p)
{
	int i, result = 0;
	
	if (p==0) return n+m;
	for (i=0; i< m; i++) result += mystery(result,n,p-1);
	return result;
}