How to make a prog more efieiant (in C)

[Jabka Atu <mashrom.head@gmail.com>]

me and friend wrote this small prog but it is still not good enugh (it 
is really long)

could you plz take a look and give me some editions ?
The prog need to take a number then test  :
find the largest digit.
is the number simatric
show the reverse number

it is released under FAL licence (F*@! all licences)
all right belong  to Debian .. :-)

------------------------------------------------------------------------

#include <stdio.h>
#include <string.h>

char get_max_digit(int num);
int get_summ_digits(int num);
char is_symmetric(int num);
int reverse_num(int num);

void menu()
{
	printf("\n1. Summ of digits.\n");
	printf("2. Max digit.\n");
	printf("3. Reverse number.\n");
	printf("4. Is symmetric.\n");
	printf("5. New number.\n");
	printf("6. Exit.\n");
}

int main(int argc, char **argv)
{
	char w = 1;
	int num = 0;

	while (w)
	{
		menu();
		while(isspace(w = getchar()));

		switch(w)
		{
		case '1':
			printf("Summ of digits is: %i\n", get_summ_digits(num));
			break;
		case '2':
			printf("Max. digit is: %i\n", get_max_digit(num));
			break;
		case '3':
			printf("Reversed number is: %d\n", reverse_num(num));
			break;
		case '4':
			printf("Is symmetric: %d\n", is_symmetric(num));
			break;
		case '5':
			printf("Enter new number: ");
			scanf("%i", &num);
			break;
		case '6':
			w = 0;
			break;
		}
	}
	return 0;
}

char get_max_digit(int num)
{
	char res = num % 10;
	char dig;

	while (num /= 10)
	{
		dig = num % 10;
		if (dig > res)
			res = dig;
	}

	return res;
}

int get_summ_digits(int num)
{
	int res = num % 10;

	while (num /= 10)
		res += num % 10;

	return res;
}

int reverse_num(int num)
{
	int res = 0;
	int multiplier = 10;

	while (num / multiplier)
		multiplier *= 10;
	multiplier /= 10;

	while(num / 10)
	{
		res += num % 10 * multiplier;
		multiplier /= 10;
		num /= 10;
	}
	res += num;

	return res;
}

char is_symmetric(int num)
{
	char dig1, dig2, res = 0;
	int multiplier1, multiplier2;

	multiplier1 = 1;
	multiplier2 = 10;

	while (num / multiplier2)
		multiplier2 *= 10;
	multiplier2 /= 10;
	
	while(
             (((num / multiplier1) % 10) == ((num / multiplier2) % 10)) &&
	      (multiplier1 <= multiplier2)
	     )
	{
		multiplier1 *= 10;
		multiplier2 /= 10;
	}

	return (multiplier1 <= multiplier2) ? 0 : 1;
}