How to make a prog more efieiant (in C)

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 .. :-)

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#include
#include

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;
}

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