if anyone can please help i have to turn in this program by 2moro night and ive been up all night trying to figure it out. i had to write a program to find all the factors of any given number typed as long as it is a possitive interger. then i have to 1. list the possitive intergers. 2. show how many intergers there are. 3 add all the intergers and 4 multiply them. i will paste what i have so far but i am having trouble with showing 2-4! please just give me advice or anything. thanx

Code:
```#include <stdio.h>

int main () {
int val, factor, sum;
int product, total;
printf("Enter a positive integer greater than one: \n");
scanf("&#37;d", &val);

factor=1;

if(val<0){
printf("Sorry, that input is not valid. \n""Enter a positive integer greater than one.");
}
else{

printf("Here is a list of the possitive intregers of %d\n", val);
while(factor<=val){

if(val%factor==0){

printf("%d ", factor);

}//if end
factor=factor+1;
}//while end

//sum
sum=1;
while(sum<=val);{
if(val%sum==0){
}
total= sum+sum;

}//while end
printf("%d\n",total);
}//else end
printf("\n");

system("pause");
return 0;
}```

2. you're code arrengement looked horrible to me... Take a look at this: http://irc.essex.ac.uk/www.iota-six...._operators.asp

Code:
```#include <stdio.h>

int main (void) {

int val, factor=1,count=0,sum=0,multip=1;

printf("Enter a positive integer greater than one: \n");
scanf("%i", &val);

if(val<0){
printf("Sorry, that input is not valid. \n""Enter a positive integer greater than one.\n");
return 0;
}

printf("\nHere is a list of the possitive integers for %i:\n", val);

while(factor<=val){
if(val%factor==0){
printf("%i\n", factor);
count++;         /*counts how many integers we found*/
sum+=factor;  /*sums them*/
multip*=factor; /*multiplies them*/
}
factor++;
}

printf("We have a total of %i integeres.\nWhich are summed together: %i and multiplied:%i\n",count,sum,multip);

return 0;
}```

3. yeah i know. this is my second program ever so i guess i can use that as an exuse. i got the program to add and multiply but i stayed up unitl 4 trying to figure out how to get it to count. i was going to organize it when i finished though, and play with it so i could have only one while loop. i just wanted it to do what i asked first no matter how sloppy it was. thanx for the help though!

4. Originally Posted by lilcoder
you're code arrengement looked horrible to me... Take a look at this: http://irc.essex.ac.uk/www.iota-six...._operators.asp

Code:
```#include <stdio.h>

int main (void) {

int val, factor=1,count=0,sum=0,multip=1;

printf("Enter a positive integer greater than one: \n");
scanf("%i", &val);

if(val<0){
printf("Sorry, that input is not valid. \n""Enter a positive integer greater than one.\n");
return 0;
}

printf("\nHere is a list of the possitive integers for %i:\n", val);

while(factor<=val){
if(val%factor==0){
printf("%i\n", factor);
count++;         /*counts how many integers we found*/
sum+=factor;  /*sums them*/
multip*=factor; /*multiplies them*/
}
factor++;
}

printf("We have a total of %i integeres.\nWhich are summed together: %i and multiplied:%i\n",count,sum,multip);

return 0;
}```

I suppose that if i have completely understand what you want to do, your strategy is wrong.
Consider a number like 8, first factor is 2, you have 8 / 2 = 4, for the next factor you set the factor to factor++ so you have the number 3 to get divided with the number 4. Thats error as the next number that factor is is still the number 2.
8 = 2 * 2 * 2;
Also should you consider creating a function to tell if is a prime number or not, if the given number is prime then say it is prime and print it.
Think of it.