Finding integer divisors of a set of numbers in C language - c

Question

I want to do this in a function: How do I find out in a C program if a number is divisible by 2, 3, 4, 5, 6, 8, 9, 25 and 125 without using the % operator and using the divisibility rules? the base should be 10*

Answer 1

To use divisibility rules, you have to work with digits. Perhaps task assumes no division (at least in explicit form - you can extract digits from string representation)
For divisibility by 2, check whether the last digit is in set 0,2,4,6,8
For divisibility by 4, check whether the last digit + doubled previous one is in set 0,4,8. If result is larger than 10, repeat (88=>2*8+8=24=>2*2+4=8)
Similar situation for 8, but sum last + 2*previous + 4*second_from_the_end (512 => 4*5+2*1+2=24=>2*2+4=8)
For divisibility by 5, check whether the last digit is in set 0,5, similar situation for 25, 125
For divisibility by 3, sum all digits, repeat process until result becomes < 10. So-called "digit root" should be in set 0,3,6,9, similar situation for divisibility by 9.
For 6 check divisibilty by both 2 and by 3
I am not strong in C, so my example perhaps is very weird (ideone check)
#include <stdio.h>
int divby3(int n) {
char s[10];
do {
sprintf(s, "%d", n); //transform 72 to char buffer "72"
n = 0;
int i = 0;
while(s[i]) //until nil (end of string) found, we can also use for loop
n += s[i++] - 0x30; //get difference of current char and char "0"
}
while (n >= 10); //until 1-digit value
return (n==0) || (n==3) || (n==6) || (n==9);
}
int divby5(int n) {
char s[10];
int len = sprintf(s, "%d", n);
n = s[len - 1] - 0x30; //last digit
return (n==0) || (n==5);
}
int main(void) {
printf("%d", divby3(72)); //try 71
return 0;
}

Answer 2

A function that uses the a - (a / b * b) implementation of the modulus operator: (credit #MikeCat)
bool isDivisible(int a, int b)
{
if((a - (a / b * b)) == 0) return true;
return false;
}
Usage:
int main(void)
{
int a = 8;
int b = 4;
int c = 3;
bool res = isDivisible(a,b);
bool res2 = isDivisible(a,c);
return 0;
}
EDIT - to address question in comments:
"how can i represent such a program with the divisibility rules? Thank you for your code, i forgott to mention that i have to use the divisibility rules in each function"
The following shows how to pass in divisibility rules as an argument...
const int div_1[] = {2, 3, 4, 5, 6, 8, 9, 25, 125};
const int div_2[] = {7, 5, 17, 12, 11};
int main()
{
size_t size = 0;
size = sizeof div_1/sizeof div_1[0];
bool res = isDivisible(2*3*4*5*6*8*9*25*125, div_1, size);
size = sizeof div_2/sizeof div_2[0];
bool res2 = isDivisible(125, div_2, size);
return 0;
}
// numerator divisor array array size
bool isDivisible(long a, long div_rules[], size_t size)
{
//divisibility rules
const int divisors[] = {2, 3, 4, 5, 6, 8, 9, 25, 125};
for(int i = 0; i<size;i++)
{
if((a - (a / div_rules[i] * div_rules[i])) != 0) return false;
}
return true;
}