MOD calculation / reversing

[wilson bibe]

Maybe this function can help you:

#include <stdio.h>
#include <conio.h>

int main(void)
{
int a,b,res

printf ("\nenter a number:");
scanf ("%d", &a);

a = a * a;
a= int a;

printf ("\nenter a number:");
scanf ("%d", &b);

b = b;

res = a / b;

a = a * res;

a = int a;

res = a / b;

printf("\nResult of division of %d / %d = %d. \n",a,b,res));

getch();
return(0);
}

[bilbo]

Hi raduga_fb,
I think you must provide us more details...

If a can be as big as 256 hex digits, as you said in your first post, using a big integers library and an exhaustive loop, the values which solve your equation

c = MOD (a^3, b)

are many and many, as te$ter pointed out: I calculated 603000 possible solutions for a in the range 0 - 0x10f1e8fab, using for c and b the values you gave!

If numbers are to be manipulated with regular C functions (as wilson bibe suggested), where the max value for a can just be 0x7FFFFFFF, then

c = MOD (a * MOD (a*a, b) , b)

is different from

c = MOD (a^3, b)

for a > 1290 (cube root of 0x7FFFFFFF)!

In any case, the MOD operator is not reversable, in the sense that many inputs can give the same output:

a = b % c

is equivalent to the expression

a + nc = b

with n which can assume all integer values...

Best regards
bilbo