''今有物不知其数，三三数之剩二，五五数之剩三，七七数之剩二。问物几何？''
This puzzle can be translated into the solution of the system of congruences
$$x \equiv 2 \pmod 3\\x \equiv 3 \pmod 5\\x \equiv 2 \pmod 7$$
/*
Description: Using Brute-force Strategy to Search a Solution
今有物不知其数,三三数之剩二,五五数之剩 三,七七数之剩二,问物几何?
x = 2 (mod 3)
x = 3 (mod 5)
x = 2 (mod 7)
Author: Liutong Xu
*/
#include <stdio.h>
#define LIMIT 100
int main(void)
{
int x;
x = 1;
while (!(x % 3 == 2 && x % 5 == 3 && x % 7 == 2)) //while not a solution
x++;
printf("x = %d\n",x);
//if you wang to find all solutions
//for (x = 1; x < LIMIT; x++)
// if (x % 3 == 2 && x % 5 == 3 && x % 7 == 2)
// printf("x = %d\n",x);
return 0;
} 