#include <stdio.h>

typedef unsigned long long int ul;


ul gcd(ul a, ul b)
{
    if (!b) return a;
    return gcd(b, a % b);
}


char recurring_decimal_digit(ul a, ul b, ul n)
{
    ul d = gcd(a, b);
    a /= d;
    b /= d;

    /* 将
     * \frac{a}{b}
     * 转化为
     * \frac{a}{2^{b2} \cdot 5^{b5} \cdot b'}
     * 的形式
     */
    ul b2 = 0;
    ul b5 = 0;
    for (; !(b % 2); b /= 2) b2 += 1;
    for (; !(b % 5); b /= 5) b5 += 1;

    /* 计算
     * \frac{1}{10^\mbox{offset}} \cdot \frac{a \cdot m}{b}
     * 其中
     * \mbox{offset} = \max(c2, c5)
     * m = \frac{10^\mbox{offset}}{2^{b2} \cdot 5^{b5}}
     * 
     * offset 表示原小数中前 offset 位不在循环体内
     */
    ul m = 1;
    ul base = 2;
    ul offset = b5;
    ul exp = b5 - b2;
    if (b2 > b5) {
        base = 5;
        offset = b2;
        exp = b2 - b5;
    }
    while (exp--) m *= base;

    /*
     * 计算
     * \frac{1}{10^\mbox{offset}} \cdot \left( \mbox{pre} + \frac{r_m \cdot r_a \mod b}{b} \right)
     * 其中
     * \frac{r_m \cdot r_a \mod b}{b}
     * 为纯循环小数，容易计算循环节
     */
    ul qm = m / b;
    ul rm = m % b;
    ul qa = a / b;
    ul ra = a % b;
    ul pre = (qa * b + ra) * qm + qa * rm;

    /* 若 n 在非循环节中，则直接返回 pre 中对应的数位 */
    if (n <= offset) {
        n = offset - n;
        while (n--) pre /= 10;
        return pre % 10;
    }

    n -= offset;
    a = rm * ra % b;
    if (!a) return 0;   // 有限小数补0

    /* 计算循环节unit，进而将 n 轮运算优化为 n % unit 轮 */
    ul unit = 1;
    ul mod = 10;
    for (; 1 != mod; mod = (10 * mod) % b) unit += 1;

    ul r = 0;
    n %= unit;
    if (!n) n = unit;
    while (n--) {
        a *= 10;
        r = a / b;
        a -= r * b;
    }

    return r % 10;
}


int main(int argc, char **argv)
{
    if (3 > argc) return -1;
    printf("%d\n", recurring_decimal_digit(atoi(argv[1]), atoi(argv[2]), atoi(argv[3])));
    return 0;
}