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#include <bits/stdc++.h>
using namespace std;
/*
由于 L 在100以内，因此可以枚举 A′,B′ 的所有组合，然后判断：
(1)、A′,B′ 是否互质；
(2)、A′B′ 是否大于等于 AB，并且最小
时间复杂度：O(L^2 LogL)
*/
int gcd(int a, int b) {
return b ? gcd(b, a % b) : a;
}
int main() {
int A, B, L;
cin >> A >> B >> L;
int a, b;
double mi = 1e9;
for (int i = 1; i <= L; i++)
for (int j = 1; j <= L; j++)
if (gcd(i, j) == 1) {
double x = i * 1.0 / j;
double X = A * 1.0 / B;
if (x >= X && x - X < mi) {
mi = x - X;
a = i, b = j;
cout << a << ' ' << b << endl;
return 0;
