#include <bits/stdc++.h>
using namespace std;
inline long long int in() {
  long long int p = 0;
  register char ch = 0;
  while (ch < '0' or ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' and ch <= '9') {
    p = (p << 1) + (p << 3) + ch - '0';
    ch = getchar();
  }
  return p;
}
bool IsPowerOfTwo(long long int x) { return (x != 0) && ((x & (x - 1)) == 0); }
long long int poer(long long int a, long long int b) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs *= a;
    a *= a;
    b /= 2;
  }
  return rs;
}
long long int mod_poer(long long int a, long long int b, long long int mo) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs = (rs * a) % mo;
    a = (a * a) % mo;
    b /= 2;
  }
  return rs % mo;
}
long long int d, x, y;
void ext_euclid(long long int a, long long int b) {
  if (b == 0) {
    d = a;
    x = 1;
    y = 0;
  } else {
    ext_euclid(b, a % b);
    long long int temp = x;
    x = y;
    y = temp - (a / b) * y;
  }
}
long long int mod_inverse1(long long int a, long long int m) {
  ext_euclid(a, m);
  return (x % m + m) % m;
}
long long int mod_inverse2(long long int a, long long int m) {
  return mod_poer(a, m - 2, m);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int a, b, n;
  cin >> a >> b >> n;
  if (n >= b) {
    cout << a / gcd(a, b) << "/" << b / gcd(a, b) << endl;
  } else {
    long double x = a * 1.0 / b * 1.0;
    long long int ans1, ans2;
    for (long long int i = 1; i <= n; ++i) {
      long long int l = 0, r = 100000;
      while (l <= r) {
        long long int mid = (l + r) / 2;
        if ((long double)(mid * 1.0 / (i * 1.0)) >= x)
          r = mid - 1;
        else
          l = mid + 1;
      }
      if (fabs((long double)((l - 1) * 1.0 / (i * 1.0) - x)) <
          fabs((long double)((ans1 * 1.0) / (ans2 * 1.0) - x)))
        ans1 = l - 1, ans2 = i;
      if (fabs((long double)((l)*1.0 / (i * 1.0) - x)) <
          fabs((long double)((ans1 * 1.0) / (ans2 * 1.0) - x)))
        ans1 = l, ans2 = i;
    }
    cout << ans1 / gcd(ans1, ans2) << "/" << ans2 / gcd(ans1, ans2) << endl;
  }
  return 0;
}
