#include <bits/stdc++.h>
using namespace std;

#define mk make_pair

int n, a, b;

int exgcd(int a, int b, int& x, int& y) {
	if (b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	int x2, y2;
	int g = exgcd(b, a % b, x2, y2);
	x = y2;
	y = x2 - (a / b) * y2;
	return g;
}

map<pair<int, pair<int, int> >, int> mem;

int dfs(int len, int a, int b) {
	if (len == 0) {
		return 0;
	}
	
	if (mem.count(mk(len, mk(a, b))))
		return mem[mk(len, mk(a, b))];
	
	// AB | ... (remaining len positions)
	int x, y;
	int g = exgcd(a, b, x, y);
	
	if (g != 1) { // never collide again
		return len / b + (len + 1) / a;
	}
	
	// we already get a * x0 + b * y0 = 1
	// the general solution is x = x0 + b * k, y = y0 - a * k, where k is any integer
	x = (x % b + b) % b;
	
	// cout << "solve " << len << " " << a << " " << b << endl;
	// cout << x << " * a" << endl;
	
	assert(x > 0);
	
	if (a * x - 1 > len) { // next collision is beyond the length
		return len / b + (len + 1) / a;
	}
	
	return mem[mk(len, mk(a, b))] = (a * x - 1) / b + x + min(dfs(len - (a * x - 1), a, b), dfs(len - (a * x - 1), b, a));
}

int main() {
	cin >> a >> b >> n;
	
	long long lcm = (long long)a * b / __gcd(a, b);
	if (lcm > n) { // the first collision is beyond the length
		cout << n / a + n / b << endl;
		return 0;
	}
	
	cout << lcm / a + lcm / b + min(dfs(n - lcm, a, b), dfs(n - lcm, b, a)) << endl;
	
	return 0;
}