problem cf 1800
由于每一位的和和顺序无关，所以枚举即可，
时间复杂度$O(nlogn)$

// Decline is inevitable,
// Romance will last forever.
#include <bits/stdc++.h>
using namespace std;
//#define mp make_pair
#define pii pair<int,int>
#define pb push_back
#define ll long long
#define LL long long
#define ld long double
#define endl '\n'
#define RE0 return 0
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)

#define int long long
const int maxn = 1e6 + 10;
const int P = 1e9 + 7;    //998244353
const int INF = 0x7f7f7f7f;
int fac[maxn];
int a, b, n;
bool judge(int c) {
    while(c) {
        int z = c % 10;
        if(z != a && z != b) return false;
        c /=10;
    }
    return true;
}
ll power(ll a, ll b) {
    ll ans = 1 % P;
    for (; b; b >>= 1) {
        if (b & 1)  ans = ans * a % P;
        a = a * a % P;
    }
    return ans;
}
inline ll inv(ll x) {
    return power(x, (ll)(P-2));
}
void solve() {
    fac[0] = 1;
    cin >> a >> b >> n;
    for(int i =1 ; i <=n; i++) {
        fac[i] = fac[i-1] * i % P;
    }
    int cnt = 0;
    for(int i = 0; i <= n; i++) {
        int c = a * i + b * (n-i);
        if(judge(c)) {
            cnt = (cnt + fac[n]*inv(fac[i])%P*inv(fac[n-i])%P)%P;
        }
    }
    cout << cnt << endl;
}
signed main() {
     ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//    int T; scanf("%d", &T); while(T--)
//    freopen("1channel00.txt","r",stdin);
//    freopen("2.txt","w",stdout);
//    int T; cin >> T; while(T--)
        solve();
    RE0;
}