【链接】 我是链接,点我呀:)
【题意】

让你找到长度为n的数字
这个数字只由a或者b组成
且这n个数码的和也是由a或者b组成的
求出满足这样要求的数字的个数

【题解】

枚举答案数字中b的个数为y,那么a出现的个数就为n-y
那么和就是na + (b-a)y;
这个数字最多就7位的样子
很容易检查是不是只包含a或者b
然后如果满足只包含a或者b
则答案加上C(n,y)
即n个位置中选择y个放b,其他的放a
组合数取余的话,预处理一下阶乘以及阶乘的逆元就好

【代码】

import java.io.*;
import java.util.*;

public class Main {
    
    
    static InputReader in;
    static PrintWriter out;
        
    public static void main(String[] args) throws IOException{
        //InputStream ins = new FileInputStream("E:\\rush.txt");
        InputStream ins = System.in;
        in = new InputReader(ins);
        out = new PrintWriter(System.out);
        //code start from here
        new Task().solve(in, out);
        out.close();
    }
    
    static int N = (int)1e6;
    static class Task{
        
        long MOD = (int)1e9+7;
        int a,b,n;
        long fac[],rfac[];
        
        long _pow(long x,long y) {
            long ans = 1;
            while (y>0) {
                if (y%2==1) ans = (ans * x)%MOD;
                x = (x*x)%MOD;
                y/=2;
            }
            return ans;
        }
        
        long C(int n,int m) {
            //n!/((n-m)!*m!)
            if (n<m) return 0;
            long temp1 = fac[n];
            temp1 = temp1*rfac[n-m]%MOD;
            temp1 = temp1*rfac[m]%MOD;
            return temp1;
        }
        
        boolean ok(int x) {
            if (x==0) return false;
            while (x>0) {
                int temp = x%10;
                if (temp!=a && temp!=b) return false;
                x = x/10;
            }
            return true;
        }
        
        public void solve(InputReader in,PrintWriter out) {
            fac = new long[N+10];
            rfac = new long[N+10];
            fac[0] = 1;
            for (int i = 1;i <= N;i++) fac[i] = fac[i-1]*i%MOD;
            rfac[N] = _pow(fac[N],MOD-2 );
            for (int i = N-1;i >= 0;i--) {
                rfac[i] = rfac[i+1]*(i+1)%MOD;
            }
            a = in.nextInt();b = in.nextInt();n = in.nextInt();
            long ans = 0;
            for (int y = 0;y <= n;y++) {
                //b有y个
                int sumdigits = n*a + (b-a)*y;
                if (ok(sumdigits)) {
                    ans = ans + C(n,y);
                    ans = ans % MOD;
                }
            }
            out.println(ans);
        }
    }

    

    static class InputReader{
        public BufferedReader br;
        public StringTokenizer tokenizer;
        
        public InputReader(InputStream ins) {
            br = new BufferedReader(new InputStreamReader(ins));
            tokenizer = null;
        }
        
        public String next(){
            while (tokenizer==null || !tokenizer.hasMoreTokens()) {
                try {
                tokenizer = new StringTokenizer(br.readLine());
                }catch(IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}