Problem 111
Primes with runs
Considering 4-digit primes containing repeated digits it is clear that they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by 22, and so on. But there are nine 4-digit primes containing three ones:
We shall say that M(n, d) represents the maximum number of repeated digits for an n-digit prime where d is the repeated digit, N(n, d) represents the number of such primes, and S(n, d) represents the sum of these primes.
So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit prime where one is the repeated digit, there are N(4, 1) = 9 such primes, and the sum of these primes is S(4, 1) = 22275. It turns out that for d = 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are N(4, 0) = 13 such cases.
In the same way we obtain the following results for 4-digit primes.
| Digit, d | M(4, d) | N(4, d) | S(4, d) |
|---|---|---|---|
| 0 | 2 | 13 | 67061 |
| 1 | 3 | 9 | 22275 |
| 2 | 3 | 1 | 2221 |
| 3 | 3 | 12 | 46214 |
| 4 | 3 | 2 | 8888 |
| 5 | 3 | 1 | 5557 |
| 6 | 3 | 1 | 6661 |
| 7 | 3 | 9 | 57863 |
| 8 | 3 | 1 | 8887 |
| 9 | 3 | 7 | 48073 |
For d = 0 to 9, the sum of all S(4, d) is 273700.
Find the sum of all S(10, d).
有重复数字的素数
考虑一个有重复数字的4位素数，显然这4个数字不能全都一样：1111被11整除，2222被22整除，依此类推；但是，有9个4位素数包含有三个一：
我们记M(n, d)是n位素数中数字d重复出现的最多次数，N(n, d)是这类素数的个数，而S(n, d)是这类素数的和。
因此M(4, 1) = 3是4位素数中数字1重复出现的最多次数，有N(4, 1) = 9个这类素数，而它们的和是S(4, 1) = 22275。还能得出，对于d = 0，在4位素数中最多重复出现M(4, 0) = 2次，但是有N(4, 0) = 13个这类素数。
同样地，我们可以得到4位素数的如下结果。
| 数字d | M(4, d) | N(4, d) | S(4, d) |
|---|---|---|---|
| 0 | 2 | 13 | 67061 |
| 1 | 3 | 9 | 22275 |
| 2 | 3 | 1 | 2221 |
| 3 | 3 | 12 | 46214 |
| 4 | 3 | 2 | 8888 |
| 5 | 3 | 1 | 5557 |
| 6 | 3 | 1 | 6661 |
| 7 | 3 | 9 | 57863 |
| 8 | 3 | 1 | 8887 |
| 9 | 3 | 7 | 48073 |
对于d = 0至9，所有S(4, d)的和为273700。
求所有S(10, d)的和。