---
id: 5900f3db1000cf542c50feee
title: 'Problem 111: Primes with runs'
challengeType: 5
forumTopicId: 301736
dashedName: problem-111-primes-with-runs
---

# --description--

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:

1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111

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).

# --hints--

`euler111()` should return 612407567715.

```js
assert.strictEqual(euler111(), 612407567715);
```

# --seed--

## --seed-contents--

```js
function euler111() {

  return true;
}

euler111();
```

# --solutions--

```js
// solution required
```