---
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--

`primesWithRuns()` should return `612407567715`.

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

# --seed--

## --seed-contents--

```js
function primesWithRuns() {

  return true;
}

primesWithRuns();
```

# --solutions--

```js
// solution required
```