2.0 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3db1000cf542c50feee | Problem 111: Primes with runs | 5 | 301736 | 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
.
assert.strictEqual(primesWithRuns(), 612407567715);
--seed--
--seed-contents--
function primesWithRuns() {
return true;
}
primesWithRuns();
--solutions--
// solution required