1.1 KiB
1.1 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f4761000cf542c50ff88 | Problem 265: Binary Circles | 5 | 301914 | problem-265-binary-circles |
--description--
2N binary digits can be placed in a circle so that all the N-digit clockwise subsequences are distinct.
For N=3, two such circular arrangements are possible, ignoring rotations:
For the first arrangement, the 3-digit subsequences, in clockwise order, are: 000, 001, 010, 101, 011, 111, 110 and 100.
Each circular arrangement can be encoded as a number by concatenating the binary digits starting with the subsequence of all zeros as the most significant bits and proceeding clockwise. The two arrangements for N=3 are thus represented as 23 and 29: 00010111 2 = 23 00011101 2 = 29
Calling S(N) the sum of the unique numeric representations, we can see that S(3) = 23 + 29 = 52.
Find S(5).
--hints--
euler265()
should return 209110240768.
assert.strictEqual(euler265(), 209110240768);
--seed--
--seed-contents--
function euler265() {
return true;
}
euler265();
--solutions--
// solution required