ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
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1.1 KiB
id, title, challengeType, forumTopicId, dashedName
| 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