1.3 KiB
1.3 KiB
id | challengeType | title | forumTopicId |
---|---|---|---|
5900f4761000cf542c50ff88 | 5 | Problem 265: Binary Circles | 301914 |
Description
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).
Instructions
Tests
tests:
- text: <code>euler265()</code> should return 209110240768.
testString: assert.strictEqual(euler265(), 209110240768);
Challenge Seed
function euler265() {
return true;
}
euler265();
Solution
// solution required