49 lines
		
	
	
		
			1.1 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			49 lines
		
	
	
		
			1.1 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
 | |
| id: 5900f4761000cf542c50ff88
 | |
| title: 'Problem 265: Binary Circles'
 | |
| challengeType: 5
 | |
| forumTopicId: 301914
 | |
| dashedName: 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.
 | |
| 
 | |
| ```js
 | |
| assert.strictEqual(euler265(), 209110240768);
 | |
| ```
 | |
| 
 | |
| # --seed--
 | |
| 
 | |
| ## --seed-contents--
 | |
| 
 | |
| ```js
 | |
| function euler265() {
 | |
| 
 | |
|   return true;
 | |
| }
 | |
| 
 | |
| euler265();
 | |
| ```
 | |
| 
 | |
| # --solutions--
 | |
| 
 | |
| ```js
 | |
| // solution required
 | |
| ```
 |