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			69 lines
		
	
	
		
			1.4 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			69 lines
		
	
	
		
			1.4 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
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| id: 5900f4761000cf542c50ff88
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| challengeType: 5
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| title: 'Problem 265: Binary Circles'
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| ---
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| 
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| ## Description
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| <section id='description'>
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| 2N binary digits can be placed in a circle so that all the N-digit clockwise subsequences are distinct.
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| 
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| For N=3, two such circular arrangements are possible, ignoring rotations:
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| 
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| 
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| For the first arrangement, the 3-digit subsequences, in clockwise order, are: 000, 001, 010, 101, 011, 111, 110 and 100.
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| 
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| 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:
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| 00010111 2 = 23
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| 00011101 2 = 29
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| 
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| Calling S(N) the sum of the unique numeric representations, we can see that S(3) = 23 + 29 = 52.
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| 
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| Find S(5).
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| </section>
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| 
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| ## Instructions
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| <section id='instructions'>
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| 
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| </section>
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| 
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| ## Tests
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| <section id='tests'>
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| 
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| ```yml
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| tests:
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|   - text: <code>euler265()</code> should return 209110240768.
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|     testString: assert.strictEqual(euler265(), 209110240768, '<code>euler265()</code> should return 209110240768.');
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| 
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| ```
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| 
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| </section>
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| 
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| ## Challenge Seed
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| <section id='challengeSeed'>
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| 
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| <div id='js-seed'>
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| 
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| ```js
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| function euler265() {
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|   // Good luck!
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|   return true;
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| }
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| 
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| euler265();
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| ```
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| 
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| </div>
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| 
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| 
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| 
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| </section>
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| 
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| ## Solution
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| <section id='solution'>
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| 
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| ```js
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| // solution required
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| ```
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| </section>
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