43 lines
		
	
	
		
			818 B
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			43 lines
		
	
	
		
			818 B
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
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| id: 5900f45d1000cf542c50ff70
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| title: 'Problem 241: Perfection Quotients'
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| challengeType: 5
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| forumTopicId: 301888
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| dashedName: problem-241-perfection-quotients
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| ---
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| 
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| # --description--
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| 
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| For a positive integer n, let σ(n) be the sum of all divisors of n, so e.g. σ(6) = 1 + 2 + 3 + 6 = 12.
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| 
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| A perfect number, as you probably know, is a number with σ(n) = 2n.
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| 
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| Let us define the perfection quotient of a positive integer asp(n)= σ(n)n . Find the sum of all positive integers n ≤ 1018 for which p(n) has the form k + 1⁄2, where k is an integer.
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| 
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| # --hints--
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| 
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| `euler241()` should return 482316491800641150.
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| 
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| ```js
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| assert.strictEqual(euler241(), 482316491800641150);
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| ```
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| 
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| # --seed--
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| 
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| ## --seed-contents--
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| 
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| ```js
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| function euler241() {
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| 
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|   return true;
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| }
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| 
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| euler241();
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| ```
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| 
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| # --solutions--
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| 
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| ```js
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| // solution required
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| ```
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