53 lines
		
	
	
		
			1.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			53 lines
		
	
	
		
			1.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
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| id: 5900f3b31000cf542c50fec6
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| title: 'Problem 71: Ordered fractions'
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| challengeType: 5
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| forumTopicId: 302184
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| dashedName: problem-71-ordered-fractions
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| ---
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| 
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| # --description--
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| 
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| Consider the fraction, `n`/`d`, where `n` and `d` are positive integers. If `n`<`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
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| 
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| If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
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| 
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| <div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, <strong>2/5</strong>, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
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| 
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| It can be seen that 2/5 is the fraction immediately to the left of 3/7.
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| 
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| By listing the set of reduced proper fractions for `d` ≤ 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.
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| 
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| # --hints--
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| 
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| `orderedFractions()` should return a number.
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| 
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| ```js
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| assert(typeof orderedFractions() === 'number');
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| ```
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| 
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| `orderedFractions()` should return 428570.
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| 
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| ```js
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| assert.strictEqual(orderedFractions(), 428570);
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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 orderedFractions() {
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| 
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|   return true;
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| }
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| 
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| orderedFractions();
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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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