56 lines
		
	
	
		
			1.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			56 lines
		
	
	
		
			1.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
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| id: 5900f3ae1000cf542c50fec1
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| challengeType: 5
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| title: 'Problem 66: Diophantine equation'
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| videoUrl: ''
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| localeTitle: 问题66:丢番图方程
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| ---
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| 
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| ## Description
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| <section id="description">考虑形式的二次丢番图方程:x2  -  Dy2 = 1例如,当D = 13时,x中的最小解是6492  -  13×1802 = 1.可以假设当D是正整数时没有解广场。通过在D中找到D = {2,3,5,6,7}的最小解,我们得到以下结果:32  -  2×22 = 1 22  -  3×12 = 192  -  5×42 = 1 52  -  6× 22 = 1 82  -  7×32 = 1因此,通过考虑D中对于D≤7的最小解,当D = 5时获得最大的x。在x的最小解中找到D≤1000的值,其中获得x的最大值。 </section>
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| 
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| ## Instructions
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| <section id="instructions">
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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>euler66()</code>应返回661。
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|     testString: 'assert.strictEqual(euler66(), 661, "<code>euler66()</code> should return 661.");'
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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 euler66() {
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|   // Good luck!
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|   return true;
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| }
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| 
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| euler66();
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| 
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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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