61 lines
		
	
	
		
			958 B
		
	
	
	
		
			Markdown
		
	
	
	
	
	
		
		
			
		
	
	
			61 lines
		
	
	
		
			958 B
		
	
	
	
		
			Markdown
		
	
	
	
	
	
|   | --- | ||
|  | id: 5900f5311000cf542c510042 | ||
|  | title: 'Problem 451: Modular inverses' | ||
|  | challengeType: 5 | ||
|  | forumTopicId: 302124 | ||
|  | dashedName: problem-451-modular-inverses | ||
|  | --- | ||
|  | 
 | ||
|  | # --description--
 | ||
|  | 
 | ||
|  | Consider the number 15. | ||
|  | 
 | ||
|  | There are eight positive numbers less than 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14. | ||
|  | 
 | ||
|  | The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14 | ||
|  | 
 | ||
|  | because | ||
|  | 
 | ||
|  | 1\*1 mod 15=1 | ||
|  | 
 | ||
|  | 2\*8=16 mod 15=1 | ||
|  | 
 | ||
|  | 4\*4=16 mod 15=1 | ||
|  | 
 | ||
|  | 7\*13=91 mod 15=1 | ||
|  | 
 | ||
|  | 11\*11=121 mod 15=1 | ||
|  | 
 | ||
|  | 14\*14=196 mod 15=1 | ||
|  | 
 | ||
|  | Let I(n) be the largest positive number m smaller than n-1 such that the modular inverse of m modulo n equals m itself. So I(15)=11. Also I(100)=51 and I(7)=1. | ||
|  | 
 | ||
|  | Find ∑I(n) for 3≤n≤2·107 | ||
|  | 
 | ||
|  | # --hints--
 | ||
|  | 
 | ||
|  | `euler451()` should return 153651073760956. | ||
|  | 
 | ||
|  | ```js | ||
|  | assert.strictEqual(euler451(), 153651073760956); | ||
|  | ``` | ||
|  | 
 | ||
|  | # --seed--
 | ||
|  | 
 | ||
|  | ## --seed-contents--
 | ||
|  | 
 | ||
|  | ```js | ||
|  | function euler451() { | ||
|  | 
 | ||
|  |   return true; | ||
|  | } | ||
|  | 
 | ||
|  | euler451(); | ||
|  | ``` | ||
|  | 
 | ||
|  | # --solutions--
 | ||
|  | 
 | ||
|  | ```js | ||
|  | // solution required | ||
|  | ``` |