49 lines
		
	
	
		
			1.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
		
		
			
		
	
	
			49 lines
		
	
	
		
			1.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
|   | --- | ||
|  | id: 5900f3b21000cf542c50fec5 | ||
|  | title: 'Problem 70: Totient permutation' | ||
|  | challengeType: 5 | ||
|  | forumTopicId: 302183 | ||
|  | dashedName: problem-70-totient-permutation | ||
|  | --- | ||
|  | 
 | ||
|  | # --description--
 | ||
|  | 
 | ||
|  | Euler's Totient function, φ(`n`) \[sometimes called the phi function], is used to determine the number of positive numbers less than or equal to `n` which are relatively prime to `n`. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6. The number 1 is considered to be relatively prime to every positive number, so φ(1)=1. | ||
|  | 
 | ||
|  | Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180. | ||
|  | 
 | ||
|  | Find the value of `n`, 1 < `n` < 10<sup>7</sup>, for which φ(`n`) is a permutation of `n` and the ratio `n`/φ(`n`) produces a minimum. | ||
|  | 
 | ||
|  | # --hints--
 | ||
|  | 
 | ||
|  | `totientPermutation()` should return a number. | ||
|  | 
 | ||
|  | ```js | ||
|  | assert(typeof totientPermutation() === 'number'); | ||
|  | ``` | ||
|  | 
 | ||
|  | `totientPermutation()` should return 8319823. | ||
|  | 
 | ||
|  | ```js | ||
|  | assert.strictEqual(totientPermutation(), 8319823); | ||
|  | ``` | ||
|  | 
 | ||
|  | # --seed--
 | ||
|  | 
 | ||
|  | ## --seed-contents--
 | ||
|  | 
 | ||
|  | ```js | ||
|  | function totientPermutation() { | ||
|  | 
 | ||
|  |   return true; | ||
|  | } | ||
|  | 
 | ||
|  | totientPermutation(); | ||
|  | ``` | ||
|  | 
 | ||
|  | # --solutions--
 | ||
|  | 
 | ||
|  | ```js | ||
|  | // solution required | ||
|  | ``` |