47 lines
		
	
	
		
			1.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			47 lines
		
	
	
		
			1.2 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
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| id: 5900f3ee1000cf542c50ff00
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| title: 'Problem 130: Composites with prime repunit property'
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| challengeType: 5
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| forumTopicId: 301758
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| dashedName: problem-130-composites-with-prime-repunit-property
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| ---
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| 
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| # --description--
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| 
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| A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
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| 
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| Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
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| 
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| You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
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| 
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| However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.
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| 
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| Find the sum of the first twenty-five composite values of n for whichGCD(n, 10) = 1 and n − 1 is divisible by A(n).
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| 
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| # --hints--
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| 
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| `euler130()` should return 149253.
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| 
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| ```js
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| assert.strictEqual(euler130(), 149253);
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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 euler130() {
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| 
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|   return true;
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| }
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| 
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| euler130();
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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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