--- id: 5900f38f1000cf542c50fea2 challengeType: 5 title: 'Problem 35: Circular primes' forumTopicId: 302009 --- ## Description
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime. There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97. How many circular primes are there below `n`, whereas 100 ≤ `n` ≤ 1000000?
Note:
Circular primes individual rotation can exceed `n`.
## Instructions
## Tests
```yml tests: - text: circularPrimes(100) should return a number. testString: assert(typeof circularPrimes(100) === 'number'); - text: circularPrimes(100) should return 13. testString: assert(circularPrimes(100) == 13); - text: circularPrimes(100000) should return 43. testString: assert(circularPrimes(100000) == 43); - text: circularPrimes(250000) should return 45. testString: assert(circularPrimes(250000) == 45); - text: circularPrimes(500000) should return 49. testString: assert(circularPrimes(500000) == 49); - text: circularPrimes(750000) should return 49. testString: assert(circularPrimes(750000) == 49); - text: circularPrimes(1000000) should return 55. testString: assert(circularPrimes(1000000) == 55); ```
## Challenge Seed
```js function circularPrimes(n) { return n; } circularPrimes(1000000); ```
## Solution
```js function rotate(n) { if (n.length == 1) return n; return n.slice(1) + n[0]; } function circularPrimes(n) { // Nearest n < 10^k const bound = 10 ** Math.ceil(Math.log10(n)); const primes = [0, 0, 2]; let count = 0; // Making primes array for (let i = 4; i <= bound; i += 2) { primes.push(i - 1); primes.push(0); } // Getting upperbound const upperBound = Math.ceil(Math.sqrt(bound)); // Setting other non-prime numbers to 0 for (let i = 3; i < upperBound; i += 2) { if (primes[i]) { for (let j = i * i; j < bound; j += i) { primes[j] = 0; } } } // Iterating through the array for (let i = 2; i < n; i++) { if (primes[i]) { let curr = String(primes[i]); let tmp = 1; // tmp variable to hold the no of rotations for (let x = rotate(curr); x != curr; x = rotate(x)) { if (x > n && primes[x]) { continue; } else if (!primes[x]) { // If the rotated value is 0 then it isn't a circular prime, break the loop tmp = 0; break; } tmp++; primes[x] = 0; } count += tmp; } } return count; } ```