chore(i8n,learn): processed translations

curriculum/challenges/espanol/10-coding-interview-prep/project-euler/problem-35-circular-primes.md

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+---
+id: 5900f38f1000cf542c50fea2
+title: 'Problem 35: Circular primes'
+challengeType: 5
+forumTopicId: 302009
+dashedName: problem-35-circular-primes
+---
+
+# --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`.
+
+# --hints--
+
+`circularPrimes(100)` should return a number.
+
+```js
+assert(typeof circularPrimes(100) === 'number');
+```
+
+`circularPrimes(100)` should return 13.
+
+```js
+assert(circularPrimes(100) == 13);
+```
+
+`circularPrimes(100000)` should return 43.
+
+```js
+assert(circularPrimes(100000) == 43);
+```
+
+`circularPrimes(250000)` should return 45.
+
+```js
+assert(circularPrimes(250000) == 45);
+```
+
+`circularPrimes(500000)` should return 49.
+
+```js
+assert(circularPrimes(500000) == 49);
+```
+
+`circularPrimes(750000)` should return 49.
+
+```js
+assert(circularPrimes(750000) == 49);
+```
+
+`circularPrimes(1000000)` should return 55.
+
+```js
+assert(circularPrimes(1000000) == 55);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```js
+function circularPrimes(n) {
+
+  return n;
+}
+
+circularPrimes(1000000);
+```
+
+# --solutions--
+
+```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;
+}
+```