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curriculum/challenges/italian/10-coding-interview-prep/project-euler/problem-77-prime-summations.md

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+---
+id: 5900f3b91000cf542c50fecc
+title: 'Problem 77: Prime summations'
+challengeType: 5
+forumTopicId: 302190
+dashedName: problem-77-prime-summations
+---
+
+# --description--
+
+It is possible to write ten as the sum of primes in exactly five different ways:
+
+<div style='margin-left: 4em;'>
+  7 + 3<br>
+  5 + 5<br>
+  5 + 3 + 2<br>
+  3 + 3 + 2 + 2<br>
+  2 + 2 + 2 + 2 + 2<br><br>
+</div>
+
+What is the first value which can be written as the sum of primes in over `n` ways?
+
+# --hints--
+
+`primeSummations(5)` should return a number.
+
+```js
+assert(typeof primeSummations(5) === 'number');
+```
+
+`primeSummations(5)` should return `11`.
+
+```js
+assert.strictEqual(primeSummations(5), 11);
+```
+
+`primeSummations(100)` should return `31`.
+
+```js
+assert.strictEqual(primeSummations(100), 31);
+```
+
+`primeSummations(1000)` should return `53`.
+
+```js
+assert.strictEqual(primeSummations(1000), 53);
+```
+
+`primeSummations(5000)` should return `71`.
+
+```js
+assert.strictEqual(primeSummations(5000), 71);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```js
+function primeSummations(n) {
+
+  return true;
+}
+
+primeSummations(5);
+```
+
+# --solutions--
+
+```js
+function primeSummations(n) {
+  function getSievePrimes(max) {
+    const primesMap = new Array(max).fill(true);
+    primesMap[0] = false;
+    primesMap[1] = false;
+    const primes = [];
+
+    for (let i = 2; i < max; i += 2) {
+      if (primesMap[i]) {
+        primes.push(i);
+        for (let j = i * i; j < max; j += i) {
+          primesMap[j] = false;
+        }
+      }
+      if (i === 2) {
+        i = 1;
+      }
+    }
+    return primes;
+  }
+
+  const MAX_NUMBER = 100;
+  const primes = getSievePrimes(MAX_NUMBER);
+
+  for (let curNumber = 2; curNumber < MAX_NUMBER; curNumber++) {
+    const combinations = new Array(curNumber + 1).fill(0);
+    combinations[0] = 1;
+    for (let i = 0; i < primes.length; i++) {
+      for (let j = primes[i]; j <= curNumber; j++) {
+        combinations[j] += combinations[j - primes[i]];
+      }
+    }
+    if (combinations[curNumber] > n) {
+      return curNumber;
+    }
+  }
+
+  return false;
+}
+```