fix(curriculum): rework Project Euler 77 (#42077)
* fix: rework challenge to use argumnet in function * fix: add solution * fix: position block evenly between paragraphspull/42129/head
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@ -15,23 +15,41 @@ It is possible to write ten as the sum of primes in exactly five different ways:
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5 + 5<br>
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5 + 3 + 2<br>
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3 + 3 + 2 + 2<br>
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2 + 2 + 2 + 2 + 2<br>
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2 + 2 + 2 + 2 + 2<br><br>
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</div>
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What is the first value which can be written as the sum of primes in over five thousand different ways?
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What is the first value which can be written as the sum of primes in over `n` ways?
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# --hints--
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`primeSummations()` should return a number.
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`primeSummations(5)` should return a number.
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```js
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assert(typeof primeSummations() === 'number');
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assert(typeof primeSummations(5) === 'number');
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```
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`primeSummations()` should return 71.
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`primeSummations(5)` should return `11`.
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```js
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assert.strictEqual(primeSummations(), 71);
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assert.strictEqual(primeSummations(5), 11);
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```
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`primeSummations(100)` should return `31`.
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```js
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assert.strictEqual(primeSummations(100), 31);
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```
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`primeSummations(1000)` should return `53`.
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```js
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assert.strictEqual(primeSummations(1000), 53);
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```
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`primeSummations(5000)` should return `71`.
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```js
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assert.strictEqual(primeSummations(5000), 71);
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```
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# --seed--
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@ -39,16 +57,54 @@ assert.strictEqual(primeSummations(), 71);
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## --seed-contents--
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```js
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function primeSummations() {
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function primeSummations(n) {
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return true;
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}
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primeSummations();
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primeSummations(5);
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```
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# --solutions--
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```js
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// solution required
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function primeSummations(n) {
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function getSievePrimes(max) {
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const primesMap = new Array(max).fill(true);
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primesMap[0] = false;
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primesMap[1] = false;
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const primes = [];
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for (let i = 2; i < max; i += 2) {
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if (primesMap[i]) {
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primes.push(i);
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for (let j = i * i; j < max; j += i) {
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primesMap[j] = false;
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}
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}
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if (i === 2) {
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i = 1;
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}
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}
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return primes;
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}
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const MAX_NUMBER = 100;
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const primes = getSievePrimes(MAX_NUMBER);
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for (let curNumber = 2; curNumber < MAX_NUMBER; curNumber++) {
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const combinations = new Array(curNumber + 1).fill(0);
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combinations[0] = 1;
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for (let i = 0; i < primes.length; i++) {
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for (let j = primes[i]; j <= curNumber; j++) {
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combinations[j] += combinations[j - primes[i]];
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}
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}
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if (combinations[curNumber] > n) {
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return curNumber;
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}
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}
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return false;
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}
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```
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