911d18bae5
* fix: rework challenge to use argument in function * fix: add solution * fix: improve look and spacing
104 lines
1.9 KiB
Markdown
104 lines
1.9 KiB
Markdown
---
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id: 5900f3ba1000cf542c50fecd
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title: 'Problem 78: Coin partitions'
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challengeType: 5
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forumTopicId: 302191
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dashedName: problem-78-coin-partitions
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---
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# --description--
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Let ${p}(n)$ represent the number of different ways in which `n` coins can be separated into piles. For example, five coins can be separated into piles in exactly seven different ways, so ${p}(5) = 7$.
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<div style='text-align: center;'>
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| Coin piles |
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| ----------------- |
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| OOOOO |
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| OOOO O |
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| OOO OO |
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| OOO O O |
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| OO OO O |
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| OO O O O |
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| O O O O O |
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</div><br>
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Find the least value of `n` for which ${p}(n)$ is divisible by `divisor`.
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# --hints--
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`coinPartitions(7)` should return a number.
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```js
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assert(typeof coinPartitions(7) === 'number');
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```
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`coinPartitions(7)` should return `5`.
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```js
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assert.strictEqual(coinPartitions(7), 5);
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```
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`coinPartitions(10000)` should return `599`.
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```js
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assert.strictEqual(coinPartitions(10000), 599);
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```
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`coinPartitions(100000)` should return `11224`.
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```js
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assert.strictEqual(coinPartitions(100000), 11224);
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```
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`coinPartitions(1000000)` should return `55374`.
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```js
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assert.strictEqual(coinPartitions(1000000), 55374);
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```
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# --seed--
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## --seed-contents--
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```js
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function coinPartitions(divisor) {
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return true;
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}
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coinPartitions(7);
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```
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# --solutions--
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```js
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function coinPartitions(divisor) {
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const partitions = [1];
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let n = 0;
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while (partitions[n] !== 0) {
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n++;
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partitions.push(0);
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let i = 0;
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let pentagonal = 1;
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while (pentagonal <= n) {
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const sign = i % 4 > 1 ? -1 : 1;
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partitions[n] += sign * partitions[n - pentagonal];
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partitions[n] = partitions[n] % divisor;
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i++;
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let k = Math.floor(i / 2) + 1;
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if (i % 2 !== 0) {
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k *= -1;
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}
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pentagonal = Math.floor((k * (3 * k - 1)) / 2);
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}
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}
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return n;
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}
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```
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