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---
id: 5900f3ba1000cf542c50fecd
title: 'Problem 78: Coin partitions'
challengeType: 5
forumTopicId: 302191
dashedName: problem-78-coin-partitions
---
# --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;' >
| Coin piles |
| ----------------- |
| OOOOO |
| OOOO O |
| OOO OO |
| OOO O O |
| OO OO O |
| OO O O O |
| 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);
```
`coinPartitions(10000)` should return `599` .
```js
assert.strictEqual(coinPartitions(10000), 599);
```
`coinPartitions(100000)` should return `11224` .
```js
assert.strictEqual(coinPartitions(100000), 11224);
```
`coinPartitions(1000000)` should return `55374` .
```js
assert.strictEqual(coinPartitions(1000000), 55374);
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```
# --seed--
## --seed-contents--
```js
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function coinPartitions(divisor) {
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return true;
}
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coinPartitions(7);
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```
# --solutions--
```js
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function coinPartitions(divisor) {
const partitions = [1];
let n = 0;
while (partitions[n] !== 0) {
n++;
partitions.push(0);
let i = 0;
let pentagonal = 1;
while (pentagonal < = n) {
const sign = i % 4 > 1 ? -1 : 1;
partitions[n] += sign * partitions[n - pentagonal];
partitions[n] = partitions[n] % divisor;
i++;
let k = Math.floor(i / 2) + 1;
if (i % 2 !== 0) {
k *= -1;
}
pentagonal = Math.floor((k * (3 * k - 1)) / 2);
}
}
return n;
}
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```