freeCodeCamp/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-69-totient-maximum.md

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---
id: 5900f3b11000cf542c50fec4
title: 'Problem 69: Totient maximum'
challengeType: 5
forumTopicId: 302181
dashedName: problem-69-totient-maximum
---
# --description--
Euler's Totient function, ${\phi}(n)$ (sometimes called the phi function), is used to determine the number of numbers less than `n` which are relatively prime to `n`. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, ${\phi}(9) = 6$.
<div style='margin-left: 4em;'>
| $n$ | $\text{Relatively Prime}$ | $\displaystyle{\phi}(n)$ | $\displaystyle\frac{n}{{\phi}(n)}$ |
| --- | ------------------------- | ------------------------ | ---------------------------------- |
| 2 | 1 | 1 | 2 |
| 3 | 1,2 | 2 | 1.5 |
| 4 | 1,3 | 2 | 2 |
| 5 | 1,2,3,4 | 4 | 1.25 |
| 6 | 1,5 | 2 | 3 |
| 7 | 1,2,3,4,5,6 | 6 | 1.1666... |
| 8 | 1,3,5,7 | 4 | 2 |
| 9 | 1,2,4,5,7,8 | 6 | 1.5 |
| 10 | 1,3,7,9 | 4 | 2.5 |
</div>
It can be seen that `n` = 6 produces a maximum $\displaystyle\frac{n}{{\phi}(n)}$ for `n` ≤ 10.
Find the value of `n``limit` for which $\displaystyle\frac{n}{{\phi(n)}}$ is a maximum.
# --hints--
`totientMaximum(10)` should return a number.
```js
assert(typeof totientMaximum(10) === 'number');
```
`totientMaximum(10)` should return `6`.
```js
assert.strictEqual(totientMaximum(10), 6);
```
`totientMaximum(10000)` should return `2310`.
```js
assert.strictEqual(totientMaximum(10000), 2310);
```
`totientMaximum(500000)` should return `30030`.
```js
assert.strictEqual(totientMaximum(500000), 30030);
```
`totientMaximum(1000000)` should return `510510`.
```js
assert.strictEqual(totientMaximum(1000000), 510510);
```
# --seed--
## --seed-contents--
```js
function totientMaximum(limit) {
return true;
}
totientMaximum(10);
```
# --solutions--
```js
function totientMaximum(limit) {
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 = i + 2) {
if (primesMap[i]) {
primes.push(i);
for (let j = i * i; j < max; j = j + i) {
primesMap[j] = false;
}
}
if (i === 2) {
i = 1;
}
}
return primes;
}
const MAX_PRIME = 50;
const primes = getSievePrimes(MAX_PRIME);
let result = 1;
for (let i = 0; result * primes[i] < limit; i++) {
result *= primes[i];
}
return result;
}
```