7a04b8977f
* fix: rework challenge to use argument in function * fix: use MathJax for display consistency * fix: add solution
119 lines
2.7 KiB
Markdown
119 lines
2.7 KiB
Markdown
---
|
|
id: 5900f3b21000cf542c50fec5
|
|
title: 'Problem 70: Totient permutation'
|
|
challengeType: 5
|
|
forumTopicId: 302183
|
|
dashedName: problem-70-totient-permutation
|
|
---
|
|
|
|
# --description--
|
|
|
|
Euler's Totient function, ${\phi}(n)$ (sometimes called the phi function), is used to determine the number of positive numbers less than or equal to `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, ${\phi}(9) = 6$. The number 1 is considered to be relatively prime to every positive number, so ${\phi}(1) = 1$.
|
|
|
|
Interestingly, ${\phi}(87109) = 79180$, and it can be seen that 87109 is a permutation of 79180.
|
|
|
|
Find the value of `n`, 1 < `n` < `limit`, for which ${\phi}(n)$ is a permutation of `n` and the ratio $\displaystyle\frac{n}{{\phi}(n)}$ produces a minimum.
|
|
|
|
# --hints--
|
|
|
|
`totientPermutation(10000)` should return a number.
|
|
|
|
```js
|
|
assert(typeof totientPermutation(10000) === 'number');
|
|
```
|
|
|
|
`totientPermutation(10000)` should return `4435`.
|
|
|
|
```js
|
|
assert.strictEqual(totientPermutation(10000), 4435);
|
|
```
|
|
|
|
`totientPermutation(100000)` should return `75841`.
|
|
|
|
```js
|
|
assert.strictEqual(totientPermutation(100000), 75841);
|
|
```
|
|
|
|
`totientPermutation(500000)` should return `474883`.
|
|
|
|
```js
|
|
assert.strictEqual(totientPermutation(500000), 474883);
|
|
```
|
|
|
|
`totientPermutation(10000000)` should return `8319823`.
|
|
|
|
```js
|
|
assert.strictEqual(totientPermutation(10000000), 8319823);
|
|
```
|
|
|
|
# --seed--
|
|
|
|
## --seed-contents--
|
|
|
|
```js
|
|
function totientPermutation(limit) {
|
|
|
|
return true;
|
|
}
|
|
|
|
totientPermutation(10000);
|
|
```
|
|
|
|
# --solutions--
|
|
|
|
```js
|
|
function totientPermutation(limit) {
|
|
function getSievePrimes(max) {
|
|
const primes = [];
|
|
const primesMap = new Array(max).fill(true);
|
|
primesMap[0] = false;
|
|
primesMap[1] = false;
|
|
|
|
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;
|
|
}
|
|
|
|
function sortDigits(number) {
|
|
return number.toString().split('').sort().join('');
|
|
}
|
|
|
|
function isPermutation(numberA, numberB) {
|
|
return sortDigits(numberA) === sortDigits(numberB);
|
|
}
|
|
|
|
const MAX_PRIME = 4000;
|
|
const primes = getSievePrimes(MAX_PRIME);
|
|
|
|
let nValue = 1;
|
|
let minRatio = Infinity;
|
|
|
|
for (let i = 1; i < primes.length; i++) {
|
|
for (let j = i + 1; j < primes.length; j++) {
|
|
const num = primes[i] * primes[j];
|
|
if (num > limit) {
|
|
break;
|
|
}
|
|
|
|
const phi = (primes[i] - 1) * (primes[j] - 1);
|
|
const ratio = num / phi;
|
|
|
|
if (minRatio > ratio && isPermutation(num, phi)) {
|
|
nValue = num;
|
|
minRatio = ratio;
|
|
}
|
|
}
|
|
}
|
|
return nValue;
|
|
}
|
|
```
|