freeCodeCamp/curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-49-prime-permutatio...

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---
id: 5900f39d1000cf542c50feb0
title: 'Problem 49: Prime permutations'
challengeType: 5
forumTopicId: 302159
dashedName: problem-49-prime-permutations
---
# --description--
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.
There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
What 12-digit number do you form by concatenating the three terms in this sequence?
# --hints--
`primePermutations()` should return a number.
```js
assert(typeof primePermutations() === 'number');
```
`primePermutations()` should return 296962999629.
```js
assert.strictEqual(primePermutations(), 296962999629);
```
# --seed--
## --seed-contents--
```js
function primePermutations() {
return true;
}
primePermutations();
```
# --solutions--
```js
function primePermutations() {
function arePermutations(num1, num2) {
const numStr1 = num1.toString();
let numStr2 = num2.toString();
if (numStr1.length !== numStr2.length) {
return false;
}
for (let i = 0; i < numStr1.length; i++) {
const index = numStr2.indexOf(numStr1[i]);
if (index === -1) {
return false;
}
numStr2 = numStr2.slice(0, index) + numStr2.slice(index + 1);
}
return true;
}
function isPrime(num) {
if (num < 2) {
return false;
} else if (num === 2) {
return true;
}
const sqrtOfNum = Math.floor(num ** 0.5);
for (let i = 2; i <= sqrtOfNum + 1; i++) {
if (num % i === 0) {
return false;
}
}
return true;
}
for (let num1 = 1000; num1 <= 9999; num1++) {
const num2 = num1 + 3330;
const num3 = num2 + 3330;
if (isPrime(num1) && isPrime(num2) && isPrime(num3)) {
if (arePermutations(num1, num2) && arePermutations(num1, num3)
&& num1 !== 1487) {
// concatenate and return numbers
return (num1 * 100000000) + (num2 * 10000) + num3;
}
}
}
return 0;
}
```