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id	title	challengeType	forumTopicId	dashedName
5900f3b11000cf542c50fec4	Problem 69: Totient maximum	5	302181	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.

`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

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.

assert(typeof totientMaximum(10) === 'number');

totientMaximum(10) should return 6.

assert.strictEqual(totientMaximum(10), 6);

totientMaximum(10000) should return 2310.

assert.strictEqual(totientMaximum(10000), 2310);

totientMaximum(500000) should return 30030.

assert.strictEqual(totientMaximum(500000), 30030);

totientMaximum(1000000) should return 510510.

assert.strictEqual(totientMaximum(1000000), 510510);

--seed--

--seed-contents--

function totientMaximum(limit) {

  return true;
}

totientMaximum(10);

--solutions--

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;
}

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