1.6 KiB
1.6 KiB
id | challengeType | title | forumTopicId |
---|---|---|---|
5900f3b11000cf542c50fec4 | 5 | Problem 69: Totient maximum | 302181 |
Description
Euler's Totient function, φ(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, φ(9)=6.
n | Relatively Prime | φ(n) | n/φ(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 n/φ(n) for n ≤ 10.
Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
Instructions
Tests
tests:
- text: <code>totientMaximum()</code> should return a number.
testString: assert(typeof totientMaximum() === 'number');
- text: <code>totientMaximum()</code> should return 510510.
testString: assert.strictEqual(totientMaximum(), 510510);
Challenge Seed
function totientMaximum() {
return true;
}
totientMaximum();
Solution
// solution required