**Problem:**

Euler's Totient function, φ(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, φ(9)=6.

The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.

Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.

Find the value of n, 1 [<] n [<] 107, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.

The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.

Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.

Find the value of n, 1 [<] n [<] 107, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.

**Solution:**

648

**Code:**

The solution may include methods that will be found here: Library.java .

The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}

/*

* Solution to Project Euler problem 20

* By Nayuki Minase

*

* http://nayuki.eigenstate.org/page/project-euler-solutions

* https://github.com/nayuki/Project-Euler-solutions

*/

public final class p020 implements EulerSolution {

public static void main(String[] args) {

System.out.println(new p020().run());

}

public String run() {

String temp = Library.factorial(100).toString();

int sum = 0;

for (int i = 0; i < temp.length(); i++)

sum += temp.charAt(i) - '0';

return Integer.toString(sum);

}

}

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