**Problem:**

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.

Let us consider repunits of the form R(10n).

Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).

Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).

Let us consider repunits of the form R(10n).

Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).

Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).

**Solution:**

100

**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 33

* By Nayuki Minase

*

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

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

*/

public final class p033 implements EulerSolution {

public static void main(String[] args) {

System.out.println(new p033().run());

}

/*

* Consider an arbitrary fraction n/d:

* Let n = 10 * n1 + n0 be the numerator.

* Let d = 10 * d1 + d0 be the denominator.

* As stated in the problem, we need 10 <= n < d < 100.

* We must disregard trivial simplifications where n0 = d0 = 0.

*

* Now, a simplification with n0 = d0 is impossible because:

* n1 / d1 = n / d = (10*n1 + n0) / (10*d1 + n0).

* n1 * (10*d1 + n0) = d1 * (10*n1 + n0).

* 10*n1*d1 + n1*n0 = 10*d1*n1 + d1*n0.

* n1*n0 = d1*n0.

* n1 = d1.

* This implies n = d, which contradicts the fact that n < d.

* Similarly, we cannot have a simplification with n1 = d1 for the same reason.

*

* Therefore we only need to consider the cases where n0 = d1 or n1 = d0.

* In the first case, check that n1/d0 = n/d; in the second case, check that n0/d1 = n/d.

*/

public String run() {

int numer = 1;

int denom = 1;

for (int d = 10; d < 100; d++) {

for (int n = 10; n < d; n++) {

int n0 = n % 10, n1 = n / 10;

int d0 = d % 10, d1 = d / 10;

if (n1 == d0 && n0 * d == n * d1 || n0 == d1 && n1 * d == n * d0) {

numer *= n;

denom *= d;

}

}

}

return Integer.toString(denom / Library.gcd(numer, denom));

}

}

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