**Problem:**

There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, 5C3 = 10.

In general,

nCr =

n!

r!(n[−]r)!

,where r [≤] n, n! = n[×](n[−]1)[×]...[×]3[×]2[×]1, and 0! = 1.

It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.

How many, not necessarily distinct, values of nCr, for 1 [≤] n [≤] 100, are greater than one-million?

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, 5C3 = 10.

In general,

nCr =

n!

r!(n[−]r)!

,where r [≤] n, n! = n[×](n[−]1)[×]...[×]3[×]2[×]1, and 0! = 1.

It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.

How many, not necessarily distinct, values of nCr, for 1 [≤] n [≤] 100, are greater than one-million?

**Solution:**

6857

**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 3

* By Nayuki Minase

*

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

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

*/

public final class p003 implements EulerSolution {

public static void main(String[] args) {

System.out.println(new p003().run());

}

/*

* Algorithm: Divide out all the smallest prime factors except the last one.

* For example, 1596 = 2 * 2 * 3 * 7 * 19. The algorithm ensures that the smallest factors will be found first.

* After dividing out the smallest factors, the last factor to be found will be equal to the quotient, so it must be the largest prime factor.

*/

public String run() {

long n = 600851475143L;

while (true) {

long p = smallestFactor(n);

if (p < n)

n /= p;

else

return Long.toString(n);

}

}

private static long smallestFactor(long n) {

for (long i = 2, end = Library.sqrt(n); i <= end; i++) {

if (n % i == 0)

return i;

}

return n; // Prime

}

}

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