Problem 45 of Project Euler required one to find numbers which are triangular, pentagonal and hexagonal together. Checking for triangularity is redundancy as all hexagonal numbers are also triangular. So it boils down to solving the equation:
n(3n – 1) / 2 = m(2m – 1)
Using the Diophantine Equation solver at http://www.alpertron.com.ar/QUAD.HTM, we get the solution as
The solution comes very easily from here.
Continue Reading ?
This was solved by brute force. I implemented a method to find the number of factors of a number by its prime factorization. Having already built up a sieve method to quickly get primes, the number of factors can be found faster this way.
PrimeNumber class:
package pba.common.number;
import java.util.ArrayList;
import java.util.BitSet;
import java.util.List;
public class PrimeNumbers {
private List<Integer> m_primeNumbers;
/**
* @param sizeOfSieve The upper bound to the determination of prime numbers.
*/
public PrimeNumbers(int sizeOfSieve) {
m_primeNumbers = new ArrayList<Integer>();
sieveOfEratosthenes(sizeOfSieve);
}
/**
* ...Continue Reading ?
Problem 42 in Project Euler provides a list of some 2000 English words and requires us to find the number of triangle words. I went about in a simple straight forward way attempting this one. A class TriangleNumbers keeps a sorted list of all the triangle numbers. A static method in EnglishWord class returns the word value for the given word. This number is checked for validity.
Obviously, new ...
Continue Reading ?