The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …

Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

public class Problem12 {
 
	static long result = 0;
	 
	public static void solve() {
		long currentnumber = 0;
		long factors = 0;
		long sqrt = 0;
		for (long i = 0; i < Long.MAX_VALUE; i++) {
			currentnumber +=i;
			sqrt = (long)(Math.sqrt(currentnumber));
			for (int j = 1; j < sqrt; j++) {
				if (currentnumber % j == 0) {
					factors+=2;
				}
			}
			if (sqrt*sqrt == currentnumber) {
				factors--;
			}			 
			System.out.println(currentnumber + " has " + factors + " factors");
			if (factors >500) {
				result = currentnumber;
				return;
			}
			factors = 0;
		}
	}
	 
	public static void main(String[] args) {	 
		solve();
		System.out.println(result);
	} 
}

Lösung: 76576500

The series, 1¹ + ²² + ³³ + … + 10¹⁰ = 10405071317.

Find the last ten digits of the series, 1¹ + 2² + 3³ + … + 1000¹⁰⁰⁰.

BigInteger number = BigInteger.ONE;
BigInteger result = BigInteger.ZERO;
 
for (long i = 1; i <=1000; i++) {
	number = BigInteger.valueOf(i);
	result = result.add(number.pow((int)i));
}
String output = result.toString();
System.out.println(output.substring(output.length()-10));

Lösung: 9110846700

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Note: as 1! = 1 and 2! = 2 are not sums they are not included.

public class Project34 {
 
	public static void main(String[] args) {
		String number = "";
		int tempresult = 0;
		int result = 0;
		for (int i = 3; i < 1000000; i++) {
			number = Integer.toString(i);
			for (int j = 0; j < number.length(); j++) {
				tempresult += fac(Integer.parseInt(String.valueOf(number.charAt(j))));
			}
			System.out.println(i + " " + tempresult);
			if (i==tempresult) {
				System.out.println(i);
				result += i;
			}
			tempresult = 0;
		}
		System.out.println("Ergebnis: " + result);	 
	}
	
	public static int fac(int number){
		if (number==1 || number == 0) {
			return 1;
		}
		number = number*fac(number-1);
		return number;	 
	} 
}

Lösung: 40730

Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
1634 = 1⁴ + 6⁴ + 3⁴ + 4⁴
8208 = 8⁴ + 2⁴ + 0⁴ + 8⁴
9474 = 9⁴ + 4⁴ + 7⁴ + 4⁴
As 1 = 1⁴ is not a sum it is not included.
The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

int sum = 0;
String temp = "";
int tempresult = 0;
for (int i = 0; i < 236196; i++) {
	temp = Integer.toString(i);
	for (int j = 0; j < temp.length(); j++) {
		tempresult += Math.pow(Integer.parseInt(String.valueOf(temp.charAt(j))), 5);
	}
	if (tempresult == i && !(i==0||i==1)) {
		System.out.println(i);
		sum += i;
	}
	tempresult = 0;
}
System.out.println("Ergebnis:" + sum);

Lösung: 443839

The Fibonacci sequence is defined by the recurrence relation:

F_n = F_n−1 + F_n−2 , where F₁ = 1 and F₂ = 1.

Hence the first 12 terms will be:

F₁ = 1
F₂ = 1
F₃ = 2
F₄ = 3
F₅ = 5
F₆ = 8
F₇ = 13
F₈ = 21
F₉ = 34
F₁₀ = 55
F₁₁ = 89
F₁₂ = 144

The 12th term, F₁₂ , is the first term to contain three digits.

What is the first term in the Fibonacci sequence to contain 1000 digits?

BigInteger number1 = BigInteger.ZERO;
BigInteger number2 = BigInteger.ONE;
BigInteger temp = BigInteger.ZERO;
int count = 1;
 
while (number2.toString().length()<1000) {
	temp = number1.add(number2);
	number1 = number2;
	number2 = temp;
	count++;
}
System.out.println(count);

Lösung: 4782

n! means n × (n − 1) × … × 3 × 2 × 1

For example, 10! = 10 × 9 × … × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

int result = 0;
BigInteger temp = BigInteger.ONE;
 
for (long i=2; i<=100; i++) {
	temp = temp.multiply(BigInteger.valueOf(i));
}
String number = temp.toString();
for (int i=0; i<number.length(); i++) {
	result += Character.digit(number.charAt(i), 10);
}
System.out.println(result);

Lösung: 648

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

long result = 0;
boolean[] gestrichen = new boolean[limit];
for (int i=2; i<gestrichen.length; i++) {
	gestrichen[i] = false;
}
 
for (int i=2; i<gestrichen.length; i++) {
	if (!gestrichen[i]) {
		System.out.println(i + " ");
		result += (long)i;
		for (int j=2*i; j<limit; j=j+i) {
			gestrichen[j] = true;
		}
	}
}
System.out.println("=============");
System.out.println(result);

Lösung: 142913828922