Project Euler: Problem 2 – Even Fibonacci numbers

Problem 2:

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
 
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
 
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Idea:

Since we don’t need to keep track of any of the Fibonacci numbers we can just keep a running total of the even numbered Fibonacci numbers as we go, until the next number generated is greater than 4 million.

long answer = 2; //starting on 3rd term = 3, previous even fib number = 2

long first = 1;
long second = 2;
long current = first+second;

while(current < 4000000) {
   first = second;
   second = current;
   current = first+second;

   if(current % 2 == 0) {
      answer += current;
   }
}

System.out.println("Answer: "+answer);

Project Euler: Problem 1 – Multiples of 3 and 5

Problem 1:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Idea:

Pretty basic problem, just use a loop and the MOD operator to keep a running tally of all the numbers that divide evenly into either 3 or 5.

long answer = 0;
for(int i = 0; i &lt; 1000; i++) {
   if(i % 3 == 0 || i % 5 == 0) {
      answer += i;
   }
}
System.out.println("Answer: "+answer);