Project Euler – Problem #’s 1, 2, 3, 4, 5, 6, 7 – Solved with Javascript

js

Project Euler – Problem # 1 – Solved with Javascript:

Project Euler – Problem # 2 – Solved with Javascript:

Project Euler – Problem # 3 – Solved with Javascript:

Project Euler – Problem # 4 – Solved with Javascript:

Project Euler – Problem # 5 – Solved with Javascript:

Project Euler – Problem # 6 – Solved with Javascript:

Project Euler – Problem # 7 – Solved with Javascript:

Advertisements

Project Euler – Problem # 1 – Solved with the Go programming language

Add all the natural numbers below one thousand that are multiples of 3 or 5.

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.

**********

Python, Java and Go all have the modulus operator. It was helpful in solving this problem.

One Possible Solution: Go

package main

import "fmt"

var (
	sum int = 0
)

func main() {
	for i := 0; i < 1000; i++ {
		if i%3 == 0 || i%5 == 0 {
			sum = sum + i
		}
	}
	fmt.Println(sum)
}

Project Euler – Problem # 1 – Solved with Java & Python

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.

**********

I have been learning Java from internet tutorials. If you see some code that can be improved on, feel free to leave a comment.

One Possible Solution: Java

public class Problem1 {
	public static void main(String[] args){
		int sum = 0;
		for(int counter = 1; counter < 1000; counter++){
			if(counter % 3 == 0 || counter % 5 == 0){
				sum = sum + counter;
			}else{
				continue;
			}
		}
		System.out.println(sum);
	}

}

One Possible Solution: Python

# Python version = 2.7.2
# Platform = win32

sum = 0
for i in range(1, 1000):
    if i % 3 == 0 or i % 5 == 0:
        sum = sum + i
print "Answer = ", sum