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

project-euler-problem-1.2.js

// Do the maths
for(
    var sum = 0, i = 1;
    i < 1000;
    !(i % 3 && i % 5) && (sum += i), i++
);
// Log the result
console.log(sum);

project-euler-problem-1.js

// Prep the array
for(
    var arr = [], i = 1;
    i < 1000;
    !(i % 3 && i % 5) && arr.push(i), i++
);
// Sum the array
console.log(
    arr.reduce(
        function(prev, current) {
            return prev + current
        }
    )
);

project-euler-problem-1.py

print sum([i for i in range(1000) if not (i % 3 and i % 5)])

project-euler-problem-1.rb

puts (1...1000).select{ |n| n % 3 == 0 || n % 5 == 0 }.reduce(:+)

Nah, that's the algebraic solution to the problem. I remember doing that at school. :D

Actually Dave, can you do it again using only 1000, 3, and 5 as parameters?

Your maths is wrong 'cos whilst 999 is valid for 3, it should be 995 for 5 and… uh… 990 for 15, I believe.

For those that are interested, problem 2 solved with JS here: https://gist.github.com/1009018

(999 / 15) * 15 = 990
(999 / 15.0) * 15 = 998.9999
Integer truncation is my friend

Here is one that solves the sum of multiples of numbers in an array less than a given limit.
Find the codein JS HERE

//test: find the sum of the multiples of 3 and 5 less than 1000
console.log(sumOfMultplesOfNumbersLessThan([3,5], 1000));

sum = 0;
for(i = 1; i<1000;i++){
if((i % 3 === 0) || (i % 5 ===0)){
sum += i;
}
}

newbie rules!

Super useful. thanks for posting this with the different code options.