# Euler Problem 001

## Mar 13, 2017 09:03 · 272 words · 2 minute read

### The problem

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.

### Solution

The first algorithm is super straightforward. We want to simply loop over each number between the start and the end checking if the number is a multiple of 3 or 5 and then adding that to the sum. Now this is going to O(n). It isn’t bad at all.

```
var isMultipleOfThree = function(number) {
return (number % 3) === 0;
}
var isMultipleOfFive = function(number) {
return (number % 5) === 0;
}
var eulerProblemOne = function(upperBound){
var sum = 0;
for(var i = 0; i < upperBound; i+=3){
if(isMultipleOfFive(i) || isMultipleOfThree(i)){
sum += i;
}
}
return sum;
}
```

But we can do a little better than this. Instead of iterating over every single number between 0 and the upper bound we can simply add up to the upper bound across 3 and 5 making sure to account for double counting.

```
var isMultipleOfThree = function(number) {
return (number % 3) === 0;
}
var isMultipleOfFive = function(number) {
return (number % 5) === 0;
}
var eulerProblemOne = function(upperBound){
var sum = 0;
for(var i = 3; i < upperBound; i+=3){
if(!isMultipleOfFive(i)){
sum += i;
}
}
for(var i=5; i < upperBound; i+=5){
sum +=i;
}
return sum;
}
```

If you’d like to see the full code please see my daily toy problem exercises that I’ve been working on. It includes tests and a README.