## DEV Community

Peter Kim Frank

Posted on • Updated on

# Project Euler #1 - Multiples of 3 and 5

I thought it would be fun to create a thread where the community could solve a problem from Project Euler. This is 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.

Look forward to seeing solutions in your language of choice! Massimo Artizzu • Edited

I'll take a mathematical approach.

The sum of the first n positive integers is n*(n + 1)/2. Multiply by k, and you get the sum of the first n multiples of k.

There are 333 multiples of 3 below 1000, and 199 multiples of 5, so we get 166833 and 99500, respectively. But we can't just sum them, as we'd count the multiples of 15 twice (15 is their least common multiple), so we have to subtract those once (total: 33165).

Result: 233168

In general, let's use JavaScript for example:

``````const limit = 1000;
const m = 3, n = 5;

const mulM = Math.floor((limit - 1) / m);
const mulN = Math.floor((limit - 1) / n);

const lcm = leastCommonMultiple(m, n);
const mulLcm = Math.floor((limit - 1) / lcm);

const result = m * mulM * (mulM + 1) / 2
+ n * mulN * (mulN + 1) / 2
- lcm * mulLcm * (mulLcm + 1) / 2;
``````

There, no iterations, pure math 👍 And blazing fast

Using JavaScript's new support of `BigInt`, it's immediate even with huge numbers like 123456789012345678901234567890: it's 3556368375755728575115582031196717989244271707186887161545. tang

yesss the Gaussian sum is such a good tool! Pierre Bouillon

I did this in LOLCODE (first program that I'm writing with it, probably not the best implementation)

``````HAI 1.2
CAN HAS STDIO?
I HAS A GOAL ITZ 1000
I HAS A TOTAL ITZ 0

I HAS A VAR ITZ 0
IM IN YR LOOP UPPIN YR VAR TIL BOTH SAEM VAR AN GOAL
BOTH SAEM 0 AN MOD OF VAR AN 3
O RLY?
YA RLY
TOTAL R SUM OF TOTAL AN VAR
NO WAI
BOTH SAEM 0 AN MOD OF VAR AN 5
O RLY?
YA RLY
TOTAL R SUM OF TOTAL AN VAR
OIC
OIC
IM OUTTA YR LOOP

VISIBLE TOTAL

KTHXBYE
`````` Phil Nash

👏👏👏👏👏 Sung M. Kim

``````Enumerable
.Range(1, 1000)
.Where(i => i % 3 == 0 || i % 5 == 0)
.Sum();
``````

Explanation

1. `Enumerable.Range` generates a number between 1 & 1000
2. `Where` filters records that matches a condition (It's like `filter` in JS)
3. `Sum` is a convenience method for summing up a sequence (instead of doing `Aggregate((acc, n) => acc + n))`, which is equivalent to `reduce` in JS)

Source & Tests on Github. Phil Nash • Edited

## Ruby

``````(1...1000).select { |n| n % 3 == 0 || n % 5 == 0 }.sum
# => 233168
``````

## JavaScript

It's a shame getting a range and a sum isn't quite as easy in JS.

``````Array.from(Array(1000).keys())
.filter(n => n % 3 === 0 || n % 5 === 0)
.reduce((acc, n) => acc + n);
// => 233168
`````` Or generate the range with the es6 spread operator:

``````[...Array(1000).keys()]
.filter(n => n % 3 === 0 || n % 5 === 0)
.reduce((acc, n) => acc + n)
``````

There's also a nasty (but shorter) `eval` hack to use in place of reduce. You can use `.join(+)` to convert the array into a string containing each number joined by the '+' sign, then evaluate that string as if it's a JavaScript expression to get the sum:

``````eval([...Array(1000).keys()].filter(n => n % 3 === 0 || n % 5 === 0).join('+'))
``````

It's a bad practice to use eval, of course, but useful for JS code golf.

Or with lodash:

``````_(_.range(1, 1000)).filter(n => n % 3 === 0 || n % 5 === 0).sum()
`````` Matheus Calegaro • Edited

Here it is ¯\_(ツ)_/¯
I'm looking forward for feedbacks

``````let sum = 0,
i = 0

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

console.log('The sum is %d', sum)
`````` Guillaume Martigny

Smallest nitpicking ever: declare `i` inside the `for` loop.

``````for (let i = 0; i < 1000; i++)
``````

It will prevent it from leaking outside the loop. Jacob Evans

I wouldn't even call that nitpicking, that is solid best practice advice to avoid memory leaks. Ali Spittel
``````def sum_natural_numbers_below_n(n):
# O(N) complexity
mult_3 = range(3, n, 3)
mult_5 = range(5, n, 5)
all_multiples = set(mult_3 + mult_5)
return sum(all_multiples)

print(sum_natural_numbers_below_n(1000))
`````` Seiei Miyagi

Ruby✨💎✨

``````require "set"

multiples_of_3 = Enumerator.new {|y| x = 0; y << x while x += 3 }
multiples_of_5 = Enumerator.new {|y| x = 0; y << x while x += 5 }

puts [multiples_of_3, multiples_of_5].each_with_object(Set.new) { |e, s|
s.merge(e.take_while(&1000.method(:>)))
}.sum
`````` Alain Van Hout

In Java

``````    final int sum = IntStream.range(0, 1000)
.filter(x -> x % 3 == 0 || x % 5 == 0)
.sum();
System.out.println(sum);
`````` ### Python

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

or Python One-Liner

``````print(sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0))
`````` Gabi • Edited

Hi, I am adding my first thought on this in Java.

private int sum(int inputValue) {
int sum = 0;
for (int i = 1; i < inputValue; i++) {
if (i % 5 == 0 || i % 3 == 0) {
sum += i;
}
}
return sum;
} Meghan (she/her) • Edited
``````new Uint16Array(1000)
.map((v,i) => i)
.filter((v) => v % 3 == 0 || v % 5 === 0)
.reduce((ac,cv) => ac + cv)

// 233168
`````` Ronaldo Peres

Hi,
I am trying to solve this one, and already solved.

I am going to ask, what is the sum of all the multiples of 3 or 5 below 100000??

I am trying to do this but is giving me a timeout, when using loops. Stephen Leyva (He/Him)

Python implementation that runs in Θ(1) time

``````def solve(n):
three = 3 * sum_all(math.floor(n/3))
fives = 5 * sum_all(math.floor(n/5))
sames = 15 * sum_all(math.floor(n/15))
return three + fives - sames

def sum_all(n):
return n*(n+1)/2

`````` Frederik 👨‍💻➡️🌐 Creemers
``````result = sum(x for x in range(1001) if x % 3 == 0 or x % 5 == 0)
``````