### Project Euler #1 - Multiples of 3 and 5

#### Peter Kim Frank on June 11, 2018

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 th... [Read Full] 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.

This is very clever and very nice!

``````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.

## 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()
``````

At last count I did that exercise in 11 different languages, so maybe I'll post some of the less common ones.

Clojure:

``````(defn problem1
[n]
(reduce + (filter #(or (= 0 (mod % 3)) (= 0 (mod % 5))) (range 1 n))))

(println (problem1 1000))
``````

``````main :: IO ()
main = print problem1

problem1 :: Integer
problem1 = sum (filter (\x -> x `mod` 3 == 0 || x `mod` 5 == 0) [1..999])
``````

Haskell (a bit more interesting, but wasteful):

``````import Data.List
problem1 = sum \$ nub \$ [3,6..999] ++ [5,10..999]
``````
``````Number dividesBy = (x):
self % x == 0.

sum = 0
1 to 999 (x):
if (x dividesBy(3) || x dividesBy(5)): sum += x
_x
(sum, "\n") join print
``````

GNU Smalltalk:

``````res := ((3 to: 999) select: [:i| (i \\ 3 = 0) | (i \\ 5 = 0)])
inject: 0 into: [:sum :i| sum+i].
res printNl
``````

Here's one in Haskell using list comprehension:

``````sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]
``````

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)
``````

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.

I wouldn't even call that nitpicking, that is solid best practice advice to avoid memory leaks.

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
``````

``````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))
``````

### 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))
``````

``````new Uint16Array(1000)
.map((v,i) => i)
.filter((v) => v % 3 == 0 || v % 5 === 0)
.reduce((ac,cv) => ac + cv)

// 233168
``````

``````result = sum(x for x in range(1001) if x % 3 == 0 or x % 5 == 0)
``````

``````let sum=0;
let start=999;

while(start){
if(start % 3 === 0 || start % 5===0){
sum +=start;
}
start--;
}

console.log(sum);
``````
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