# [Codewars - 6kyu解題] Multiples of 3 or 5 加總3或5的倍數

###### tags: `Codewars`,`6kyu`,`Javascript`,`Array`,`for-loop`

> Javascript菜鳥紀錄Codewars解題過程

## Instructions 題目
:link: https://www.codewars.com/kata/514b92a657cdc65150000006

:pushpin: **Instructions:**
接收一個數字，回傳小於該數字的所有自然數中是3或5的倍數的數字總和。
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.

Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in.

:bulb: **Notice:**

若同時是3和5的倍數，只要計算一次即可。
If the number is a multiple of both 3 and 5, only count it once.

## My Solution 我的解法

function solution(number){
  var arr = []
  var res = 0
  for(i=0;i<number;i++){
    if(i%3==0||i%5==0){arr.push(i);}
  }
  arr.forEach(i => res+=i);
  return res
}

## Solutions 其他更精簡的寫法

function solution(number){
  var sum = 0;
  
  for(var i = 1;i< number; i++){
    if(i % 3 == 0 || i % 5 == 0){
      sum += i
    }
  }
  return sum;
}