---
# System prepended metadata

title: '[Codewars - 6kyu解題] Multiples of 3 or 5 加總3或5的倍數'
tags: [Javascript, 6kyu, for-loop, Codewars, Array]

---

# [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

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