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In this article we will solve together the Multiples of 3 or 5 challenge from CodeWars, you can find it at this link. The difficulty of this challenge is easy.
Let’s read the task together:
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.
Note: If the number is a multiple of both 3 and 5, only count it once. Also, if a number is negative, return 0(for languages that do have them)
This challenge is very simple and we can achieve the expected result using the remainder operator (%
).
What this oparator does is is return the remainder left over when one operand is divided by a second operand.
Let’s look at some examples:
6%3; // 0 6%2; // 0 6%4; // 2 6%5; // 1 6%7; // 6
Let’s go over each example:
Knowing this, we can easily determine if a number is a multiple of 3 or 5 and then perform the sum we need;
function solution(number){ let sum = 0; for (var i = 0; i < number; i++) { if (i % 3 === 0 || i % 5 === 0) { sum += i; } } return sum; }
sum
variable that will hold the total sum of numbers%
(remainder) operator that we saw aboveThere are many other ways of solving this problem, let me know yours in the comment.
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