# Project Euler 1: Multiples of 3 and 5

## Problem

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.

Official Problem

## Solution Approach

In this problem, we have to find the sum of elements of 3 or 5 which are less than 1000. The solution is pretty straightforward. You can apply a single loop over the multiples of the given numbers which will do the trick.

The catch here is if we take the multiples of 3 and multiples of 5 separately then we have to handle the cases of 15, 30, 45 β¦. Separately as they are both the multiples of 3 as well as 5, hence it’s better to take the two conditions together using an βorβ condition which will take multiples of 15 only once and hence our problem is solved.

Coming to the code part of the solution, first, initialize the two variables sum as 0 and limit as 1000 as given in the question. Then we iterate from 3 to the limit and use or condition to check if the element is divisible by either 3 or 5 or not. If it is divisible then we just add that element to the sum and return that as the answer.

## Solution Code

### Java Solution

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