Project Euler #001 in Python: Multiples of 3 or 5

Deniz — Mon, 11 May 2026 07:12:50 +0000

Problem

Project Euler Problem #001 asks us to find the sum of all natural numbers below 1000 that are multiples of 3 or 5.

This is a great beginner-friendly problem because it teaches one of the most important ideas in programming:

Check every number, apply a condition, and accumulate the result.

A number should be added to the total if it is divisible by 3 or 5.

In Python, divisibility can be checked with the modulo operator:

n % 3 == 0

This means:

n leaves no remainder when divided by 3.

So the condition becomes:

n % 3 == 0 or n % 5 == 0

total = 0

for n in range(1000):
    if n % 3 == 0 or n % 5 == 0:
        total += n

print(total)

We start with:

total = 0

Then we loop through every number below 1000:

for n in range(1000):

For each number, we check whether it is a multiple of 3 or 5:

if n % 3 == 0 or n % 5 == 0:

If the condition is true, we add that number to the total:

total += n

The final answer is:

Time complexity: O(n)

Space complexity: O(1)

This solution checks each number once and only keeps one running total.

I also made a short visual version of this solution as part of my EulerCodeLab series:

▶️ YouTube Short:

I’m building a daily Project Euler with Python visual series.

Each problem includes: