1 - Multiples of 3 or 5

3 minute read Published: 2026-04-28

Problem statement on Project Euler's Website

Solution in the Uiua pad

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 $1000$.

Foundations and research before we begin

See if you know these concepts before moving on

Modulo

Modulo is a standard operation in most programming languages. It allows you to ask what is the remainder when I divide a number by another number?

In Uiua, writing mod a b will ask the question what is the remainder of b when divided by a?

Let's see some examples:

# Hint: mod gets formatted to ◿
    ◿ 5 14
4
    ◿ 6 13
1
    ◿ 2 14
0
    ◿ 3 15
0
    ◿ 5 15
0

What does it mean? Take for example mod 5 14. Why does it result in 4? Personally, I reason about it like so. See how close you can get without going over, and then calculate the difference.

Target: 14
5 * 1 = 5   <- not enough
5 * 3 = 15  <- Too much, 15 > 14
5 * 2 = 10  <- Just right

Then: target - closest multiple without going over
14-10 = 4

You can also think about it in this way. What does the fractional part of a division represent? It represents the remainder, does it not? Let's keep with mod 5 14, but let's try another programming language:

> 14.0 / 5
=> 2.8

# The whole part
> 5 * 2
=> 10

# The rational part
# The remainder
> 5 * 0.8
=> 4.0

# The modulo
> 14 % 5
=> 4

Key insight: When mod a b results in 0, then it means b is fully divisible, with no remainder, by a.

Further reading:

FizzBuzz

If you are unsure on what way to go to solve this problem, try looking at discussions on FizzBuzz. This is (or was?) a common interview question for entry level programmers and considerable literature exists.

Towards the answer

We'll build up the answer from what we know.

We'll check a range of numbers and take their 3 and 5 modulo. According to the key insight, we want to find all numbers whose 3 or 5 modulo results in 0, meaning that they are a multiple of either.

Let's create a range, starting at 1 (mod 0 X is dividing by 0) and ending at 9, like the problem example.

    range,1 9
    ⇡₁9
[1 2 3 4 5 6 7 8 9]

Let's take the 3 and 5 modulo. Let's also preserve the original range with by.

    by fork mod,3 mod,5 range,1 9
    ⊸⊃◿₃◿₅ ⇡₁9
[1 2 3 4 5 6 7 8 9]
[1 2 3 4 0 1 2 3 4]
[1 2 0 1 2 0 1 2 0]

Now we want all instances of 0s in either modulo. We can take the min of these two arrays and then ask where it is 0 with eq 0.

    =0↧⊸⊃◿₃ ◿₅ ⇡₁9
[1 2 3 4 5 6 7 8 9]
[0 0 1 0 1 1 0 0 1]

Now that we have a masking array we simply keep it.

    ▽=0↧⊸⊃◿₃ ◿₅ ⇡₁9
[3 5 6 9]

and finally we can get the sum with reduce add

    /+▽=0↧⊸⊃◿₃ ◿₅ ⇡₁9
23

Solution

Given what we worked out above, we simply extend the range to 999 to check all numbers under 1000. The solution is then

    /+▽=0↧⊸⊃◿₃ ◿₅ ⇡₁999
233168