Among the natural numbers below 10, the multiples of 3 or 5 are $3,5,6,$ and $9$. Their sum is $23$.

Find the sum of all multiples of 3 or 5 below 1000.

2×3×5×7

An ordinary loop works. The more interesting solution asks how to remove the loop entirely.

Hints

Open one at a time

Sum multiples of 3 and multiples of 5 separately.

Correct for numbers counted twice.

Solution

Best opened after a real attempt

Arithmetic series and inclusion–exclusion

The sum of positive multiples of $k$ below a limit $N$ is

$$k\frac{m(m+1)}{2},m=\lfloor \frac{N-1}{k}\rfloor .$$

Add the multiples of 3 and 5, then subtract the multiples of 15 that were counted twice:

$$3\frac{333\cdot 334}{2}+5\frac{199\cdot 200}{2}-15\frac{66\cdot 67}{2}=\boxed{{\displaystyle 233168}}.$$

This small problem introduces a durable habit: identify overlap explicitly before adding counts or sums.