Review question

# What is the sum of the multiples of 5 or 8 but not both? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8249

## Solution

Considering only the integers from $1$ to $1000$ inclusive, find the sum of those integers which are either multiples of $5$ (i.e., $5$, $10$, $15$, …) or multiples of $8$ (i.e., $8$, $16$, $24$, …) but are not multiples of both $5$ and $8$.

We need $(5 + 10 + 15 \ldots + 1000) + (8+16+24+\ldots +1000) - 2(40 + 80 + \ldots + 1000).$

We’ll use the formula that the sum of the first $n$ terms of an arithmetic progression is $\dfrac{n}{2}(u_1 + u_n),$

where $u_n$ is the $n$th term. So here we have

$\dfrac{200}{2}(5+1000) + \dfrac{125}{2}(8+1000)- 2\times \dfrac{25}{2}(40+1000) = 137 500.$

Alternatively, we have

$(5 + 10 + 15 \ldots + 1000) + (8+16+24+\ldots +1000) - 2(40 + 80 + \ldots + 1000)$

$= 5(1+2+3\ldots+200) + 8(1+2+3+\ldots+125) - 2\times 40(1+2+3+\ldots + 25).$

Now we know $1 + 2 + \ldots + n = \dfrac{n(n+1)}{2}$, so this becomes

$5\dfrac{200(201)}{2}+8\dfrac{125(126)}{2}-80\dfrac{25(26)}{2} = 137 500.$