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Maths — Pure Mathematics 2

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Maths — Pure Mathematics 2

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Home/IAL/Maths/Underground Mathematics/What is the sum of the multiples of 5 or 8 but not both?

What is the sum of the multiples of 5 or 8 but not both?

Review question1.11.21.34.14.24.3

© University of Cambridge (Underground Mathematics); © UCLES / Oxford MAT / NRICH for sourced exam questions. Source: undergroundmathematics.org.

What is the sum of the multiples of 5 or 8 but not both?

Question

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., $5$ , $8$ , $24$ , …) but are not multiples of both $5$ and $8$ .

Solution

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.

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