Added 8 then multiplied by 5, got 2015

After adding 8: 2015$\div$5 = 403

Original number: 8 less than 403, which is 395

Multiply by 8: 395$\times$8 = 3160

Add 5: 3160 + 5 = 3165

If Amy started with $n$,

Add $8$ then multiply by $5$ and get $2015$, so $(n+8)\times5=2015$

$$\begin{align}n+8&=2015\div5\\n+8&=403\\n&=403-8\\n&=395.\end{align}$$

Amy was supposed to multiply $n$ by $8$ and then add $5$, so she should have found $8n+5$.

$$8n+5=8\times395+5=3160+5=3165.$$

When Amy added $8$ to the number and then multiplied by $5$, she got $2015$, so $(n+8)\times5=2015$, where $n$ is the number that Amy started with.

Amy was supposed to multiply $n$ by $8$ and then add $5$, so she should have found $8n+5$.

We want to get from $(n+8)\times5$ to $8n+5$. Knowing $(n+8)\times8$ would be helpful, because $$\begin{align}(n+8)\times8&=8n+64\\&=8n+(5+59)\\&=(8n+5)+59\end{align}$$

If $(n+8)\times5=2015$, then $n+8=2015\div5=403$, so $(n+8)\times8=403\times8=3224$.

So $3224$ is $59$ more than $8n+5$, so $8n+5=3224-59=3165$. So Amy would have got $3165$.

You can find more short problems, arranged by curriculum topic, in our short problems collection.