### Pyramids

What are the missing numbers in the pyramids?

### Paving the Way

A man paved a square courtyard and then decided that it was too small. He took up the tiles, bought 100 more and used them to pave another square courtyard. How many tiles did he use altogether?

### Chess

What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?

##### Age 11 to 14 Short Challenge Level:

Working out what number Amy started with
When Amy added 8 to the number and then multiplied by 5, she got 2015.
So after Amy had added 8 to the number but before she had multiplied by 5, she had 2015$\div$5 = 403.
So the number must have been 8 less than 403, which is 395.

Amy was supposed to multiply the number by 8 and then add 5.
395$\times$8 = 3160, and 3160 + 5 = 3165.
So Amy would have got 3165.

Using algebra to find what number Amy started with
When Amy added $8$ to the number and then multiplied by $5$, she got $2015$.
That means that $(n+8)\times5=2015$, where $n$ is the number that Amy started with.
If $(n+8)\times5=2015$, then \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 $8n+5$.
$$8n+5=8\times395+5=3160+5=3165.$$

Using algebra to find the value of the correct expression
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.