# Sunday Times Teaser 3007 – Paving Stones

*by Howard Williams*

#### Published Sunday May 10 2020 (link)

James has decided to lay square block paving stones on his rectangular patio. He has calculated that starting from the outside and working towards the middle that he can lay a recurring concentric pattern of four bands of red stones, then three bands of grey stones, followed by a single yellow stone band. By repeating this pattern and working towards the centre he is able to finish in the middle with a single line of yellow stones to complete the patio.

He requires 402 stones to complete the first outermost band. He also calculates that he will require exactly 5 times the number of red stones as he does yellow stones.

How many red stones does he require?`

2 Comments
Leave one →

Here is a manual solution. Working out from the centre of the patio, let the number of bricks in the single line of yellow bricks be \(n\). The bands of bricks proceed through a repeated cycle of length eight in the colours yellow, grey and red as follows: (y, g, g, g, r, r, r, r). Let \(k\) be the number of completed cycles. After some analysis, the following formulas give the situation at the end of each complete colour cycle:

\[\begin{array}{|l|r|}\hline \text{patio height } (h) & 16k-1\\ \hline \text{patio width } (w) & n+16k-2\\ \hline \text{total yellow bricks } (y) & n+2(k-1)(n+16k-1)\\ \hline \text{total grey bricks } (g )& 6k(n+16k-9) \\ \hline \text{total red bricks } (r) & 8k(n+16k+5)\\ \hline\text{bricks in last red band } (l) & 2(n+32k-5)\\ \hline \end{array}\]We have two equations relating these quantities:\[2(n+ 32k – 5)=402\]\[5\{n + 2(k-1)(n+16k-1)\}=8k(n+16k+5)\]After eliminating \(n\) and simplifying we obtain:\[16k^2-181k+510=0\]

Solving this gives the two roots: \(k=6\) and \(k=85/16\) of which we need the integer valued one: \(k =6\) and \(n = 14\). Plugging into the equation for the number of red bricks used gives the teaser solution as \(5520\).