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 →

Brian Gladman permalink


from itertools import count

# map the position in the slab colour cycle:
# (y, g, g, g, r, r, r, r) to a counter index
clr2ctr = {0:0, 1:1, 2:1, 3:1, 4:2, 5:2, 6:2, 7:2}

# the number of yellow bricks at the centre 
# (build from the centre outwards)
for n in count(1):
  # the patio width and height as it grows
  w, h, = n, 1
  
  # counters for the yellow, grey and red counts 
  # and the area of the previous rectangle
  ygr, a0 = [0, 0, 0], 0
  for cy in count(0):
    
    # the current built area (a) and the number
    # of bricks (nb) in its outer edges 
    a = w * h
    nb, a0 = a - a0, a
    
    # add the number of slabs in the current
    # band to the appropriate colour counter
    ygr[clr2ctr[cy % 8]] += nb
    
    # are we at the end of four red bands?    
    if cy % 8 == 7:
      # finish if the last red band used
      # 402 or more bricks
      if nb >= 402:
        break
    
    # proceed to the next band      
    w, h = w + 2, h + 2

  # there are 402 bricks in the final red band
  if nb == 402:
    y, g, r = ygr
    print(f"red={r}, grey={g}, yellow={y} [height={h}, width=({w}, centre={n})].")
    break

from itertools import count

# map the position in the slab colour cycle:

# (y, g, g, g, r, r, r, r) to a counter index

clr2ctr = {0:0, 1:1, 2:1, 3:1, 4:2, 5:2, 6:2, 7:2}

# the number of yellow bricks at the centre

# (build from the centre outwards)

for n in count(1):

# the patio width and height as it grows

w, h, = n, 1

# counters for the yellow, grey and red counts

# and the area of the previous rectangle

ygr, a0 = [0, 0, 0], 0

for cy in count(0):

# the current built area (a) and the number

# of bricks (nb) in its outer edges

a = w * h

nb, a0 = a - a0, a

# add the number of slabs in the current

# band to the appropriate colour counter

ygr[clr2ctr[cy % 8]] += nb

# are we at the end of four red bands?

if cy % 8 == 7:

# finish if the last red band used

# 402 or more bricks

if nb >= 402:

break

# proceed to the next band

w, h = w + 2, h + 2

# there are 402 bricks in the final red band

if nb == 402:

y, g, r = ygr

print(f"red={r}, grey={g}, yellow={y} [height={h}, width=({w}, centre={n})].")

break

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 \(5520\).

Erling Torkildsen permalink


# Below, rl and rh are the length and height of the covered patio
# and at the same time outermost layer of red stones.
# m is the number of times the pattern is repeated and n is the
# number of stones in the ending line of yellow stones.
# sumR and sumY are the number of red and yellow stones used.

for m in range(1, 7):
  for n in range(2, 175):
    rl, rh = 15 + 16 * (m - 1) + (n - 1), 15 + 16 * (m - 1)
    if 2 * rl + 2 * rh - 4 == 402:
      sumR = sum(8 * (32 * i + n - 11) for i in range(1, m + 1))
      sumY = sum(2 * (32 * j + n - 33) for j in range(2, m + 1)) + n
      if sumR == 5 * sumY:
        print(f"{sumR} red stones were used.")

# Below, rl and rh are the length and height of the covered patio

# and at the same time outermost layer of red stones.

# m is the number of times the pattern is repeated and n is the

# number of stones in the ending line of yellow stones.

# sumR and sumY are the number of red and yellow stones used.

for m in range(1, 7):

for n in range(2, 175):

rl, rh = 15 + 16 * (m - 1) + (n - 1), 15 + 16 * (m - 1)

if 2 * rl + 2 * rh - 4 == 402:

sumR = sum(8 * (32 * i + n - 11) for i in range(1, m + 1))

sumY = sum(2 * (32 * j + n - 33) for j in range(2, m + 1)) + n

if sumR == 5 * sumY:

print(f"{sumR} red stones were used.")

Sunday Times Teaser 3007 – Paving Stones

by Howard Williams

Published Sunday May 10 2020 (link)

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