Sunday Times Teaser 3272 – Festive Visitors

by BRG on 8 June 2025

by Danny Roth

Published Sunday June 08 2025 (link)

George and Martha’s five daughters left home some years ago but have all moved into the same road, Square Avenue, which has 25 normally numbered houses. At Christmas last year, the parents drove to the road to visit but, with age rolling on, George’s memory was beginning to fail him. “Can’t remember the house numbers!” he moaned. “Quite easy!” retorted Martha, “if you remember three things. If you take the house numbers of Andrea, Bertha and Caroline, square each of them and add up the three squares, you get the square of Dorothy’s house number. Furthermore, the average of Andrea’s and Dorothy’s house numbers equals the average of Bertha’s and Caroline’s house numbers. Finally, Elizabeth’s house number is the average of the other four and could not be smaller.”

Which five houses got a knock on the door?

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BRG permalink


from itertools import combinations

min_e, nbrs = 26, None
# choose the house numbers for Andrea, Bertha and Caroline
for a, b, c in combinations(range(1, 26), 3):
  # calculate Dorothy's house number
  d2 = a ** 2 + b ** 2 + c ** 2
  if (d := round(d2 ** (1 / 2))) ** 2 == d2:
    # the average of Andrea and Dorothy's house numbers 
    # is the same as the average of Bertha and Caroline's     
    if a + d == b + c:
      # calculate Elizabeth's house number ...
      e, r = divmod(a + b + c + d, 4)
      if r == 0:
        # ... which is the smallest possible
        if e < min_e:
          min_e = e
          nbrs = (a, b, c, d, e)
          
print(f"House numbers: {sorted(nbrs)}")

from itertools import combinations

min_e, nbrs = 26, None

# choose the house numbers for Andrea, Bertha and Caroline

for a, b, c in combinations(range(1, 26), 3):

# calculate Dorothy's house number

d2 = a ** 2 + b ** 2 + c ** 2

if (d := round(d2 ** (1 / 2))) ** 2 == d2:

# the average of Andrea and Dorothy's house numbers

# is the same as the average of Bertha and Caroline's

if a + d == b + c:

# calculate Elizabeth's house number ...

e, r = divmod(a + b + c + d, 4)

if r == 0:

# ... which is the smallest possible

if e < min_e:

min_e = e

nbrs = (a, b, c, d, e)

print(f"House numbers: {sorted(nbrs)}")

John Z. permalink


from itertools import combinations

sqs = tuple(i * i for i in range(26))

# Maximum possible value of E
sol = (26,)

# get combinations of three square
for A, B, C in combinations(sqs[1:], 3):

  D2  = A + B + C

  if D2 not in sqs:
    continue
  # Dorothy's house number
  D = sqs.index(D2)

  # Get square roots
  (A, B, C) = tuple(sqs.index(i) for i in (A, B, C))
  
  if A + D !=  B + C:
    continue

  if not (E4 := A + B + D + C) % 4:
    sol = min(sol, (E4 // 4, A, B, C, D))

print('Addresses:', str(sorted(sol))[1:-1])

from itertools import combinations

sqs = tuple(i * i for i in range(26))

# Maximum possible value of E

sol = (26,)

# get combinations of three square

for A, B, C in combinations(sqs[1:], 3):

D2 = A + B + C

if D2 not in sqs:

continue

# Dorothy's house number

D = sqs.index(D2)

# Get square roots

(A, B, C) = tuple(sqs.index(i) for i in (A, B, C))

if A + D != B + C:

continue

if not (E4 := A + B + D + C) % 4:

sol = min(sol, (E4 // 4, A, B, C, D))

print('Addresses:', str(sorted(sol))[1:-1])

Frits permalink

Processing three (e, b) combinations.


# e = (a + b + c + d) / 4
# e = (b + c) / 2 as a + d = b + c
# d^2 = a^2 + b^2 + c^2
# d = b + c - a
# 2.(b.c - a.b - a.c) = 0 so a = b.c / (b + c)

# we can assume b < c 
# b + c is even so b.c must also be even and thus both b and c must be even
# b != 2 as 2.c / (2 + c) is not a whole number for c > b

# start e from 5 as b >= 4
for e in range(5, 26):
  e2, par = e + e, e % 2
  # a = b.c / e2 = b - (b^2 / e2)
  # b (even) = 2.x ==> 2.x.x / e must be an integer, if e odd then x odd
  for b in range(4 + 2 * par, e, 2 + 2 * par):
    c = e2 - b
    a, r = divmod(b * c, e2)
    if r: continue
    d = e2 - a
    if d > 25 or len({a, b, c, d, e}) < 5: continue
    # it can be proven that a < b < e < c < d
    print(f"answer: {(a, b, e, c, d)}")
    exit(0)

# e = (a + b + c + d) / 4

# e = (b + c) / 2 as a + d = b + c

# d^2 = a^2 + b^2 + c^2

# d = b + c - a

# 2.(b.c - a.b - a.c) = 0 so a = b.c / (b + c)

# we can assume b < c

# b + c is even so b.c must also be even and thus both b and c must be even

# b != 2 as 2.c / (2 + c) is not a whole number for c > b

# start e from 5 as b >= 4

for e in range(5, 26):

e2, par = e + e, e % 2

# a = b.c / e2 = b - (b^2 / e2)

# b (even) = 2.x ==> 2.x.x / e must be an integer, if e odd then x odd

for b in range(4 + 2 * par, e, 2 + 2 * par):

c = e2 - b

a, r = divmod(b * c, e2)

if r: continue

d = e2 - a

if d > 25 or len({a, b, c, d, e}) < 5: continue

# it can be proven that a < b < e < c < d

print(f"answer: {(a, b, e, c, d)}")

exit(0)

Frits permalink

Processing two (b, a) combinations.


# 2.(b.c - a.b - a.c) = 0 so b = a.c / (c - a)
#                         or c = a.b / (b - a)
# and 4.e^2 = b^2 + c^2 + 2.b.c  lead to
#
# e = b^2 / 2 * (b - a)
min_e, max_b, b = 26, 25 - 3, 4
while b <= max_b:
  b2 = b * b
  # e > b so 2.a > b
  for a in range(b // 2 + 1, b):
    e, r = divmod(b2, 2 * (b - a))
    if r or e <= b: continue
    c, r = divmod(a * b, b - a)
    if r or c <= e: continue
    d = b + c - a
    if d > 25: continue
    if e < min_e:
      min_e = e
      max_b = e - 1
      sol = (a, b, e, c, d) # a < b < e < c < d
    break

  b += 2 # b is even

print(f"answer: {sol}")

# 2.(b.c - a.b - a.c) = 0 so b = a.c / (c - a)

# or c = a.b / (b - a)

# and 4.e^2 = b^2 + c^2 + 2.b.c lead to

# e = b^2 / 2 * (b - a)

min_e, max_b, b = 26, 25 - 3, 4

while b <= max_b:

b2 = b * b

# e > b so 2.a > b

for a in range(b // 2 + 1, b):

e, r = divmod(b2, 2 * (b - a))

if r or e <= b: continue

c, r = divmod(a * b, b - a)

if r or c <= e: continue

d = b + c - a

if d > 25: continue

if e < min_e:

min_e = e

max_b = e - 1

sol = (a, b, e, c, d) # a < b < e < c < d

break

b += 2 # b is even

print(f"answer: {sol}")

Frits permalink

Brian rewrote the formula for “a” to a list of reciprocals.
Here is a program using Jim Randell’s reciprocals().


from math import ceil

# express a / b as 1/x + 1/y
# m, M, g = minimum/maximum, minimum allowed gap between numbers    
def reciprocals(b=1, a=1, m=1, M=1e9, g=0):      
  dmin = max(m, ceil(M * b / (M * a - b)))
  dmax = min(M, (2 * b - (g > 0)) // a)
  for x in range(dmin, dmax + 1):
    (y, r) = divmod(x * b, x * a - b)
    if r == 0 and (x + g <= y <= M):
      yield [x, y]

min_e, sol = 26, tuple()
# a = b.c / (b + c), e = (b + c) / 2,  b + c is even so
# b.c must also be even and thus both b and c must be even
# a = b.c / (b + c) rewritten: 1/a = 1/b + 1/c
# 1/a = 1/2.x + 1/2.y or 2/a = 1/x + 1/y
for a in range(3, 22):
  # solve 2/a = 1/x + 1/y
  for x, y in reciprocals(a, 2, m=a // 2 + 1, M=12, g=1):
    b, c, e = 2 * x, 2 * y, x + y    # e = (b + c) // 2
    if e < min_e and (d := b + c - a) <= 25:
      sol = (a, b, e, c, d)
      min_e = e
  else: continue    
  break
 
if sol:      
  print(f"answer: {sol}")

from math import ceil

# express a / b as 1/x + 1/y

# m, M, g = minimum/maximum, minimum allowed gap between numbers

def reciprocals(b=1, a=1, m=1, M=1e9, g=0):

dmin = max(m, ceil(M * b / (M * a - b)))

dmax = min(M, (2 * b - (g > 0)) // a)

for x in range(dmin, dmax + 1):

(y, r) = divmod(x * b, x * a - b)

if r == 0 and (x + g <= y <= M):

yield [x, y]

min_e, sol = 26, tuple()

# a = b.c / (b + c), e = (b + c) / 2, b + c is even so

# b.c must also be even and thus both b and c must be even

# a = b.c / (b + c) rewritten: 1/a = 1/b + 1/c

# 1/a = 1/2.x + 1/2.y or 2/a = 1/x + 1/y

for a in range(3, 22):

# solve 2/a = 1/x + 1/y

for x, y in reciprocals(a, 2, m=a // 2 + 1, M=12, g=1):

b, c, e = 2 * x, 2 * y, x + y # e = (b + c) // 2

if e < min_e and (d := b + c - a) <= 25:

sol = (a, b, e, c, d)

min_e = e

else: continue

break

if sol:

print(f"answer: {sol}")

Sunday Times Teaser 3272 – Festive Visitors

by Danny Roth

Published Sunday June 08 2025 (link)

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