Sunday Times Teaser 1952 – Naturally Enough

by BRG on August 31, 2023

by Victor Bryant

Published 13th February 2000

_ | 5 6 7 8 9 1 | ? ? ? ? ? 2 | ? ? ? ? ? 3 | ? ? ? ? ? 4 | ? ? ? ? ?

Imagine putting a digit into each of the 20 boxes of this grid so that, reading across, there are four five-figure numbers (labelled 1–4) and, reading down, five four-figure numbers (labelled 5–9). Do this in such a way that (naturally) the resulting numbers 1–9 have the following properties:

1, 4 and 9 are squares;
1 and 8 are cubes;
2, 3, 5 and 7 are primes;
6 is the product of two primes;
the average of 1–9 is more than 4.

You should then be able to answer the question:

What is the five-figure number forming 4 across?

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BRG permalink


from number_theory import Primes, is_prime, factor
from itertools import product

solve_one = True

# index digits of 4-digit cubes on their first digits
ms2cb4 = dict()
for sqr in range(10, 22):
  cb = sqr ** 3
  i = cb // 1000
  ms2cb4[i] = ms2cb4.get(i, []) + [[int(d) for d in str(cb)]]

# index digits of 4-digit squares on their first digits
ms2sq4 = dict()
for sqr in range(32, 100):
  sq = sqr ** 2
  i = sq // 1000
  ms2sq4[i] = ms2sq4.get(i, []) + [[int(d) for d in str(sq)]]

# index 5-digit primes on their last two digits
l22pr5 = dict()
for pr in Primes().range(10000, 100000):
  i = pr % 100
  l22pr5[i] = l22pr5.get(i, []) + [[int(d) for d in str(pr)]]

# index 5-digit squares on their last two digits
l22sq5 = dict()
for sr in range(100, 317):
  sq = sr ** 2
  i = sq % 100
  l22sq5[i] = l22sq5.get(i, []) + [[int(d) for d in str(sq)]]

# array of digits to an integer
a2i = lambda s: int(''.join(str(d) for d in s))   

cnt = 0
# row 1 is a 6th power
for r1r in range(5, 7):
  r1 = [int(d) for d in str(r1r ** 6)]
  
  # find column 8 cube values matching row 1
  for c8 in ms2cb4[r1[3]]:

    # find column 9 square values matching row 1
    for c9 in ms2sq4[r1[4]]:

      try:
        # consider rows 2, 3 and 4 indexed from colums 8 and 9
        for r2, r3, r4 in product(l22pr5[10 * c8[1] + c9[1]],
                                  l22pr5[10 * c8[2] + c9[2]],
                                  l22sq5[10 * c8[3] + c9[3]]):
          cols = tuple((zip(r1, r2, r3, r4)))
          
          # colums 5, and 7 are primes 
          if (is_prime(a2i(cols[0])) and is_prime(a2i(cols[2]))):
            
            # column 6 is a product of two primes
            f = factor(a2i(cols[1]))
            if len(f) == 2 and f[0][1] == f[1][1] == 1:
              
              # average of all nine row/column values is greater 
              # than the row 4 value
              rw = tuple(a2i(d) for d in (r1, r2, r3, r4))
              cl = tuple(a2i(d) for d in cols) 
              if sum(n for n in (rw + cl)) > 9 * a2i(r4):
                fs = (f"{rw[0]}\n{rw[1]}\n{rw[2]}\n{rw[3]}\n")
                if solve_one:
                  print(f"{fs}Solution: {rw[3]}")
                  exit()
                else:
                  cnt += 1
                  print(f"{cnt:3}:\n{fs}")
      except KeyError:
        continue

from number_theory import Primes, is_prime, factor

from itertools import product

solve_one = True

# index digits of 4-digit cubes on their first digits

ms2cb4 = dict()

for sqr in range(10, 22):

cb = sqr ** 3

i = cb // 1000

ms2cb4[i] = ms2cb4.get(i, []) + [[int(d) for d in str(cb)]]

# index digits of 4-digit squares on their first digits

ms2sq4 = dict()

for sqr in range(32, 100):

sq = sqr ** 2

i = sq // 1000

ms2sq4[i] = ms2sq4.get(i, []) + [[int(d) for d in str(sq)]]

# index 5-digit primes on their last two digits

l22pr5 = dict()

for pr in Primes().range(10000, 100000):

i = pr % 100

l22pr5[i] = l22pr5.get(i, []) + [[int(d) for d in str(pr)]]

# index 5-digit squares on their last two digits

l22sq5 = dict()

for sr in range(100, 317):

sq = sr ** 2

i = sq % 100

l22sq5[i] = l22sq5.get(i, []) + [[int(d) for d in str(sq)]]

# array of digits to an integer

a2i = lambda s: int(''.join(str(d) for d in s))

cnt = 0

# row 1 is a 6th power

for r1r in range(5, 7):

r1 = [int(d) for d in str(r1r ** 6)]

# find column 8 cube values matching row 1

for c8 in ms2cb4[r1[3]]:

# find column 9 square values matching row 1

for c9 in ms2sq4[r1[4]]:

try:

# consider rows 2, 3 and 4 indexed from colums 8 and 9

for r2, r3, r4 in product(l22pr5[10 * c8[1] + c9[1]],

l22pr5[10 * c8[2] + c9[2]],

l22sq5[10 * c8[3] + c9[3]]):

cols = tuple((zip(r1, r2, r3, r4)))

# colums 5, and 7 are primes

if (is_prime(a2i(cols[0])) and is_prime(a2i(cols[2]))):

# column 6 is a product of two primes

f = factor(a2i(cols[1]))

if len(f) == 2 and f[0][1] == f[1][1] == 1:

# average of all nine row/column values is greater

# than the row 4 value

rw = tuple(a2i(d) for d in (r1, r2, r3, r4))

cl = tuple(a2i(d) for d in cols)

if sum(n for n in (rw + cl)) > 9 * a2i(r4):

fs = (f"{rw[0]}\n{rw[1]}\n{rw[2]}\n{rw[3]}\n")

if solve_one:

print(f"{fs}Solution: {rw[3]}")

exit()

else:

cnt += 1

print(f"{cnt:3}:\n{fs}")

except KeyError:

continue

Frits permalink

Faster for getting all solutions. It turned out that building dictionaries with a single comprehension statement slowed things down.


from number_theory import Primes, is_prime, factor
from itertools import product

# array of digits to an integer
a2i = lambda s: int(''.join(str(d) for d in s))   

#      P  f  P  C  S
#      5  6  7  8  9
#  [1] A  B  C  D  E   S+C
#  [2] F  G  H  I  J    P  
#  [3] K  L  M  N  P    P
#  [4] Q  R  S  T  U    S
 
solve_one = False

# index digits of 4-digit cubes on their first digit
d1cb4 = dict()
for sqr in range(10, 22):
  cb = sqr ** 3
  i = cb // 1000
  d1cb4[i] = d1cb4.get(i, []) + [[int(d) for d in str(cb)]]
  
# index digits of 4-digit squares on their first digit
d1sq4 = dict()
for sqr in range(32, 100):
  sq = sqr ** 2
  i = sq // 1000
  d1sq4[i] = d1sq4.get(i, []) + [[int(d) for d in str(sq)]]              

# index 5-digit primes on 1st, 4th and 5th digit             
d145pr5 = dict()
for pr in Primes().range(10000, 100000):
  i = 100 * (pr // 10000) + pr % 100
  d145pr5[i] = d145pr5.get(i, []) + [[int(d) for d in str(pr)]]

# index 4-digit primes on their first and last digits (and without digit zero)
d14pr4 = dict()
for pr in Primes().range(1000, 10000):
  i = int((s := str(pr))[0] + s[-1])
  if '0' not in s:
    d14pr4[i] = d14pr4.get(i, []) + [[int(d) for d in str(pr)]]              

# index 4-digit primes on 1st, 2nd and 4th digit
d124pr4 = dict()              
for pr in Primes().range(1000, 10000):
  i = int((s := str(pr))[:2] + s[-1])
  d124pr4[i] = d124pr4.get(i, []) + [[int(d) for d in str(pr)]]

# index 5-digit squares on their last two digits 
d45sq5 = dict()
for sr in range(100, 317):
  sq = sr ** 2
  # 1st and 3rd digit must be odd 
  if int((s := str(sq))[0]) % 2  and int(s[2]) % 2:
    i = sq % 100
    d45sq5[i] = d45sq5.get(i, []) + [[int(d) for d in str(sq)]]

cnt = 0
# row 1 is a 6th power
for r1r in range(5, 7):
  r1 = [int(d) for d in str(r1r ** 6)]
   
  # find column 9 square values matching row 1
  for c9 in d1sq4[r1[4]]:
    # row 2 and row 3 are primes
    if any(c9[i] % 2 == 0 for i in [1, 2]): continue
    
    # find column 8 cube values matching row 1
    for c8 in d1cb4[r1[3]]:
      try:
        # consider row 4 indexed from colums 8 and 9
        for r4 in d45sq5[10 * c8[3] + c9[3]]:
        
          # consider column 5 indexed from rows 1 and 4
          for c5 in d14pr4[10 * r1[0] + r4[0]]:
            '''
            # average of all nine row/column values is greater 
            # than the row 4 value
        
            sum3r5c = sum([11111 * x for x in c5[:-1]]) + a2i(c5) + a2i(c9) + 1000 * sum(r1[1:4]) + 3 * 999
            if c5[-1] > sum3r5c // 30000: continue 
            '''          
            # consider rows 2 and 3 indexed from colums 5, 8 and 9
            for r2, r3 in product(d145pr5[100 * c5[1] + 10 * c8[1] + c9[1]],
                                  d145pr5[100 * c5[2] + 10 * c8[2] + c9[2]]): 
              cols = tuple((zip(r1, r2, r3, r4)))                     
              if not is_prime(a2i(cols[2])): continue
              
              # column 6 is a product of two primes
              f = factor(a2i(cols[1]))
              if len(f) == 2 and f[0][1] == f[1][1] == 1:
                # average of all nine row/column values is greater 
                # than the row 4 value
                rw = tuple(a2i(d) for d in (r1, r2, r3, r4))
                cl = tuple(a2i(d) for d in cols) 
                if sum(n for n in (rw + cl)) > 9 * a2i(r4):
                  fs = (f"{rw[0]}\n{rw[1]}\n{rw[2]}\n{rw[3]}\n")
                  if solve_one:
                    print(f"{fs}Solution: {rw[3]}")
                    exit()
                  else:
                    cnt += 1
                    #print(f"{cnt:3}:\n{fs}")
      except KeyError:
        continue    

if not solve_one:
  if cnt: print(fs)
  print(f"{cnt:3} solutions")

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from number_theory import Primes, is_prime, factor

from itertools import product

# array of digits to an integer

a2i = lambda s: int(''.join(str(d) for d in s))

# P f P C S

# 5 6 7 8 9

# [1] A B C D E S+C

# [2] F G H I J P

# [3] K L M N P P

# [4] Q R S T U S

solve_one = False

# index digits of 4-digit cubes on their first digit

d1cb4 = dict()

for sqr in range(10, 22):

cb = sqr ** 3

i = cb // 1000

d1cb4[i] = d1cb4.get(i, []) + [[int(d) for d in str(cb)]]

# index digits of 4-digit squares on their first digit

d1sq4 = dict()

for sqr in range(32, 100):

sq = sqr ** 2

i = sq // 1000

d1sq4[i] = d1sq4.get(i, []) + [[int(d) for d in str(sq)]]

# index 5-digit primes on 1st, 4th and 5th digit

d145pr5 = dict()

for pr in Primes().range(10000, 100000):

i = 100 * (pr // 10000) + pr % 100

d145pr5[i] = d145pr5.get(i, []) + [[int(d) for d in str(pr)]]

# index 4-digit primes on their first and last digits (and without digit zero)

d14pr4 = dict()

for pr in Primes().range(1000, 10000):

i = int((s := str(pr))[0] + s[-1])

if '0' not in s:

d14pr4[i] = d14pr4.get(i, []) + [[int(d) for d in str(pr)]]

# index 4-digit primes on 1st, 2nd and 4th digit

d124pr4 = dict()

for pr in Primes().range(1000, 10000):

i = int((s := str(pr))[:2] + s[-1])

d124pr4[i] = d124pr4.get(i, []) + [[int(d) for d in str(pr)]]

# index 5-digit squares on their last two digits

d45sq5 = dict()

for sr in range(100, 317):

sq = sr ** 2

# 1st and 3rd digit must be odd

if int((s := str(sq))[0]) % 2 and int(s[2]) % 2:

i = sq % 100

d45sq5[i] = d45sq5.get(i, []) + [[int(d) for d in str(sq)]]

cnt = 0

# row 1 is a 6th power

for r1r in range(5, 7):

r1 = [int(d) for d in str(r1r ** 6)]

# find column 9 square values matching row 1

for c9 in d1sq4[r1[4]]:

# row 2 and row 3 are primes

if any(c9[i] % 2 == 0 for i in [1, 2]): continue

# find column 8 cube values matching row 1

for c8 in d1cb4[r1[3]]:

try:

# consider row 4 indexed from colums 8 and 9

for r4 in d45sq5[10 * c8[3] + c9[3]]:

# consider column 5 indexed from rows 1 and 4

for c5 in d14pr4[10 * r1[0] + r4[0]]:

'''

# average of all nine row/column values is greater

# than the row 4 value

sum3r5c = sum([11111 * x for x in c5[:-1]]) + a2i(c5) + a2i(c9) + 1000 * sum(r1[1:4]) + 3 * 999

if c5[-1] > sum3r5c // 30000: continue

'''

# consider rows 2 and 3 indexed from colums 5, 8 and 9

for r2, r3 in product(d145pr5[100 * c5[1] + 10 * c8[1] + c9[1]],

d145pr5[100 * c5[2] + 10 * c8[2] + c9[2]]):

cols = tuple((zip(r1, r2, r3, r4)))

if not is_prime(a2i(cols[2])): continue

# column 6 is a product of two primes

f = factor(a2i(cols[1]))

if len(f) == 2 and f[0][1] == f[1][1] == 1:

# average of all nine row/column values is greater

# than the row 4 value

rw = tuple(a2i(d) for d in (r1, r2, r3, r4))

cl = tuple(a2i(d) for d in cols)

if sum(n for n in (rw + cl)) > 9 * a2i(r4):

fs = (f"{rw[0]}\n{rw[1]}\n{rw[2]}\n{rw[3]}\n")

if solve_one:

print(f"{fs}Solution: {rw[3]}")

exit()

else:

cnt += 1

#print(f"{cnt:3}:\n{fs}")

except KeyError:

continue

if not solve_one:

if cnt: print(fs)

print(f"{cnt:3} solutions")

Sunday Times Teaser 1952 – Naturally Enough

by Victor Bryant

Published 13th February 2000

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