Sunday Times Teaser 3081 – Connect Four

by BRG on October 8, 2021

by Howard Williams

Published Sunday October 10 2021 (link)

I have four different two-digit numbers, each having at least one digit which is a three. When I multiply any three of these numbers together I get a product that, with the inclusion of a leading zero, is one or more repetitions of the repetend of the reciprocal of the fourth two-digit number. A repetend is the repeating or recurring decimal of a number. For example 1 divided by 27 is 0.037037……, giving a repetend of 037; in that case, the product would be 37 or 37037 or 37037037 etc.

What, in ascending order, are the four two-digit numbers?

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from itertools import combinations
from functools import cache
from math import prod

# return the leading and recurring digit
# sequences for decimal reciprocal (1/d)
@cache
def reciprocal_to_decimal(d):
  n, l_r, ds = 1, [], ''
  while n and n not in l_r:
    l_r += [n]
    i, n = divmod(10 * n, d)
    ds += str(i)
  p = l_r.index(n) if n else 0
  # return the initial and the recurring part
  return ds[:p], ds[p:]
  
# two digit numbers containing at least one '3' digit
nbrs = set(n for n in range(10, 100) if '3' in str(n))

# consider all combinations of four such numbers
for n4 in combinations(nbrs, 4):
  sol = []
  # choose the one for the reciprocal
  for rc in n4:
    # form the product of the other three (with a leading zero)
    pr = '0' + str(prod(n4) // rc)
    # find the recurring part of the reciprocal
    s, rs = reciprocal_to_decimal(rc)
    # if it has a recurring part
    if rs:
      # check if the product string is a repeating sequence 
      # of the reciprocal's recurring part
      d, r = divmod(len(pr), len(rs))
      if not r and rs * d == pr:
        t = ' * '.join(str(x) for x in n4 if x != rc)
        sol.append((t, pr[1:], rc, s, rs))
  if sol:
    print(tuple(sorted(n4)))
    for t, p, rc, s, rs in sol:
      print(f"  {t} = {p}; 1/{rc} = 0.{s}({rs})...")

from itertools import combinations

from functools import cache

from math import prod

# return the leading and recurring digit

# sequences for decimal reciprocal (1/d)

@cache

def reciprocal_to_decimal(d):

n, l_r, ds = 1, [], ''

while n and n not in l_r:

l_r += [n]

i, n = divmod(10 * n, d)

ds += str(i)

p = l_r.index(n) if n else 0

# return the initial and the recurring part

return ds[:p], ds[p:]

# two digit numbers containing at least one '3' digit

nbrs = set(n for n in range(10, 100) if '3' in str(n))

# consider all combinations of four such numbers

for n4 in combinations(nbrs, 4):

sol = []

# choose the one for the reciprocal

for rc in n4:

# form the product of the other three (with a leading zero)

pr = '0' + str(prod(n4) // rc)

# find the recurring part of the reciprocal

s, rs = reciprocal_to_decimal(rc)

# if it has a recurring part

if rs:

# check if the product string is a repeating sequence

# of the reciprocal's recurring part

d, r = divmod(len(pr), len(rs))

if not r and rs * d == pr:

t = ' * '.join(str(x) for x in n4 if x != rc)

sol.append((t, pr[1:], rc, s, rs))

if sol:

print(tuple(sorted(n4)))

for t, p, rc, s, rs in sol:

print(f" {t} = {p}; 1/{rc} = 0.{s}({rs})...")

Here is an analytical solution


from math import prod, log10

# If the reciprocal of the integer X is a decimal fraction
# which repeats every N digits and the recurring sequence 
# of digits when written as an integer is R then:
#
#   1 / X = R / 10^N + R / 10^(2.N) + R / 10^(3.N) + ...
#
# Summing this geometric progression gives X.R = 10^N - 1
#
# If R factors into three two digit integers and X is itself
# a two digit integer then we are looking for an N such that
# 10^N - 1 factors into four two digit integers, each of
# which contains at least one '3' digit

# two digit numbers containing at least one '3' digit
nbrs = tuple(n for n in range(10, 100) if '3' in str(n))

# represent the intger <x> as the product of <n> different 
# values from <nbrs> in sorted order
def decompose(x, n, seq=()):
  if not n:
    if x == 1:
      yield seq
  else:
    for i in nbrs:
      if x % i == 0 and (not seq or i > seq[-1]):
        yield from decompose(x // i, n - 1, seq + (i,))

min_p, max_p = prod(nbrs[:4]), prod(nbrs[-4:])
for N in range(int(log10(min_p)), int(log10(max_p)) + 1):
  for s in decompose(10 ** N - 1, 4):
    print(s)

from math import prod, log10

# If the reciprocal of the integer X is a decimal fraction

# which repeats every N digits and the recurring sequence

# of digits when written as an integer is R then:

# 1 / X = R / 10^N + R / 10^(2.N) + R / 10^(3.N) + ...

# Summing this geometric progression gives X.R = 10^N - 1

# If R factors into three two digit integers and X is itself

# a two digit integer then we are looking for an N such that

# 10^N - 1 factors into four two digit integers, each of

# which contains at least one '3' digit

# two digit numbers containing at least one '3' digit

nbrs = tuple(n for n in range(10, 100) if '3' in str(n))

# represent the intger <x> as the product of <n> different

# values from <nbrs> in sorted order

def decompose(x, n, seq=()):

if not n:

if x == 1:

yield seq

else:

for i in nbrs:

if x % i == 0 and (not seq or i > seq[-1]):

yield from decompose(x // i, n - 1, seq + (i,))

min_p, max_p = prod(nbrs[:4]), prod(nbrs[-4:])

for N in range(int(log10(min_p)), int(log10(max_p)) + 1):

for s in decompose(10 ** N - 1, 4):

print(s)

Sunday Times Teaser 3081 – Connect Four

by Howard Williams

Published Sunday October 10 2021 (link)

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