Sunday Times Brainteaser 1702 – And Now Not One Is …

by BRG on 10 October 2023

By Victor Bryant

Published 30th April 1995

In what follows, digits have been replaced by letters, a different letter being used consistently for each different digit. There are no zeros.

Each of the recurring decimals:

.ANDANDAND …
.NOWNOWNOW …
.NOTNOTNOT …
.ONEONEONE …
.ISISISIS …

equals a fraction with a denominator less than 50. And now not one is easy to find.

Please find what is DONE.

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BRG permalink


from itertools import permutations
from math import gcd

rfs = 'AND NOW NOT ONE IS'.split()

# check if the fraction given by the repertend <vs>
# has a denominator less than 50; if so return the
# numerator and denominator; otherwise return None
def check(vs):
  d = 10 ** len(vs) - 1
  g = gcd(int(vs), d)
  if d < 50 * g:
    return int(vs) // g, d // g
  return None

# NOW, NOT and ONE all contain N and O;  hence consider 
# all three digit values with different digits and then
# map the 1st two digits (in either order) to all third  
# digits that give fractions with denominators below 50
d21 = {}
for a, b, c in permutations('123456789', 3):
    n = int(a + b + c)
    if 999 < 50 * gcd(n, 999):
      d21[a + b] = d21.get(a + b, set()) | {c}
      d21[b + a] = d21.get(b + a, set()) | {c}

# consider possible values for N, O and WET
for (N, O), s_WET in d21.items():
  for W, E, T in permutations(s_WET, 3):
    # check that the fractions gived by NOW, NOT and ONE are valid
    if not all(check(x) for x in (N + O + W, N + O + T, O + N + E)):
      continue
    
    # the remaining numeric digits
    s4 = set('123456789').difference({N, O} | {W, E, T})
    # now consider AND
    for A in s4:
      if A + N in d21:
        for D in d21[A + N]:
          if D in s4 and check(A + N + D):
            
            # now consider IS
            for I, S in permutations(s4.difference(A + D)):
              if check(I + S):
                
                # output possible solutions
                l2d = dict(zip('NOWETADIS', (N, O, W, E, T, A, D, I, S)))
                s = []
                for rf in rfs:
                  n, d = check(''.join(l2d[c] for c in rf))
                  s.append(f".({rf}) = {n}/{d}")
                DONE = int(D + O + N + E)
                print(f"{DONE = } [{'; '.join(s)}]")

from itertools import permutations

from math import gcd

rfs = 'AND NOW NOT ONE IS'.split()

# check if the fraction given by the repertend <vs>

# has a denominator less than 50; if so return the

# numerator and denominator; otherwise return None

def check(vs):

d = 10 ** len(vs) - 1

g = gcd(int(vs), d)

if d < 50 * g:

return int(vs) // g, d // g

return None

# NOW, NOT and ONE all contain N and O; hence consider

# all three digit values with different digits and then

# map the 1st two digits (in either order) to all third

# digits that give fractions with denominators below 50

d21 = {}

for a, b, c in permutations('123456789', 3):

n = int(a + b + c)

if 999 < 50 * gcd(n, 999):

d21[a + b] = d21.get(a + b, set()) | {c}

d21[b + a] = d21.get(b + a, set()) | {c}

# consider possible values for N, O and WET

for (N, O), s_WET in d21.items():

for W, E, T in permutations(s_WET, 3):

# check that the fractions gived by NOW, NOT and ONE are valid

if not all(check(x) for x in (N + O + W, N + O + T, O + N + E)):

continue

# the remaining numeric digits

s4 = set('123456789').difference({N, O} | {W, E, T})

# now consider AND

for A in s4:

if A + N in d21:

for D in d21[A + N]:

if D in s4 and check(A + N + D):

# now consider IS

for I, S in permutations(s4.difference(A + D)):

if check(I + S):

# output possible solutions

l2d = dict(zip('NOWETADIS', (N, O, W, E, T, A, D, I, S)))

s = []

for rf in rfs:

n, d = check(''.join(l2d[c] for c in rf))

s.append(f".({rf}) = {n}/{d}")

DONE = int(D + O + N + E)

print(f"{DONE = } [{'; '.join(s)}]")

Frits permalink

Nice.

Line 23 and 24 can be replaced by a single permutation line.

Reply
- BRG permalink
  
  Good point!
  
  Reply

Sunday Times Brainteaser 1702 – And Now Not One Is …

By Victor Bryant

Published 30th April 1995

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