Sunday Times Teaser 2496 – No Title

by DJT Hogg

Published July 25 2010 (link)

I have chosen 10 different letters of the alphabet before U to represent the digits 0 to 9. I can tell you that 0’s letter is in the word NOUGHT, 1’s letter is in the word ONE, 2’s letter is in TWO, and so on up to 9’s letter, being in the word NINE.

Furthermore, I can write down a particular five-figure number that is divisible by 8, and replacing the digits by their letters I get EIGHT. Similarly, a four-figure number divisible by 9 translates to NINE.

What number is represented by TEN?

One Comment Leave one →

Brian Gladman permalink


from functools import reduce

nbrs = ('NOUGHT', 'ONE', 'TWO', 'THREE', 'FOUR', 'FIVE', 'SIX', 'SEVEN',
        'EIGHT', 'NINE')

# (digit, letters) pairs giving the possible letters for each digit 
d2ls = tuple((i, set(x).difference('UVWXYZ')) for i, x in enumerate(nbrs))

# map letters to digits consistent with the (digit letter) pairs in d2ls
def map_d2l(d2ls, l2d=dict()):
  # is the map complete?
  if len(l2d.keys()) == 10:
    yield l2d
  else:
    # find the next digit with a minimum number of possible letters
    dl_min = min(d2ls, key=lambda x: len(x[1]))
    dgt = dl_min[0]
    # try each possible letter
    for ltr in dl_min[1]:
      # create and update the mapping frrom letters to digits
      n_l2d = l2d.copy()
      n_l2d[ltr] = dgt
      # remove the allocated letter from the possibilities for the
      # remaining digits
      n_d2ls = tuple((i, x - {ltr}) for i, x in d2ls if i != dgt)
      # check that all remaining digits still have possible letters 
      if all(x for i, x in n_d2ls):
        # and proceed to the next digit if they do
        yield from map_d2l(n_d2ls, n_l2d)

# convert a string of letters to a number according to the map in l2d
s2n = lambda s: reduce(lambda x, y: 10 * x + y, (l2d[c] for c in s))

for l2d in map_d2l(d2ls):
  # find the numeric values for EIGHT, NINE and TEN
  eight, nine, ten = s2n('EIGHT'), s2n('NINE'), s2n('TEN')
  # check that EIGHT and NINE are divisible by 8 and 9 respectively
  if eight % 8 == 0 and nine % 9 == 0:
    print(f'TEN = {ten} (EIGHT = {eight}, NINE = {nine}).')

from functools import reduce

nbrs = ('NOUGHT', 'ONE', 'TWO', 'THREE', 'FOUR', 'FIVE', 'SIX', 'SEVEN',

'EIGHT', 'NINE')

# (digit, letters) pairs giving the possible letters for each digit

d2ls = tuple((i, set(x).difference('UVWXYZ')) for i, x in enumerate(nbrs))

# map letters to digits consistent with the (digit letter) pairs in d2ls

def map_d2l(d2ls, l2d=dict()):

# is the map complete?

if len(l2d.keys()) == 10:

yield l2d

else:

# find the next digit with a minimum number of possible letters

dl_min = min(d2ls, key=lambda x: len(x[1]))

dgt = dl_min[0]

# try each possible letter

for ltr in dl_min[1]:

# create and update the mapping frrom letters to digits

n_l2d = l2d.copy()

n_l2d[ltr] = dgt

# remove the allocated letter from the possibilities for the

# remaining digits

n_d2ls = tuple((i, x - {ltr}) for i, x in d2ls if i != dgt)

# check that all remaining digits still have possible letters

if all(x for i, x in n_d2ls):

# and proceed to the next digit if they do

yield from map_d2l(n_d2ls, n_l2d)

# convert a string of letters to a number according to the map in l2d

s2n = lambda s: reduce(lambda x, y: 10 * x + y, (l2d[c] for c in s))

for l2d in map_d2l(d2ls):

# find the numeric values for EIGHT, NINE and TEN

eight, nine, ten = s2n('EIGHT'), s2n('NINE'), s2n('TEN')

# check that EIGHT and NINE are divisible by 8 and 9 respectively

if eight % 8 == 0 and nine % 9 == 0:

print(f'TEN = {ten} (EIGHT = {eight}, NINE = {nine}).')

Sunday Times Teaser 2496 – No Title

by DJT Hogg

Published July 25 2010 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta