Sunday Times Teaser 2929 – My Pin Numbers

by Graham Smithers

Published November 11 2018 (link)

Using all but one of the digits 0 to 9, and systematically replacing them by a letter, my pin numbers become ONE, TWO, FIVE and SEVEN.

These numbers are such that:

ONE is odd

TWO is even

FIVE is odd and divisible by 5

SEVEN is divisible by 7

ONE + TWO + TWO = FIVE

If I told you which digit was not used, you should be able to work out my pin numbers.

What pin number is represented by SEVEN?

One Comment Leave one →

Brian Gladman permalink

This solution uses my partition_unique and alphasum packages.


from functools import reduce

from partition_unique import partition_unique
from alphasum import AlphaSum

# convert a word (w) to a number using a map of letters to digits (l2d) 
w2n = lambda w, l2d: reduce(lambda x, y: 10 * x + y, (l2d[c] for c in w))

# for saving solutions
sols = []
# solve the alphametic ONE + TWO + TWO = FIVE with zero leading digits allowed
for l2d in AlphaSum(['ONE', 'TWO', 'TWO'], 'FIVE', d2nv={}).solve():
  
  #  ONE and FIVE are odd, TWO is even and FIVE is divisible by 5 
  if l2d['E'] == 5 and not l2d['O'] & 1:

    # now try the unused digits for S 
    for S in set(range(10)).difference(l2d.values()): 
      l2d['S'] = S
      SEVEN = w2n('SEVEN', l2d)

      # SEVEN is divisible by 7
      if SEVEN % 7 == 0:
        # find the unused digit and save this solution
        ud, = set(range(10)).difference(l2d.values())
        sols.append((ud, tuple(l2d.items())))

# look for solution that has a unique unused digit
for ud, sol in partition_unique(sols, f=lambda ud, sol: ud)[0]:
  w, d = ('ONE', 'TWO', 'FIVE', 'SEVEN'), dict(sol)
  t = [w2n(x, d) for x in w]
  print(f'PIN ({w[3]}) = {t[3]:05d} '
        f'({w[0]} = {t[0]:03d}, {w[1]} = {t[1]:03d}, {w[2]} = {t[2]:04d}).')

from functools import reduce

from partition_unique import partition_unique

from alphasum import AlphaSum

# convert a word (w) to a number using a map of letters to digits (l2d)

w2n = lambda w, l2d: reduce(lambda x, y: 10 * x + y, (l2d[c] for c in w))

# for saving solutions

sols = []

# solve the alphametic ONE + TWO + TWO = FIVE with zero leading digits allowed

for l2d in AlphaSum(['ONE', 'TWO', 'TWO'], 'FIVE', d2nv={}).solve():

# ONE and FIVE are odd, TWO is even and FIVE is divisible by 5

if l2d['E'] == 5 and not l2d['O'] & 1:

# now try the unused digits for S

for S in set(range(10)).difference(l2d.values()):

l2d['S'] = S

SEVEN = w2n('SEVEN', l2d)

# SEVEN is divisible by 7

if SEVEN % 7 == 0:

# find the unused digit and save this solution

ud, = set(range(10)).difference(l2d.values())

sols.append((ud, tuple(l2d.items())))

# look for solution that has a unique unused digit

for ud, sol in partition_unique(sols, f=lambda ud, sol: ud)[0]:

w, d = ('ONE', 'TWO', 'FIVE', 'SEVEN'), dict(sol)

t = [w2n(x, d) for x in w]

print(f'PIN ({w[3]}) = {t[3]:05d} '

f'({w[0]} = {t[0]:03d}, {w[1]} = {t[1]:03d}, {w[2]} = {t[2]:04d}).')

Sunday Times Teaser 2929 – My Pin Numbers

by Graham Smithers

Published November 11 2018 (link)

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