Sunday Times Teaser 2718 – By the Dozen

by Victor Bryant

The five different, non-zero digits in a number have been replaced by letters to give the word DOZEN.

This number is divisible by at least twelve of the fifteen numbers represented by words formed by taking one, two, three, four and five letters in sequence from the word DOZEN.

What number does DOZEN represent?

2 Comments Leave one →

brian gladman permalink


from itertools import permutations
from functools import reduce

# the numbers (except DOZEN)
nbrs = ('D', 'O', 'Z', 'E', 'N', 'DO', 'OZ', 'ZE', 'EN', 
        'DOZ', 'OZE', 'ZEN','DOZE', 'OZEN')

f = lambda x, y: 10 * x + y

# permute five non-zero digits
for p in permutations(range(1, 10), 5):
  # compute the value of DOZEN
  dozen = reduce(f, p)
  # map letters to digits
  d = dict(zip('DOZEN', p))
  # count the numbers from the words listed above that divide DOZEN
  cnt = sum(dozen % reduce(f, (d[x] for x in s)) == 0 for s in nbrs)
  # if we have at least 11 without DOZEN itself
  if cnt >= 11:
    print('DOZEN = {}'.format(dozen))

from itertools import permutations

from functools import reduce

# the numbers (except DOZEN)

nbrs = ('D', 'O', 'Z', 'E', 'N', 'DO', 'OZ', 'ZE', 'EN',

'DOZ', 'OZE', 'ZEN','DOZE', 'OZEN')

f = lambda x, y: 10 * x + y

# permute five non-zero digits

for p in permutations(range(1, 10), 5):

# compute the value of DOZEN

dozen = reduce(f, p)

# map letters to digits

d = dict(zip('DOZEN', p))

# count the numbers from the words listed above that divide DOZEN

cnt = sum(dozen % reduce(f, (d[x] for x in s)) == 0 for s in nbrs)

# if we have at least 11 without DOZEN itself

if cnt >= 11:

print('DOZEN = {}'.format(dozen))

geoffrounce permalink

Another method:


# We can only afford 3 failures in divisibility in the fifteen
# numbers for DOZEN to be divisible by at least twelve numbers 

from itertools import permutations

for p1 in permutations('123456789', 5):
  d, o, z, e, n = p1
  dozen = int(d + o + z + e + n)

  # single digit numbers
  cnt = sum(dozen % int(x) != 0 for x in p1)
  if cnt > 3:
    continue
  # two  digit numbers
  p2 = d + o, o + z, z + e, e + n
  cnt += sum(dozen % int(x) != 0 for x in p2)
  if cnt > 3:
    continue
  # three digit numbers
  p3 = d + o + z, o + z + e, z + e + n
  cnt += sum(dozen % int(x) != 0 for x in p3)
  if cnt > 3:
    continue
  # four digit numbers
  p4 = d + o + z + e, o + z + e + n
  cnt += sum(dozen % int(x) != 0 for x in p4)
  if cnt > 3:
    continue
  print('DOZEN = {}'.format(dozen))

# We can only afford 3 failures in divisibility in the fifteen

# numbers for DOZEN to be divisible by at least twelve numbers

from itertools import permutations

for p1 in permutations('123456789', 5):

d, o, z, e, n = p1

dozen = int(d + o + z + e + n)

# single digit numbers

cnt = sum(dozen % int(x) != 0 for x in p1)

if cnt > 3:

continue

# two digit numbers

p2 = d + o, o + z, z + e, e + n

cnt += sum(dozen % int(x) != 0 for x in p2)

if cnt > 3:

continue

# three digit numbers

p3 = d + o + z, o + z + e, z + e + n

cnt += sum(dozen % int(x) != 0 for x in p3)

if cnt > 3:

continue

# four digit numbers

p4 = d + o + z + e, o + z + e + n

cnt += sum(dozen % int(x) != 0 for x in p4)

if cnt > 3:

continue

print('DOZEN = {}'.format(dozen))

Sunday Times Teaser 2718 – By the Dozen

by Victor Bryant

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