Sunday Times Teaser 3249 – Test to Twelve

by BRG on December 30, 2024

by Victor Bryant

Published Sunday December 29 2024 (link)

I have been testing for divisibility by numbers up to twelve. I wrote down three numbers and then consistently replaced digits by letters (with different letters used for different digits) to give

TEST TO TWELVE

Then I tested each of these numbers to see if they were divisible by any of 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12. It turned out that each of these eleven divisors was a factor of exactly one of the three numbers.

What are the three numbers?

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6 Comments Leave one →

BRG permalink

This is slow but I am short of time at the moment. I might speed it up later.


from itertools import permutations

words = ['TEST', 'TO', 'TWELVE']

# consider the possible digts for ELOSTVW
for p7 in permutations('0123456789', 7):
  l2d = dict(zip('ELOSTVW', p7))
  # T cannot be zero
  if l2d['T'] == '0':
    continue
  n3 = [int(''.join(l2d[c] for c in w)) for w in words]
  # check that each divisor only divides one of the three numbers
  if all(sum(n % f == 0 for n in n3) == 1 for f in range(2, 13)):
    print(n3)

from itertools import permutations

words = ['TEST', 'TO', 'TWELVE']

# consider the possible digts for ELOSTVW

for p7 in permutations('0123456789', 7):

l2d = dict(zip('ELOSTVW', p7))

# T cannot be zero

if l2d['T'] == '0':

continue

n3 = [int(''.join(l2d[c] for c in w)) for w in words]

# check that each divisor only divides one of the three numbers

if all(sum(n % f == 0 for n in n3) == 1 for f in range(2, 13)):

print(n3)

A lot faster with some analysis (including suggestions from John below):


# 1. There is only one even number (otherwise 2 would divide two of them)
# 2. Since even numbers can't divide odd ones, the even number must be 
#    divisible by 2, 4, 6, 8, 10 and 12 (and 3, 5 and 9 as well); so it 
#    is a multiple of 360 (the LCM of its divisors); so TWELVE must be 
#    the even  number since TO is too small and the T in TEST can't be 0
# 3. TEST (1001.T + 100.E + 10.S) (mod 11) = E - S (mod 11) != 0  since 
#    E != S; TO (10.T + O) (mod 11) == O - T (mod 11) != 0 since O != T
# 4. Hence TWELVE is divisible by both 11 and 360 and is hence a multiple 
#    of 3960
# 5. T and O cannot be divisible by 5 since TWELVE is  

words = ['TEST', 'TO', 'TWELVE']

for twelve in range(102960, 1000000, 3960):
  st = str(twelve)
  if st[2] == '0' and len(set(st)) == 5 and st[0] != '5':
    l2d = dict(zip('TWELVE', st))
    
    for O in set('1379').difference(st):
      l2d.update(O=O)
      
      for S in set('123456789').difference(st, O):
        l2d.update(S=S)
        
        n3 = tuple(int(''.join(l2d[c] for c in w)) for w in words)
        # all divisors from 2 to 12 inclusive divide exactly one number
        if all(sum(n % f == 0 for n in n3) == 1 for f in range(2, 13)):
          print(n3)

# 1. There is only one even number (otherwise 2 would divide two of them)

# 2. Since even numbers can't divide odd ones, the even number must be

# divisible by 2, 4, 6, 8, 10 and 12 (and 3, 5 and 9 as well); so it

# is a multiple of 360 (the LCM of its divisors); so TWELVE must be

# the even number since TO is too small and the T in TEST can't be 0

# 3. TEST (1001.T + 100.E + 10.S) (mod 11) = E - S (mod 11) != 0 since

# E != S; TO (10.T + O) (mod 11) == O - T (mod 11) != 0 since O != T

# 4. Hence TWELVE is divisible by both 11 and 360 and is hence a multiple

# of 3960

# 5. T and O cannot be divisible by 5 since TWELVE is

words = ['TEST', 'TO', 'TWELVE']

for twelve in range(102960, 1000000, 3960):

st = str(twelve)

if st[2] == '0' and len(set(st)) == 5 and st[0] != '5':

l2d = dict(zip('TWELVE', st))

for O in set('1379').difference(st):

l2d.update(O=O)

for S in set('123456789').difference(st, O):

l2d.update(S=S)

n3 = tuple(int(''.join(l2d[c] for c in w)) for w in words)

# all divisors from 2 to 12 inclusive divide exactly one number

if all(sum(n % f == 0 for n in n3) == 1 for f in range(2, 13)):

print(n3)

John Z permalink

Brian, I think that in line 17 the

+ ‘0’

is extraneous and results in an inconsequential length mismatch.
Rewriting line 17 as

Python

l2d = dict(zip('TWELV', st ))

1
2
3

l2d = dict(zip('TWELV', st ))

produces a different inconsequential length mismatch with ‘st’ now being longer.
I suspect that you already noticed this.

Reply
- BRG permalink
  
  Thanks John,
  
  I corrected this locally (and made another change) but forgot to upload the change.
  
  Reply
John Z permalink

Is there any advantage to using the ‘update’ method in lines 20 and 23 as opposed to making a direct assignment such as:

Python

l2d['O'] = O

1
2
3

l2d['O'] = O

Also, neither T nor O can be 5 because 5 would divide both TWELVE and one of TEST or TO. Modifying lines 16 and 19 accordingly will remove most candidate values.

Reply
- BRG permalink
  
  Hi John,
  
  Answering your first question, there is no advantage in my code above but more
  generally there can be as it allows more than one change without using multiple
  dictionary assignments or the dict(zip(a, b)) approach. For example, had I needed
  to update O and S together, I could use:
  
  Python
  
  l2d.update(O=O, S=S)
  
  1
  2
  3
  
  l2d.update(O=O, S=S)
  
  I feel that this has an advantage in readability terms when a few updates are
  needed.
  
  Thanks for the further speed up suggestions that I will now look at adopting.
  
  Reply

John Z permalink

Brute force:


from itertools import permutations

for T, E, S, O, W, L, V in permutations('0123456789', 7):

  if T != '0':

    TE = int(T + E + S + T)
    TO = int(T + O)
    TW = int(T + W + E + L + V + E)

    for d in range(2, 13):
      if (TE % d == 0) + (TO % d == 0) + (TW % d == 0) != 1:
        break
    else:
      print(TE, TO, TW)

from itertools import permutations

for T, E, S, O, W, L, V in permutations('0123456789', 7):

if T != '0':

TE = int(T + E + S + T)

TO = int(T + O)

TW = int(T + W + E + L + V + E)

for d in range(2, 13):

if (TE % d == 0) + (TO % d == 0) + (TW % d == 0) != 1:

break

else:

print(TE, TO, TW)

Sunday Times Teaser 3249 – Test to Twelve

by Victor Bryant

Published Sunday December 29 2024 (link)

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