Sunday Times Teaser 2624 – Is It 2013?

Brian Gladman permalink

Here is my version:


from itertools import permutations
from collections import defaultdict

# a dictionary for each of the four digit positions that will
# list the numbers that have each digit in this position
d4 = [defaultdict(list) for x in range(4)]

# the lowest digit of the sequence of digits 
for b in range(7):
  # permute the four consecutive digits
  for p in permutations(range(b, b + 4)):
    # form the number
    n = int(''.join(str(x) for x in p))
    # ensure it has four digits and is not divisible 
    # by any number from 2 to 11 inclusive
    if p[0] and all(n % q for q in (2, 3, 5, 7, 11)):
      # now put this number into the four dictionaries
      # indexed on the digits in each position
      for d_pos, digit in enumerate(p):
        d4[d_pos][digit] += [n]

# we now need to find one of the four dictionaries in
# which only one digit value points to a single number
for dct in d4:
  # find dictionaries with only one entry of length one
  if sum(1 for x in dct.values() if len(x) == 1) == 1:
    # find the entry that points to a single number
    t = [x[0] for x in dct.values() if len(x) == 1]
    print('The number is', t[0])

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from itertools import permutations

from collections import defaultdict

# a dictionary for each of the four digit positions that will

# list the numbers that have each digit in this position

d4 = [defaultdict(list) for x in range(4)]

# the lowest digit of the sequence of digits

for b in range(7):

# permute the four consecutive digits

for p in permutations(range(b, b + 4)):

# form the number

n = int(''.join(str(x) for x in p))

# ensure it has four digits and is not divisible

# by any number from 2 to 11 inclusive

if p[0] and all(n % q for q in (2, 3, 5, 7, 11)):

# now put this number into the four dictionaries

# indexed on the digits in each position

for d_pos, digit in enumerate(p):

d4[d_pos][digit] += [n]

# we now need to find one of the four dictionaries in

# which only one digit value points to a single number

for dct in d4:

# find dictionaries with only one entry of length one

if sum(1 for x in dct.values() if len(x) == 1) == 1:

# find the entry that points to a single number

t = [x[0] for x in dct.values() if len(x) == 1]

print('The number is', t[0])

Reply

Brian Gladman permalink

Here is an alternative version (the latter part of this is similar to that in Arthur’s version below):


from itertools import permutations

# a list for potential solutions
l = []
# the lowest digit of the sequence of digits 
for b in range(7):
  # if b is divisible by 3, then all numbers formed from the
  # digits b .. b + 3 in any order are also divisible by 3
  if b % 3:
    # permute the four consecutive digits
    for p in permutations('0123456789'[b : b + 4]):
      # form the number
      ns = ''.join(p)
      n = int(ns)
      # ensure that it is not divisible by 
      # any number from 2 to 11 inclusive
      if all(n % q for q in (2, 5, 7, 11)):
        l += [ns]

for p in range(4):
  # for each digit value, compile the list of 
  # numbers with this digit in position 'p'
  t = [[int(x) for x in l if x[p]==str(d)] for d in range(10)]
  # now list any numbers that are unique for any digit values
  t = [x[0] for x in t if len(x) == 1]
  if len(t) == 1:
    print('The number is', t[0])

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from itertools import permutations

# a list for potential solutions

l = []

# the lowest digit of the sequence of digits

for b in range(7):

# if b is divisible by 3, then all numbers formed from the

# digits b .. b + 3 in any order are also divisible by 3

if b % 3:

# permute the four consecutive digits

for p in permutations('0123456789'[b : b + 4]):

# form the number

ns = ''.join(p)

n = int(ns)

# ensure that it is not divisible by

# any number from 2 to 11 inclusive

if all(n % q for q in (2, 5, 7, 11)):

l += [ns]

for p in range(4):

# for each digit value, compile the list of

# numbers with this digit in position 'p'

t = [[int(x) for x in l if x[p]==str(d)] for d in range(10)]

# now list any numbers that are unique for any digit values

t = [x[0] for x in t if len(x) == 1]

if len(t) == 1:

print('The number is', t[0])

Reply

geoffrounce permalink

Another version (with some input from Brian):


nbr_list = []
for p in permutations('0123456789', 4):
  if p[0] == '0':
    continue
  ns = ''.join(p)
  n, ns = int(ns), sorted(ns)
  if int(ns[3]) == int(ns[0]) + 3:
    if all(n % a for a in (2, 3, 5, 7, 11)):
      nbr_list += [n]

for pos in range(4):
  cnt = list(sum(1 for x in nbr_list if str(x)[pos]
                 == str(v)) for v in range(10))
  if cnt.count(1) == 1:
    print(nbr_list[cnt.index(1)])

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nbr_list = []

for p in permutations('0123456789', 4):

if p[0] == '0':

continue

ns = ''.join(p)

n, ns = int(ns), sorted(ns)

if int(ns[3]) == int(ns[0]) + 3:

if all(n % a for a in (2, 3, 5, 7, 11)):

nbr_list += [n]

for pos in range(4):

cnt = list(sum(1 for x in nbr_list if str(x)[pos]

== str(v)) for v in range(10))

if cnt.count(1) == 1:

print(nbr_list[cnt.index(1)])

Reply

Naim Uygun permalink

Hi Brian,
How could we deduce the number from the output of this program if we knew the digit in a particular position?


from itertools import permutations

# possible consecutive numbers
L = [1234, 2345, 3456, 4567, 5678, 6789]

# there are 6 elements in the list L 
for c in range(0, 6):
    
    for p in permutations(L[c]):
      n = int(''.join(p))
      durum = True 
     # checking the divisibility
      for i in  range(2, 12):
          if n % i == 0:
              durum = False
              break
      if durum == True:
          print(n)

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from itertools import permutations

# possible consecutive numbers

L = [1234, 2345, 3456, 4567, 5678, 6789]

# there are 6 elements in the list L

for c in range(0, 6):

for p in permutations(L[c]):

n = int(''.join(p))

durum = True

# checking the divisibility

for i in range(2, 12):

if n % i == 0:

durum = False

break

if durum == True:

print(n)

Reply

Brian Gladman permalink

Hi Naim,

If we list the numbers according to their top digit we have:

1423,
2143, 2341, 2413, 2543
4231, 4253, 4321, 4523, 4567, 4657
5647, 5867, 6547, 6857,
8567

So if we knew that the top digit is 1 we would then know the number is 1423. Similarly, if the top digit is known to be 8 we would then know that the number is 8567.

But since, in this case, there are two answers, we can’t tell which is correct unless we know the top digit.

But if we now list the numbers with the same second digit, maybe there will be only one answer so we don;t need to know the digit. And we can build lists for the third and the fourth digits as well.

And, hopefully, one of these lists will have only one possible answer.

Reply

Naim Uygun permalink

Dear Brian,

For your explanations, 2013 times “Thank you”
and I wish a happy new year to you.

Reply

arthurvause permalink

Using the amended version of the question:


from itertools import permutations as perm
from fractions import gcd

candidates = [p for start in range(7)
              for p in perm(range(start, start+4))
              if gcd(sum([p[i]*10**(3-i) for i in range(4)]),2*3*5*7*11)==1]

for pos in range(4):
  solutions=[]  
  for i in range(10):
    s = [c for c in candidates if c[pos]==i]
    if len(s)==1:
      solutions.append(s)
  if len(solutions)==1:
    print solutions[0][0]

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from itertools import permutations as perm

from fractions import gcd

candidates = [p for start in range(7)

for p in perm(range(start, start+4))

if gcd(sum([p[i]*10**(3-i) for i in range(4)]),2*3*5*7*11)==1]

for pos in range(4):

solutions=[]

for i in range(10):

s = [c for c in candidates if c[pos]==i]

if len(s)==1:

solutions.append(s)

if len(solutions)==1:

print solutions[0][0]

Reply

dave permalink

It is possible to solve many of these ST problems without a computer program. I help students prepare for numerical aptitiude tests, no computer needed, just a mix of O & A level maths.

Reply

brian gladman permalink

Hi David,

You are right but this site is about computer solutions using Python. The site at https://sites.google.com/site/sundaytimesteasers/ is one where non-computer solutions are discussed.

Reply

Sunday Times Teaser 2624 – Is It 2013?

by Victor Bryant

Published January 6 2013 (link)

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