Sunday Times Teaser 2811 – Making Arrangements

by Graham Smithers

Published: 7 August 2016 (link)

Beth wrote down a three-figure number and she also listed the five other three-figure numbers that could be made using those same three digits. Then she added up the six numbers: it gave a total whose digits were all different, and none of those digits appeared in her original number.

If you knew whether her original number was prime or not, and you knew whether the sum of the three digits of her original number was prime or not, then it would be possible to work out her number.

What was it?

2 Comments Leave one →

Brian Gladman permalink


from itertools import permutations
from collections import defaultdict
from number_theory import is_prime

d = defaultdict(list)
for a, b, c in permutations(range(1, 10), 3):
  
  # form the number and its digit sum
  nbr, digit_sum = 100 * a + 10 * b + c, a + b + c
  
  # form the sum of the six numbers formed from digits a, b and c
  s = [int(x) for x in str(222 * digit_sum)]
  set_s = set(s)
  # its digits are all different and disjoint from a, b and c
  if len(s) == len(set_s) and not set_s.intersection([a ,b, c]):
      # now store the number in one of four lists depending on
      # whether the number and its digit sum are/are not prime
      d[(is_prime(nbr), is_prime(digit_sum))].append(nbr)

# now look one of the lists that uniquely identifies a number
for k, v in d.items():
  if len(v) == 1:
    print('Her number is {}.'.format(v[0]))

from itertools import permutations

from collections import defaultdict

from number_theory import is_prime

d = defaultdict(list)

for a, b, c in permutations(range(1, 10), 3):

# form the number and its digit sum

nbr, digit_sum = 100 * a + 10 * b + c, a + b + c

# form the sum of the six numbers formed from digits a, b and c

s = [int(x) for x in str(222 * digit_sum)]

set_s = set(s)

# its digits are all different and disjoint from a, b and c

if len(s) == len(set_s) and not set_s.intersection([a ,b, c]):

# now store the number in one of four lists depending on

# whether the number and its digit sum are/are not prime

d[(is_prime(nbr), is_prime(digit_sum))].append(nbr)

# now look one of the lists that uniquely identifies a number

for k, v in d.items():

if len(v) == 1:

print('Her number is {}.'.format(v[0]))

Jim Randell permalink

Here is my solution.


# if the number is ABC, then, in order for the different orderings
# of (A, B, C) to make six different numbers (A, B, C) must be distinct
# and non-zero.
#
#   ABC
#   ACB
#   BAC
#   BCA
#   CAB
# + CBA
# -----
#  PQRS
#
# so we have:
#
# PQRS = 222 * (A + B + C) 

from itertools import combinations, permutations
from collections import defaultdict
from enigma import irange, nsplit, nconcat, is_prime, printf

# record results by (is_prime(total), is_prime(number))
r = defaultdict(list)

# consider values for A, B, C
for (A, B, C) in combinations(irange(1, 9), 3):
  # the sum f the digits
  s = A + B + C
  # calculate PQRS
  (P, Q, R, S) = nsplit(222 * s)

  # check all digits in ABC, PQRS are distinct
  if len(set((A, B, C, P, Q, R, S))) != 7: continue

  printf("[ABC = {A}{B}{C}, s={s}, PQRS = {P}{Q}{R}{S}]")

  # now look for possible inital numbers
  ps = is_prime(s)
  for (X, Y, Z) in permutations((A, B, C)):
    n = nconcat(X, Y, Z)
    pn = is_prime(n)
    r[(ps, pn)].append((s, n))

    printf("[s={s} ({ps}), n={n} ({pn})]")

# look for unique solutions to the puzzle
for ((ps, pn), vs) in r.items():
  if len(vs) != 1: continue
  (s, n) = vs[0]
  printf("number = {n} ({pn}), sum = {s} ({ps})")

# if the number is ABC, then, in order for the different orderings

# of (A, B, C) to make six different numbers (A, B, C) must be distinct

# and non-zero.

# ABC

# ACB

# BAC

# BCA

# CAB

# + CBA

# -----

# PQRS

# so we have:

# PQRS = 222 * (A + B + C)

from itertools import combinations, permutations

from collections import defaultdict

from enigma import irange, nsplit, nconcat, is_prime, printf

# record results by (is_prime(total), is_prime(number))

r = defaultdict(list)

# consider values for A, B, C

for (A, B, C) in combinations(irange(1, 9), 3):

# the sum f the digits

s = A + B + C

# calculate PQRS

(P, Q, R, S) = nsplit(222 * s)

# check all digits in ABC, PQRS are distinct

if len(set((A, B, C, P, Q, R, S))) != 7: continue

printf("[ABC = {A}{B}{C}, s={s}, PQRS = {P}{Q}{R}{S}]")

# now look for possible inital numbers

ps = is_prime(s)

for (X, Y, Z) in permutations((A, B, C)):

n = nconcat(X, Y, Z)

pn = is_prime(n)

r[(ps, pn)].append((s, n))

printf("[s={s} ({ps}), n={n} ({pn})]")

# look for unique solutions to the puzzle

for ((ps, pn), vs) in r.items():

if len(vs) != 1: continue

(s, n) = vs[0]

printf("number = {n} ({pn}), sum = {s} ({ps})")

The enigma.py library is available at [ https://www.magwag.plus.com/jim/enigma.html ].

Sunday Times Teaser 2811 – Making Arrangements

by Graham Smithers

Published: 7 August 2016 (link)

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