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Jim Randell 5:53 pm on 1 May 2020 Permalink | Reply
Tags: by: Andrew Skidmore ( 89 )

Teaser 3006: Raffle tickets

From The Sunday Times, 3rd May 2020 [link] [link]

At our local bridge club dinner we were each given a raffle ticket. The tickets were numbered from 1 to 80. There were six people on our table and all our numbers were either prime or could be expressed as the product of non-repeating primes (e.g. 18 = 2×3×3 is not allowed). In writing down the six numbers you would use each of the digits 0 to 9 once only. If I told you the sum of the six numbers (a perfect power) you should be able to identify the numbers.

List the numbers (in ascending order).

[teaser3006]

Jim Randell 6:14 pm on 1 May 2020 Permalink | Reply

This Python program runs in 172ms.

Run: [ @repl.it ]

from enigma import irange, inf, is_duplicate, factor, seq_all_different as all_different, filter_unique, iroot, printf

# find numbers with different prime factors, and no repeated digits
ns = list()
for n in irange(1, 80):
  if is_duplicate(n): continue
  fs = factor(n)
  if fs and all_different(fs):
    ns.append(n)

# find sets of k numbers, that use the digits 0-9 exactly once
def generate(ns, k, ds=set(), s=[]):
  # are we done?
  if k == 0:
    if len(ds) == 10:
      yield tuple(s)
  else:
    # add in another number
    for (i, n) in enumerate(ns, start=1):
      t = str(n)
      if ds.intersection(t): continue
      yield from generate(ns[i:], k - 1, ds.union(t), s + [n])

# find 6-sets that are unique by sum
(ss, _) = filter_unique(generate(ns, 6), sum)

# find powers for n
def powers(n, p=1):
  for k in irange(p, inf):
    x = iroot(n, k)
    if x ** k == n:
      yield (x, k)
    if x == 1: break

# output solution
for s in ss:
  t = sum(s)
  ps = list(powers(t, 2))
  printf("{s} -> {t} {ps}")

Solution: The numbers are: 2, 3, 41, 58, 69, 70.

The sum is 243 (= 3^5).

We can find 78 different sets of 6 numbers that use the digits 0-9 exactly once, but only the set given above has a unique sum (which is the largest possible sum).

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GeoffR 11:25 am on 5 May 2020 Permalink | Reply

The programme found four solutions, but only the first is a correct unique solution
ie Tickets: 2, 3, 41, 58, 69, 70.

There is anpother duplicate solution with a total of 216, but the programme did not find it. Two other potential solutions also had a total of 225, so do not give a unique solution.

The programme would only run under the Chuffed solver in a reasonable time.


% A Solution in MiniZinc
include "globals.mzn";
 
var 0..9:A; var 0..9:B; var 0..9:C; var 0..9:D;
var 0..9:E; var 0..9:F; var 0..9:G; var 0..9:H;
var 0..9:I; var 0..9:J;

constraint all_different ([A,B,C,D,E,F,G,H,I,J]);

% total of six factors
var 100..400: total;

% perfect powers possible between 100 and 400
constraint total == pow(2,8) \/ total == pow(3,4) \/ total == pow(3,5) 
\/ total == pow(5,3) \/ total == pow(6,3) \/ total == pow(7,3)
\/ total == pow(12,2) \/ total == pow(13,2) \/ total = pow(14,2)
\/ total == pow(15,2) \/ total == pow(17,2) \/ total = pow(18,2);

% numbers are A, B, CD, EF, GH, IJ
total = A + B + CD + EF + GH + IJ;
var 11..80: CD = 10*C + D;
var 11..80: EF = 10*E + F;
var 11..80: GH = 10*G + H;
var 11..80: IJ = 10*I + J;

% Possible prime factors for CD, EF, GH and IJ
var 2..79: a1; var 2..79: a2; var 2..79: a3;
var 2..79: b1; var 2..79: b2; var 2..79: b3;
var 2..79: c1; var 2..79: c2; var 2..79: c3;
var 2..79: d1; var 2..79: d2; var 2..79: d3;

predicate is_prime(var int: x) = 
x > 1 /\ forall(i in 2..1 + ceil(sqrt(int2float(ub(x))))) 
((i < x) -> (x mod i > 0));

constraint is_prime (A) /\ is_prime(B);

% CD is prime or has two or three different prime factors
constraint (is_prime(CD))
\/ 
(a1 != a2 /\ CD == a1 * a2 /\ is_prime(a1) /\ is_prime(a2))
\/  
(a1 != a2 /\ a1 != a3 /\ a2 != a3 /\ CD == a1 * a2 * a3 /\ 
is_prime(a1) /\ is_prime(a2) /\ is_prime(a3));

% EF is prime or has two or three different prime factors
constraint (is_prime(EF))
\/ 
(b1 != b2 /\ EF == b1 * b2 /\ is_prime(b1) /\ is_prime(b2))
\/  
(b1 != b2 /\ b1 != b3 /\ b2 != b3 /\ EF == b1 * b2 * b3 /\ 
is_prime(b1) /\ is_prime(b2) /\ is_prime(b3));

% GH is prime or has two or three different prime factors
constraint (is_prime(GH))
\/ 
(c1 != c2 /\ GH == c1 * c2 /\ is_prime(c1) /\ is_prime(c2))
\/  
(c1 != c2 /\ c1 != c3 /\ c2 != c3 /\ GH == c1 * c2 * c3 /\ 
is_prime(c1) /\ is_prime(c2) /\ is_prime(c3));

% IJ is prime or has two or three different prime factors
constraint (is_prime(IJ))
\/ 
(d1 != d2 /\ IJ == d1 * d2 /\ is_prime(d1) /\ is_prime(d2))
\/  
(d1 != d2 /\ d1 != d3 /\ d2 != d3 /\ IJ == d1 * d2 * d3 /\ 
is_prime(d1) /\ is_prime(d2) /\ is_prime(d3));

constraint increasing([A ,B, CD, EF, GH, IJ]);

solve satisfy;

output ["Tickets: " ++ show(A) ++ ", " ++ show(B) ++ ", " 
++ show(CD) ++ ", " ++ show(EF) ++ ", " ++ show(GH) ++ ", "
++ show(IJ) ++ ", total = " ++ show(total) ];


% Tickets: 2, 3, 41, 58, 69, 70, total = 243  <<< Correct Solution
% time elapsed: 0.03 s
% ----------
% Tickets: 2, 3, 14, 58, 69, 70, total = 216 << A duplicate solution exists
% time elapsed: 0.03 s
% ----------
% Tickets: 2, 5, 38, 41, 69, 70, total = 225 - duplicate 
% time elapsed: 0.03 s
% ----------
% Tickets: 2, 5, 30, 41, 69, 78, total = 225 _ duplicate
% time elapsed: 0.03 s
% ----------
% ==========

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Jim Randell 11:56 am on 30 April 2020 Permalink | Reply
Tags: by: Andrew Skidmore ( 89 )
Teaser 2876: Bonfire toffee
From The Sunday Times, 5th November 2017 [link] [link]

In this subtraction sum I have consistently replaced digits with letters, different letters being used for different digits:

BONFIRE – TOFFEE = TREATS

What is the number of ENTRIES?

[teaser2876]
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- Jim Randell 11:56 am on 30 April 2020 Permalink | Reply
  We can solve this puzzle directly using the [[ SubstitutedExpression() ]] solver from the enigma.py library.
```
>>> python -m enigma SubstitutedExpression "TREATS + TOFFEE = BONFIRE" --answer="ENTRIES"
(TREATS + TOFFEE = BONFIRE) (ENTRIES)
(769470 + 758899 = 1528369) (9276390) / A=4 B=1 E=9 F=8 I=3 N=2 O=5 R=6 S=0 T=7
(769870 + 754499 = 1524369) (9276390) / A=8 B=1 E=9 F=4 I=3 N=2 O=5 R=6 S=0 T=7
ENTRIES = 9276390 [2 solutions]
```
  Solution: ENTRIES = 9276390.
  
  There are two possible sums, the values for A and F being interchangeable:
  
  769470 + 758899 = 1528369
  769870 + 754499 = 1524369
  
  LikeLike
Jim Randell 8:29 am on 28 April 2020 Permalink | Reply
Tags: by: Victor Bryant ( 148 )
Teaser 2593: Spell it out!
From The Sunday Times, 3rd June 2012 [link] [link]

I have written down a very large number (but with fewer than twenty-five digits). If you spell out each of its digits as a word, then for each digit its last letter is the same as the first letter of the next digit (the last letter of the last digit being the same as the first letter of the first digit). One example of such a number would be 83821.

Neither my number nor the sum of its digits is a palindromic number (but in fact the original number is one more than a palindrome).

What is my number?

[teaser2593]
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- Jim Randell 8:30 am on 28 April 2020 Permalink | Reply
  This Python program runs in 99ms.
  
  Run: [ @repl.it ]
```
from enigma import (int2words, irange, is_palindrome, nsplit, printf)

# digits as words
words = list(int2words(d) for d in irange(0, 9))

# possible next numbers
succ = dict()
for (d1, w1) in enumerate(words):
  succ[d1] = list(d2 for (d2, w2) in enumerate(words) if w1[-1] == w2[0])

# find numbers that form a loop according to relationship r
def solve(ns, k, r=succ):
  # check the numbers form a loop
  if ns[0] in r[ns[-1]]:
    yield ns
  # can we add another number
  if len(ns) < k:
    for n in r[ns[-1]]:
      yield from solve(ns + [n], k)

# consider possible initial digits
for n in succ.keys():
  # find loops with length less than 25
  for ns in solve([n], 24):
    # the number itself is 1 more than a palindrome
    if not (ns[0] + 1 == ns[-1] and is_palindrome(ns[1:-1])): continue
    # the sum of the digits is not a palindrome
    t = sum(ns)
    if is_palindrome(nsplit(t)): continue
    # output solution
    printf("{ns} -> {t}")
```
  Solution: The number is 183838383838383838382.
  
  The number is 21 digits long. And the digit sum is (8 + 3) × 9 + (1 + 8 + 2) = 110.
  
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Jim Randell 9:19 am on 26 April 2020 Permalink | Reply
Tags: by: Jonathan Welton
Brainteaser 1814: Clear heads
From The Sunday Times, 22nd June 1997 [link]

“I have affixed a card to each of your foreheads”, explained the professor of logic to his three most able students. All three were perfect logicians capable of instantly deducing the logical consequences of any statement.

He continued: “You can see the others’ cards, but none of you can see your own. On each card is written a positive whole number and one of the numbers is the sum of the other two”.

The professor turned to the first student, Albert: “From what you can see and what you have heard, can you deduce the number on your card?”. Albert could not.

The professor then asked Bertrand the same question. He could not deduce the number on his card.

Likewise, when asked next, Charles could not deduce the number on his card.

The professor then asked Albert the same question again and this time Albert was able to deduce that the number on his card was 50.

What number was on Bertrand’s card, and what number was on Charles’s card?

This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

[teaser1814]
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- Jim Randell 9:20 am on 26 April 2020 Permalink | Reply
  Each person can see two numbers, and knows their own number must be either the sum or difference of these two numbers.
  
  So when A is asked, the only way he would be able to narrow this choice down to a single number is if B and C’s numbers are the same, as the difference would be zero, and this is not allowed, so he would know his own number was the sum of B and C’s numbers. So A will be able to declare at the first question only if the set numbers is of the form (2n, n, n).
  
  Similarly, when B is asked, he will be able to declare if the set of numbers is of the form (n, 2n, n).
  
  But we also know that if B sees (2n, ?, n) that his own number must be 3n or n, but it cannot be n, as A would have declared if the numbers were (2n, n, n). So B can also declare if the sets of numbers are of the form (2n, 3n, n).
  
  When C is asked, he can declare if the numbers of the form: (n, n, 2n), (2n, n, 3n), (n, 2n, 3n), (2n, 3n, 5n).
  
  A is then asked again, and this time declares. So the numbers must fit one of the following patterns:
  
  (3n, 2n, n)
  (4n, 3n, n)
  (3n, n, 2n)
  (4n, n, 3n)
  (5n, 2n, 3n)
  (8n, 3n, 5n)
  
  Of these patterns, only one gives integer values when the first element is 50:
  
  (5n, 2n, 3n) = (50, 20, 30)
  
  Solution: Bertrand’s number was 20. Charles’s number was 30.
  
  Here is a Python program that generates the appropriate patterns for each question, and then determines what sequences of numbers match the patterns for the 4th question. It runs in 73ms.
  
  Run: [ @repl.it ]
```
from enigma import irange, inf, flatten, div, printf

# update pattern p, element i becomes the sum of the other elements
def update(p, i):
  p = list(p)
  p[i] = sum(p) - p[i]
  return p

# produce patterns that lead to declarations for A, B, C, A, B, C, ...
def declare():
  ds = list()
  for i in irange(0, inf):
    ps = list()
    # initial seeds
    if i < 3: ps.append(update([1, 1, 1], i))
    # add in alternatives from the previous rounds
    j = i % 3
    for p in flatten(ds):
      ps.append(update(p, j))
    yield ps
    # keep the previous 2 rounds
    ds.append(ps)
    if len(ds) > 2: ds.pop(0)

# find declaration patterns that correspond to the 4th question
for (i, ps) in enumerate(declare(), start=1):
  printf("[{i} -> {ps}]")
  if i == 4: break

# given A calculate B and C from the patterns
A = 50
for (a, b, c) in ps:
  (B, C) = (div(b * A, a), div(c * A, a))
  if B and C:
    printf("A={A} B={B} C={C}")
```
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Jim Randell 5:07 pm on 24 April 2020 Permalink | Reply
Tags: by: Stephen Hogg ( 65 )
Teaser 3005: Tubular bales
From The Sunday Times, 26th April 2020 [link] [link]

Ten equal-length, rigid tubes, each a different prime-valued external radius from 11 mm to 43 mm, were baled, broadside, by placing the 43 mm and 11 mm tube together and the third tube, not the largest remaining, touching both of these. Each subsequent tube touched the previous tube placed and the 43 mm tube. A sub-millimetre gap between the final tube placed and the 11 mm tube, made a near perfect fit.

The radius sum of the first three tubes placed against the 43 mm tube was a multiple of one of the summed radii. Curiously, that statement remains true when each of “four”, “five”, “seven” and “eight” replaces “three”. For “two” and “six” tubes placed their radius sum was a multiple of an as yet unplaced tube’s radius.

What radius tube, in mm, was placed last?

[teaser3005]
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- Jim Randell 5:30 pm on 24 April 2020 Permalink | Reply
  This Python program runs in 78ms.
```
from enigma import (primes, printf)

# numbers available to to use
rs = list(primes.between(11, 44))
assert len(rs) == 10

# check n is a multiple of some element of ms
check = lambda n, ms: any(n % m == 0 for m in ms)

# solve for the specified numbers
# ns = unused numbers
# ss = used numbers
def solve(ns, ss):
  # are we done?
  if not ns:
    yield ss
  else:
    # what number placement is this?
    k = len(ss) + 1
    t = sum(ss)
    # choose the next number to use
    for (i, n) in enumerate(ns):
      # 2nd: not the largest available
      if k == 2 and n == ns[-1]: continue
      t_ = t + n
      # 3rd, 4th, 5th, 7th, 8th: multiple of a used number
      ss_ = ss + [n]
      if k in (3, 4, 5, 7, 8) and not check(t_, ss_): continue
      # 2nd, 6th: multiple of an unused number
      ns_ = ns[:i] + ns[i + 1:]
      if k in (2, 6) and not check(t_, ns_): continue
      # solve for the remaining numbers
      yield from solve(ns_, ss_)

# collect possible final numbers
fs = set()
for ss in solve(rs[1:-1], rs[:1]):
  fs.add(ss[-1])
  printf("{ss}")

# output solution
printf("final number = {fs}")
```
  Solution: The 37 mm tube was placed last.
  
  The conditions placed on the ordering of the numbers means that although there are four possible orders that the tubes could be assembled in, the answer to the puzzle (the radius of the final tube) is the same in all cases.
  
  So it is not necessary to check that the placement results in a gap of less than 1 mm, but if you do, you find all 4 orderings result in a gap less than 1 mm (and one of them is an almost perfect fit).
  
  Here is a diagram of the packing (11, 23, 17, 41, 29, 31, 19, 13, 37) around the 43 tube. This has the largest gap of 0.66 mm.
  
  Here is a list of 4 packings, and the corresponding gaps:
  
  (11, 23, 17, 41, 29, 31, 13, 19, 37) → gap = 0.36 mm
  (11, 23, 17, 41, 29, 31, 19, 13, 37) → gap = 0.66 mm
  (11, 23, 17, 41, 31, 29, 13, 19, 37) → gap = 0.000055 mm
  (11, 23, 17, 41, 31, 29, 19, 13, 37) → gap = 0.41 mm
  
  The third one has a gap of just 55 nm, which is about half the diameter of a single coronavirus particle, and probably quite a lot smaller than the tolerance of the tubes.
  
  Perhaps the puzzle was intended to be set with the radii of the tubes measured in centimetres, then there would be only one arrangement that had a sub-millimetre gap.
  
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Jim Randell 9:24 am on 23 April 2020 Permalink | Reply
Tags: by: Danny Roth ( 89 )

Teaser 2591: Shilly Chalet

From The Sunday Times, 20th May 2012 [link] [link]

George and Martha are running a holiday camp and their four daughters are staying there. To keep the peace they have been given chalets in different areas. Amelia’s chalet number is between 1 and 99, Bertha’s is between 100 and 199, Caroline’s between 200 and 299, and Dorothy’s between 300 and 399.

George commented that the difference between any two of the four chalet numbers is either a square or a cube. Martha added that the same could be said for the sum of the chalet numbers of the three youngest daughters.

Who is the eldest daughter and what is her chalet number?

[teaser2591]

Jim Randell 9:25 am on 23 April 2020 Permalink | Reply

Programatically it is straightforward to check all the possibilities.

This Python program runs in 141ms.

Run: [ @repl.it ]

from enigma import (is_square, is_cube, irange, cached, printf)

# check for a square or a cube
check = cached(lambda n: is_square(n) or is_cube(n))

# choose a value for A
for A in irange(1, 99):
  # choose a value for B
  for B in irange(100, 199):
    if not check(B - A): continue
    # choose a value for C
    for C in irange(200, 299):
      if not (check(C - A) and check(C - B)): continue
      # choose a value for D
      for D in irange(300, 399):
        if not (check(D - A) and check(D - B) and check(D - C)): continue
        # check the total excluding the eldest
        T = A + B + C + D
        for (n, x) in zip("ABCD", (A, B, C, D)):
          if not check(T - x): continue
          # output solution
          printf("A={A} B={B} C={C} D={D}; eldest={n}")

Solution: Amelia is the eldest daughter. Her chalet is number 93.

The chalet numbers are:

A=93, B=193, C=218, D=318

And: B + C + D = 729 = 27^2 = 9^3.

This sum is both a square and a cube.

So, if you interpret “either … or …” to be an exclusive or (one or the other, but not both), then the puzzle has no solutions.

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GeoffR 10:45 am on 24 April 2020 Permalink | Reply


% A Solution in MiniZinc
include "globals.mzn";
 
var 1..99:A;       % Amelia
var 100..199:B;    % Bertha
var 200..299:C;    % Caroline
var 300..399:D;    % Dorothy

set of int: sq = {n * n | n in 2..32};  
set of int: cb = {n * n * n | n in 2..10};

% difference between any chalet numbers is either a square or a cube
constraint (B-A) in sq \/ (B-A) in cb;
constraint (C-A) in sq \/ (C-A) in cb;
constraint (C-B) in sq \/ (C-B) in cb;
constraint (D-C) in sq \/ (D-C) in cb;
constraint (D-B) in sq \/ (D-B) in cb;
constraint (D-A) in sq \/ (D-A) in cb;

% sum of three youngest daughters chalet numbers is a square or a cube
constraint ((A+B+C) in sq \/ (A+B+C) in cb)
\/  ((A+B+D) in sq \/ (A+B+D) in cb)
\/  ((B+C+D) in sq \/ (B+C+D) in cb);

solve satisfy;

% A = 93;
% B = 193;
% C = 218;
% D = 318;
% % time elapsed: 0.06 s
% ----------
% ==========
% sum of three youngest daughters chalets is a square or a cube
% A + B + C = 93 + 193 + 218 = 504 - not a square or a cube
% A + B + D = 93 + 193 + 318 = 604 - not a square or a cube
% B + C + D = 193 + 218 + 318 = 729 - is both a square and a cube

% Therefore, Ameila is the eldest daughter with chalet number 93

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GeoffR 11:46 am on 24 April 2020 Permalink | Reply

sq = [x * x for x in range(2,32)]
cb = [x * x * x for x in range(2,10)]

# form a set of square and cube numbers
sc = set(sq) | set(cb)

for a in range(1,100):
  for b in range(100,200):
    if (b-a) in sc:
      for c in range(200,300):
        if (c-a) in sc and (c-b) in sc:
          for d in range(300,400):
            if (d-c) in sc and (d-b) in sc and (d-a) in sc:
              if (a+b+d) in sc:
                print('Eldest daughter Caroline has chalet', c)
              if (b+c+d) in sc:
                print('Eldest daughter Ameila has chalet', a)
              if (a+c+d) in sc:
                  print('Eldest daughter Bertha has chalet', b)
              if (a+b+c) in sc:
                  print('Eldest daughter Dorothy has chalet', d)

# Eldest daughter Ameila has chalet 93

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Jim Randell 8:21 am on 21 April 2020 Permalink | Reply
Tags: by: Brian Young
Brain-Teaser 518: [Prophecies]
From The Sunday Times, 16th May 1971 [link]

An interesting fragment tells us what took place at the meeting of the Five Prophets:

“If Hosea hath prophesied truth, Micah and Obadiah will die in the same city; if Micah hath prophesied truth, Joel and Obadiah will die in the same city. It is the saying of Joel that three of the five prophets shall die in Babylon; and Nahum declareth to Hosea that the two of them shall die in different cities.”

It is well known that those who prophesy correctly die in Jerusalem, whereas those who make false prophecies die in Babylon.

So how many of these five will die in Jerusalem?

This puzzle was originally published with no title.

[teaser518]
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John Crabtree and Jim Randell are discussing. Toggle Comments
- Jim Randell 8:22 am on 21 April 2020 Permalink | Reply
  There are only 2^5 (= 32) possibilities, so we can examine them all.
  
  This Python program runs in 79ms.
```
from enigma import subsets, printf

# check statement made by X
check = lambda X, s: (X == "J") == bool(s)

# consider where each will die
for (H, J, M, N, O) in subsets("BJ", size=5, select="M"):

  # how many die in J?
  n = (H, J, M, N, O).count('J')

  # "H: M and O will die in the same city"
  if not check(H, M == O): continue

  # "M: J and O will die in the same city"
  if not check(M, J == O): continue

  # "J: [exactly] three of the five prophets shall die in B"
  if not check(J, n == 2): continue

  # "N: N and H shall die in different cities"
  if not check(N, N != H): continue

  # output solution
  printf("n={n} [H={H} J={J} M={M} N={N} O={O}]")
```
  Solution: One of them will die in Jerusalem.
  
  There are two solutions:
  
  H: false, B
  J: false, B
  M: false, B
  N: false, B
  O: [true], J
  
  H: false, B
  J: false, B
  M: true, J
  N: false, B
  O: [false], B
  
  We are not given a prophesy by O, to verify whether it is true or not, but we can infer where he must die in order for the prophecies we are given to be consistent, and from that whether he makes prophecies that are true or false.
  
  In the first case he must die in Jerusalem, so must make true prophecies. And in the second case he must die in Babylon, and so must make false prophecies.
  
  LikeLike
  - John Crabtree 8:18 pm on 22 April 2020 Permalink | Reply
    
    Let the prophesies be #1, #2, #3 and #4 by H, M, J and N respectively.
    From #4, H dies in Babylon.
    Then from #1, one, and only one, of M and O die in Babylon.
    Then from #2, J dies in Babylon.
    Then from #3, four prophets, including N, die in Babylon.
    And so one prophet (Micah or Obadiah) dies in Jerusalem.
    
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Jim Randell 4:27 pm on 17 April 2020 Permalink | Reply
Tags: by: John Owen ( 34 )
Teaser 3004: Going up
From The Sunday Times, 19th April 2020 [link] [link]

In our football league, the teams all play each other once, with three points for a win and one for a draw. At the end of the season, the two teams with most points are promoted, goal difference being used to separate teams with the same number of points.

Last season’s climax was exciting. With just two games left for each team, there were several teams tied at the top of the league with the same number of points. One of them, but only one, could be certain of promotion if they won their two games. If there had been any more teams on the same number of points, then none could have guaranteed promotion with two wins.

How many teams were tied at the top of the league, and how many of the remaining matches involved any of those teams?

[teaser3004]
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- Jim Randell 9:33 am on 18 April 2020 Permalink | Reply
  I supposed there were several teams, A, B, C, …, tied at the top. And we are looking for situations where team A is looking at a guaranteed promotion, if they win both their games.
  
  Obviously any of the other teams tied at the top of the table would also be in with a chance of promotion if they win both their games (as they will have the same maximum number of points as A).
  
  But if one of the teams were to win a match by an enormous number of goals, it would give them an unassailable goal difference. So A can only be guaranteed a win if there are fewer than three teams tied at the top after the games are played (so the goal difference rule doesn’t come into it).
  
  So we need to look at numbers of teams, such that there is an arrangement of remaining matches, where A (and only A) is guaranteed a promotion if they win both their matches, but if there were one more team then there would be no such arrangement of matches.
  
  This Python program is a bit slow (it takes 8.9s), but it does find what seems to be a reasonable answer. I may see if I can come up with a more efficient program later.
```
from enigma import (cproduct, multiset, irange, join, printf)

# generate matches for teams <ts> tied at the top
# ts = teams tied at the top
# ss = number of matches to be played by teams in <ts>
# team Z is used to denote any other team not in <ts>
def matches(ts, ss, ms=[]):
  # are we done?
  if not ss:
    yield ms
  else:
    # choose a team
    ks = sorted(ss.keys())
    t = ks.pop(0)
    # choose an opponent
    ks.append('Z')
    for x in ks:
      m = (t, x)
      if x == 'Z' or not (ms and m < ms[-1]):
        yield from matches(ts, ss.difference(m), ms + [m])

# find teams that can be guaranteed promotion
# ts = teams tied at the top
# ms = remaining matches
def solve(ts, ms):
  # find wins where 2 wins guarantees A a place in the top 2
  (inc, exc) = (set(), set())
  # choose winning teams for each match
  for wins in cproduct(ms):
    # collect teams with 2 wins
    m = multiset.from_seq(wins)
    ws = list(t for (t, n) in m.items() if n == 2 and t != 'Z')
    if len(ws) < 3:
      # a guaranteed win
      inc.update(ws)
    else:
      # not a guaranteed win
      exc.update(ws)
      if exc.issuperset(ts): return set()
  # return those teams that are guaranteed a win in all situations
  return inc.difference(exc)

# format a sequence of matches
fmt = lambda ms: join((x + "v" + y for (x, y) in ms), sep=", ", enc="()")

# consider teams labelled A, B, C, ... (Z is used to denote "other" teams)
# record teams where there is a set of matches that guarantees only team A a promotion
r = dict()
for n in irange(2, 25):
  teams = "ABCDEFGHIJKLMNOPQRSTUVWXY"[:n]
  ss = multiset.from_seq(teams * 2)
  for ms in matches(teams, ss):
    # does this set of matches guarantee a promotion for only A?
    ws = solve(teams, ms)
    if len(ws) == 1 and 'A' in ws:
      printf("n={n}: {ms} -> {ws}", ms=fmt(ms), ws=join(sorted(ws)))
      r[n] = ms
      break # we only need one set of matches
  if n not in r:
    printf("n={n}: <none>")
    # are we done?
    k = n - 1
    if k in r:
      ms = r[k]
      printf("{k} -> {ms}; {m} matches", m=len(ms), ms=fmt(ms))
      break
```
  Solution: There are 6 teams tied at the top. 7 of the remaining matches involve at least one of those teams.
  
  For six teams at the top:
  
  If A plays B and C and wins both matches, then neither B nor C can achieve the maximum number of points, so they are out of contention.
  
  And there are three more teams at the top, D, E, F, and they all play each other, then only one of them can achieve 2 wins, to tie with A at the top of the table.
  
  So A is guaranteed to finish in the top 2 if they win both their games, and get promotion.
  
  Introducing a seventh team, G, would mean that a third team could get two wins, so A‘s promotion would not be guaranteed.
  
  LikeLike
Jim Randell 12:02 pm on 16 April 2020 Permalink | Reply
Tags: by: Victor Bryant ( 148 )
Brainteaser 1808: Next choice
From The Sunday Times, 11th May 1997 [link]

Much nonsense is written about the lottery. It is true that there are about 14 million different choices of six numbers from 1 to 49, but often surprise is expressed when the winning six include at least two consecutive numbers. In common with a lot of people, when I choose my numbers I make sure that they are well spread out, thus fitting in with the usual conception that it is unlikely that consecutive numbers will be drawn.

This set me thinking about what proportion of the 14 million possible choices of 6 numbers actually contained no two consecutive numbers. The answer is surprising and neat. So, in percentages, to the nearest whole number, what proportion of the choices include no consecutive numbers?

This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

[teaser1808]
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- Jim Randell 12:02 pm on 16 April 2020 Permalink | Reply
  
  We have already tackled a very similar puzzle with Enigma 1029 (also set by Victor Bryant under the pseudonym Susan Denham).
  
  The number of ways to choose 6 numbers from 49 is:
  
  C(49, 6) = 49! / (6! (49 − 6)!) = 13,983,816
  
  The number of ways to choose 6 non-consecutive numbers from 49 is:
  
  C(49 − 5, 6) = 44! / (6! (44 − 6)!) = 7,059,052
  
  This is approximately 50.48% of the total number of choices. So there are slightly more non-consecutive choices than consecutive choices.
  
  Solution: Approximately 50% of choices are non-consecutive.
  
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Jim Randell 8:49 am on 14 April 2020 Permalink | Reply
Tags: by: Des MacHale ( 12 )
Teaser 2650: Squares of oblongs
From The Sunday Times, 7th July 2013 [link] [link]

I gave to each of Ruby and Annie a set of “Oblongs”. Each set consisted of nine pieces of card of sizes 1×2, 2×3, 3×4, 4×5, 5×6, 6×7, 7×8, 8×9 and 9×10. The idea was to use some or all of the cards to make various shapes in jigsaw fashion. I asked them to make a square using some of their own pieces. Ruby made the smallest square possible with her set and Annie made the largest square possible with hers. Then I collected up all the unused pieces of card.

What (in increasing order) were the sizes of these unused pieces?

[teaser2650]
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- Jim Randell 8:49 am on 14 April 2020 Permalink | Reply
  See also: Enigma 10, Enigma 1491, Enigma 1251.
  
  I packaged up the rectangle packer I wrote for Enigma 1491 into a separate module: rectpack.py.
  
  This Python program looks for subsets of the tiles that have an area that can be made into a square, and then tries to pack them into the square. It runs in 110ms.
  
  Run: [ @replit ]
```
from rectpack import pack
from enigma import (irange, subsets, is_square, Accumulator, diff, printf)

# available tiles
tiles = list((i, i + 1) for i in irange(1, 9))

# find the smallest and largest squares
rs = [ Accumulator(fn=min), Accumulator(fn=max) ]

# choose subsets of the tiles
for ts in subsets(tiles, min_size=2):
  # is the area a square number
  t = sum(a * b for (a, b) in ts)
  x = is_square(t)
  if x is None: continue

  # attempt to fit the rectangles into the square
  for ps in pack(x, x, ts):
    printf("{t} = {x}^2 -> {ts}")
    printf("-> {ps}")
    for r in rs: r.accumulate_data(x, ts)
    break  # only need one packing

# output solution
printf("min = {rs[0].value}^2, {rs[0].data}")
printf("max = {rs[1].value}^2, {rs[1].data}")

# unused tiles
us = diff(tiles, rs[0].data) + diff(tiles, rs[1].data)
printf("unused = {us}", us=sorted(us))
```
  Solution: The unused pieces are: 1×2, 2×3, 8×9.
  
  The smallest square that can be made is a 16×16 square using all the pieces except 1×2 and 8× 9.
  
  The largest square that can be made is a 18×18 square using all the pieces except 2×3.
  
  These are in fact the only two sets of pieces that can be assembled into squares.
  
  LikeLike

Jim Randell 11:43 am on 12 April 2020 Permalink | Reply
Tags: by: R Postill ( 20 )

Brain-Teaser 517: [Rugby table]

From The Sunday Times, 9th May 1971 [link]

The final table of the inter-zone Rugby championships of Pongoland reads:

Each tribe used only one goal-kicker; C’s kicked most points and A’s least. C had only one penalty scored against them.

The only types of score were:

try converted to goal (5 points)
try unconverted (3 points)
penalty goal (3 points)

How were the two sides’ scores made up in the match Aboma vs. Bwaga?

This puzzle was originally published with no title.

[teaser517]

Jim Randell 11:44 am on 12 April 2020 Permalink | Reply

It seems that the points scored by the kicker are: 2 points for a conversion; 0 points for a try; 3 points for a penalty goal.

I wrote a MiniZinc model to solve the problem:

%#! minizinc -a

% xYZ = number of score type x (= c (5), t (3), p (3)) by Y against Z

var 0..2: cAB;
var 0..4: tAB;
var 0..4: pAB;
var 0..2: cBA;
var 0..4: tBA;
var 0..4: pBA;

var 0..3: cAC;
var 0..5: tAC;
var 0..5: pAC;
var 0..1: cCA;
var 0..3: tCA;
var 0..3: pCA;

var 0..2: cBC;
var 0..4: tBC;
var 0..4: pBC;
var 0..1: cCB;
var 0..3: tCB;
var 0..3: pCB;

% total points
function var int: points(var int: c, var int: t, var int: p) = 5 * c + 3 * t + 3 * p;

% kicked points
function var int: kicks(var int: c, var int: t, var int: p) = 2 * c + 3 * p;

% point values in the table
constraint points(cAB, tAB, pAB) + points(cAC, tAC, pAC) = 18;
constraint points(cBA, tBA, pBA) + points(cCA, tCA, pCA) = 12;

constraint points(cBA, tBA, pBA) + points(cBC, tBC, pBC) = 14;
constraint points(cAB, tAB, pAB) + points(cCB, tCB, pCB) = 13;

constraint points(cCA, tCA, pCA) + points(cCB, tCB, pCB) = 9;
constraint points(cAC, tAC, pAC) + points(cBC, tBC, pBC) = 16;

% A won both their games, C lost both their games
constraint points(cAB, tAB, pAB) > points(cBA, tBA, pBA); % A beat B
constraint points(cAC, tAC, pAC) > points(cCA, tCA, pCA); % A beat C
constraint points(cBC, tBC, pBC) > points(cCB, tCB, pCB); % B beat C

% kicked points: C > B > A
constraint kicks(cAB, tAB, pAB) + kicks(cAC, tAC, pAC) < kicks(cBA, tBA, pBA) + kicks(cBC, tBC, pBC);
constraint kicks(cBA, tBA, pBA) + kicks(cBC, tBC, pBC) < kicks(cCA, tCA, pCA) + kicks(cCB, tCB, pCB);

% C had 1 penalty scored against them
constraint pAC + pBC = 1;

solve satisfy;

Solution: Aboma scored 2 conversions (for 10 points). Bwaga scored 1 try and 1 penalty (for 6 points)

The results are:

A vs. B: 10 pts (2c) vs. 6 pts (t + p)
A vs. C: 8 pts (c + t) vs. 6 pts (2p)
B vs. C: 8 pts (c + p) vs. 3 pts (p)

A’s kicker made 6 points.
B’s kicker made 8 points.
C’s kicker made 9 points.

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Jim Randell 5:27 pm on 9 April 2020 Permalink | Reply
Tags: by: Nick MacKinnon ( 60 )
Teaser 3003: All that glitters
From The Sunday Times, 12th April 2020 [link] [link]

My aunt has a collection of sovereigns, and she set me a challenge:

“You can have the coins if you can work out the dates, which (in increasing order) are equally spaced and all in the 20th century. The number of coins is an odd prime. The highest common factor of each pair of dates is an odd prime. The sum of the number of factors of each of the dates (including 1 and the date itself) is an odd prime.”

I worked out the dates, though the gift was much less valuable than I’d hoped.

What were the dates?

[teaser3003]
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- Jim Randell 5:46 pm on 9 April 2020 Permalink | Reply
  I assumed the dates we are looking for are the years in the 20th century for each coin.
  
  This Python program runs in 93ms.
```
from enigma import (primes, subsets, irange, inf, gcd, tau, printf)

primes.extend(100)  # primes up to 100

# check a number is an odd prime
check = lambda n: n > 2 and n in primes

# choose years for the first 2 coins
for (a, b) in subsets(irange(1901, 1999), size=2):
  if not check(gcd(a, b)): continue

  # consider a sequence with n terms
  d = b - a
  for n in primes.irange(3, inf):
    z = a + (n - 1) * d
    if z > 2000: break
    s = list(irange(a, z, step=d))

    # gcd of each pair is an odd prime
    if not all(check(gcd(x, y)) for (x, y) in subsets(s, size=2)): break

    # sum of the number of divisors of each year is an odd prime
    if not check(sum(tau(x) for x in s)): continue

    # output solution
    printf("a={a} b={b}, d={d} -> n={n}: {s}")
```
  Solution: The dates of the coins were: 1903, 1936, 1969.
  
  Manually (as suggested by Robert):
  
  Most number have divisors that come in pairs, so have an even number of divisors. The exception is the square numbers, which have an odd number of divisors (see: Puzzle #08).
  
  So, in order for the sum of the divisors of the dates to be odd, the list of dates must include an odd number of square numbers. And in the range 1901 – 2000 there is only one square number, 1936. So that must be one of the dates.
  
  1936 factorises as: (2^4)(11^2), so the other dates must have a GCD with 1936 of 11.
  
  For numbers less than 1936, we get: 1925, 1903. For numbers greater than 1936 we get: 1947, 1969, 1991.
  
  Looking for arithmetic sequences containing 1936, with a number of elements that is an odd prime we get:
  
  d=11: (1925, 1936, 1947); divisor sum = 35
  d=33: (1903, 1936, 1969); divisor sum = 23
  
  Only the second of these has a divisor sum that is prime, and gcd(1903, 1969) = 11 so this satisfies all the required conditions and gives the answer.
  
  LikeLike
- Robert Brown 9:08 pm on 9 April 2020 Permalink | Reply
  
  The only numbers with odd numbers of factors are perfect squares. There is only one of these in the 20th century, and that date has only has one factor >1 that’s an odd prime. Quite easy to find the answer by inspection.
  
  LikeLike
  - Jim Randell 7:53 am on 10 April 2020 Permalink | Reply
    
    @Robert: That’s a neat insight which quickly leads to a manual solution.
    
    LikeLike
Jim Randell 12:29 pm on 9 April 2020 Permalink | Reply
Tags: by: Stephen Hogg ( 65 )
Teaser 2862: Algebria’s standard
From The Sunday Times, 30th July 2017 [link] [link]

The Algebrian rectangular flag is highly symbolic. Each of its sides is an even number of inches long and a diagonal divides it into two triangles, one blue and one green, representing its two founding tribes. The length of the diagonal (in inches) is the number of states in Algebria, and the area of the flag (in square inches) is the twentieth century year in which the country obtained independence.

How many states are there in Algebria, and in which year did the country obtain independence?

[teaser2862]
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- Jim Randell 12:30 pm on 9 April 2020 Permalink | Reply
  If the dimensions of the flag are X and Y (and the diagonal is Z), then the area XY is between 1901 and 2000.
  
  X and Y are even (say X = 2x, Y = 2y), which limits xy to be between 476 and 500.
  
  Originally I looked at the possible areas to find viable x, y, z values, but the following Python program uses the [[ pythagorean_triples() ]] generator from the enigma.py library to find x, y, z values that can be scaled up to give an appropriate flag. It runs in 76ms.
```
from enigma import (pythagorean_triples, irange, inf, printf)

# consider primitive pythagorean triples
for (x, y, z) in pythagorean_triples(501, primitive=1):
  # and scale up to an appropriate size
  for k in irange(2, inf, step=2):
    (X, Y, Z) = (k * x, k * y, k * z)
    A = X * Y
    if A < 1901: continue
    if A > 2000: break
    printf("X={X} Y={Y} Z={Z} A={A} [x={x} y={y} z={z} k={k}]")
```
  Solution: There are 68 states in Algebria. It gained independence in 1920.
  
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Jim Randell 3:34 pm on 7 April 2020 Permalink | Reply
Tags: by: WT Blunt
Brainteaser 1805: Find the deuce
From The Sunday Times, 20th April 1997 [link]

Adam, Bill, Charlie and Dave play a game using the ace (counting as 1) and 2 to 10 of spades. These cards are shuffled and placed face down on the table. Each player in turn draws one card and declares whether it is odd or even. Each then in turn draws a further card and must declare a number being either the sum or the product of the two cards held. The winner is the first to deduce which cards remain on the table. They are all perfect logicians, draw every possible inference, and rely on the fact that the others do too.

In a recent game the calls went as follows:

Adam: “odd”
Bill: “odd”
Charlie: “odd”
Dave: “odd”
Adam: “6”
Bill: “12”
Charlie: “15”

But as Dave reached for his second card, a player claimed the game.

Who was the winner and where was the 2?

This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

[teaser1805]
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- Jim Randell 3:34 pm on 7 April 2020 Permalink | Reply
  They could just have easily played the game with 10 cards numbered 1 to 10.
  
  There are 5 odd cards (1, 3, 5, 7, 9) and only four players, so no one can win from the first round.
  
  This Python program collects the possible sequences of cards, and then works out it any of the players can determine the cards in the sequence given that they know their own cards. (If you can determine the cards in the sequence, you know which cards remain on the table). And then repeats this until the set of possible sequences remains unchanged (so all deductions have been made). It then moves on to the next play described in the text, with no-one winning until after C has drawn his second card. D is about to draw his second card when someone else claims victory.
  
  This Python program runs in 88ms.
  
  Run: [ @repl.it ]
```
from enigma import irange, subsets, filter_unique, ordered, intersect, printf

# cards
cards = set(irange(1, 10))

# collect hand for player k from a sequence of cards
def hand(s, k, n=4):
  return ordered(*(s[i] for i in irange(k, len(s) - 1, step=n)))

# find wins from sequences ss
def win(ss):
  r = dict()
  us = list()
  # can player k win?
  for (i, k) in enumerate("ABCD"):
    (rk, uk) = filter_unique(ss, (lambda s: hand(s, i)), (lambda s: ordered(*s)))
    if rk: r[k] = rk
    us.append(uk)
  # return wins (dict), and sequences that do not give a win (set)
  return (r, intersect(us))

# analyse sequences ss
def analyse(ss):
  n = len(ss)
  while True:
    (r, ss) = win(ss)
    yield (r, ss)
    m = len(ss)
    if m == n: break
    n = m

# play a card for player k, who delares d
def play(ss, k, d):
  for s in ss:
    for x in cards.difference(s):
      if x + s[k] == d or x * s[k] == d:
        yield s + (x,)

# the first four plays are all odd cards
ss = list(subsets((1, 3, 5, 7, 9), size=4, select="P"))

# no-one wins
for (r, ss) in analyse(ss): pass

# A (player 0) declares 6
ss = list(play(ss, 0, 6))

# no-one wins
for (r, ss) in analyse(ss): pass

# B (player 1) declares 12
ss = list(play(ss, 1, 12))

# no-one wins
for (r, ss) in analyse(ss): pass

# C (player 2) declares 15
ss = list(play(ss, 2, 15))

# someone (other than D) wins
for (r, ss) in analyse(ss):
  if r and 'D' not in r:
    for k in sorted(r.keys()):
      for v in r[k]:
        printf("{k} wins: {v}")
    break
```
  Solution: Charlie won. The 2 remained on the table.
  
  I agree with the published answer, but not with the full solution given at the back of the book.
  
  My solution is that C claims victory and the final situation when C wins is:
  
  A: 1 then 6
  B: 3 then 4
  C: 5 then 10 (or: 7 then 8)
  D: 9
  table: 2, 7, 8 (or: 2, 5, 10)
  
  It makes sense that C is the one who can claim victory as he could be holding (5, 10) or (7, 8), and only C knows for sure.
  
  The solution given in the book [ link ] says that the final situation when C wins is:
  
  A: 1 then 5 (or: 5 then 1)
  B: 3 then 4
  C: 9 then 6
  D: 7
  table: 2, 8 [and, presumably, 10]
  
  However the first four cards are odd, and if A drew the last remaining odd card, then he would know that the cards that remained on the table at this point were the 5 even cards and would claim victory.
  
  So this cannot be the situation (and given that A did not claim victory as soon as he drew his second card everyone would know that A cannot have 1 and 5 (so must have 3 and 2, or 1 and 6)).
  
  LikeLike
  - John Crabtree 6:16 am on 10 April 2020 Permalink | Reply
    
    D does not win. Summarizing the successive elimination:
    After A’s 2nd call, if A = 1 + 5, A wins. Everybody knows this. This eliminates (a) and (b).
    After B’s 2nd call, if B = 3 + 9, B wins. Everybody knows this. This eliminates (c) and (d).
    After C’s 2nd call, if D = 1, 5 or 7, D wins. This eliminates (e), (g) and (i).
    This leaves two cases, (f) and (h) as noted by Jim.
    And so C wins and the 2 is on the table.
    
    It is ironic that the published solution was correct, but for the wrong reasons.
    
    LikeLike
Jim Randell 4:52 pm on 3 April 2020 Permalink | Reply
Tags: by: Victor Bryant ( 148 )
Teaser 3002: Short-cut
From The Sunday Times, 5th April 2020 [link] [link]

To demonstrate a bit of geometry and trigonometry to my grandson, I took a rectangular piece of paper whose shorter sides were 24 cm in length. With one straight fold I brought one corner of the rectangle to the midpoint of the opposite longer side. Then I cut the paper along the fold, creating a triangle and another piece. I then demonstrated to my grandson that this other piece had double the area of the triangle.

How long was the cut?

[teaser3002]
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- Jim Randell 5:40 pm on 3 April 2020 Permalink | Reply
  
  If we assume the fold goes from one of the corners of the rectangle, then we get a nice answer. (See: Enigma 1402). I don’t think other constructions are possible. [This assumption is justified below].
  
  Brainteaser 1798 is a puzzle with a similar theme.
  
  The triangles X, Y, Z are all right-angled. We need to find the value of h, the hypotenuse of triangle X.
  
  The area of triangle X must be the same as the sum of the areas of triangles Y and Z, so:
  
  (24 − a)b = 12b + ab/2
  12b = 3ab/2
  a = 8
  
  From triangle Z:
  
  b² = 16² − 8² = 192
  
  (So the long side of the rectangle is 2b = 16√3, approximately 27.71 cm).
  
  And from triangle X:
  
  h² = 16² + 2²×192 = 1024
  h = 32
  
  Solution: The length of the cut is 32 cm.
  
  And we can then demonstrate that X = Y + Z by constructing X from Y and Z:
  
  X, Y, Z are all (30°, 60°, 90°) triangles. The same shape as one of the standard set squares.
  
  LikeLike
  - Jim Randell 3:39 pm on 5 April 2020 Permalink | Reply
    
    Adding a 24-by-2x strip on the left-hand side of the diagram (to account for the fold not going from a corner), and adjusting the bases of triangles Y and Z to (b − x) and (b + x) leads to the following 2 equations:
    
    [X = Y + Z + 48x] ⇒ 3b(8 − a) = (72 + a)x
    [Y and Z are similar] ⇒ (24 − a)x = 3b(8 − a)
    
    These can only be satisfied (for positive a, b) if x = 0 and a = 8 (i.e. a is 1/3 the height of the rectangle), as above.
    
    So the fold must go from one of the corners, and the assumption above is justified.
    
    LikeLike
- Benet Allen 7:49 pm on 5 April 2020 Permalink | Reply
  
  There’s only one shape where you can fold a corner to the midpoint of the opposite side and be left with a triangle. That’s a 2×1 rectangle. And surely, the remaining piece will have three times the area of the triangle? Am befuddled.
  
  LikeLike
  - Jim Randell 9:23 pm on 5 April 2020 Permalink | Reply
    
    My diagram above [ link ] is actually to scale. If you print it out and cut out the rectangle you will find that you can fold the white X onto the grey X, and then fold Y and Z on top (or behind) to make another copy of X, neatly demonstrating that X = Y + Z as required.
    
    LikeLike
Jim Randell 12:12 pm on 2 April 2020 Permalink | Reply
Tags: by: Dr V W Bryant ( 5 )
Brain-Teaser 516: [Cabinet reshuffle]
From The Sunday Times, 2nd May 1971 [link]

In Northminster the Premier is planning a reshuffle of five positions in his Cabinet. They are (in decreasing order of rank):

Deputy Premier,
Chancellor of the Purse-strings,
Minister of Mails,
Secretary of Science,
Head Whip.

At the moment these jobs are held (not respectively) by:

Attler,
Balfy,
Chamberton,
Disreel,
Edane.

Soon these five jobs are to be re-allocated among these five people and each person will get a different job from the one he holds at the moment.

In the reshuffle the Premier is going to promote Chamberton to Minister of the Mails, or above. Edane is also to be promoted. Attler has made it clear that if he is going to be junior to Chamberton, then the only Deputy Premier he would work under would be Disreel.

Balfy is going to be demoted, but one consolation is that after the reshuffle he will be senior to someone to whom at the moment he is junior. Finally, Disreel will become Deputy Premier only if Balfy becomes Minister of Mails.

The Premier tries to avoid a party split by arranging his reshuffle to satisfy all the above demands. Also he arranges it with the least number of demotions possible in the circumstances.

Who is the present Secretary of Science? And, who is to become Secretary of Science?

This puzzle was originally published with no title.

[teaser516]
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John Crabtree and Jim Randell are discussing. Toggle Comments
- Jim Randell 12:12 pm on 2 April 2020 Permalink | Reply
  In order to get a unique solution I had to add a couple of extra “implied” conditions (the kind that got me into trouble in Enigma 99).
  
  So: “… is going to promote Chamberton to Minister of the Mails, or above”, implies that C’s current position is below MoM.
  
  and: “Disreel will become Deputy Premier only if …”, implies that D is not currently Deputy Premier. (Also the fact that A would serve under D as Deputy Premier implies that D does not currently hold this position).
  
  and (maybe): “… Attler has made it clear that if he is going to be junior to Chamberton …”, implies that A is not currently junior to C (although this extra condition is not required to get a unique solution).
  
  With these conditions made explicit we find there is a single answer to the puzzle.
  
  This Python program runs in 86ms.
  
  Run: [ @repl.it ]
```
from enigma import (defaultdict, irange, subsets, printf)

# "p -> q" = "if p then q", "q if p", "p only if q"
implies = lambda p, q: not(p) or q

# record (by number of demotions) who occupies position 4
r = defaultdict(set)

# consider "before" positions
for before in subsets(irange(1, 5), size=5, select="P"):
  (bA, bB, bC, bD, bE) = before

  # [implied] D is not currently 1
  if bD == 1: continue

  # [implied] A is not currently junior to C
  if (bC < bA): continue

  # [implied] C is currently below 3
  if not (bC > 3): continue

  # consider "after positions"
  for after in subsets(irange(1, 5), size=5, select="P"):
    (aA, aB, aC, aD, aE) = after

    # nobody keeps the same post
    if any(bX == aX for (bX, aX) in zip(before, after)): continue

    # if A is junior to C, then D must be 1
    if not implies(aA > aC, aD == 1): continue

    # D becomes 1 only if B becomes 3
    if not implies(aD == 1, aB == 3): continue

    # C is to be promoted to 3 or above
    if not (aC < 4 and aC < bC): continue

    # E is to be promoted
    if not (aE < bE): continue

    # B is to be demoted ...
    if not (aB > bB): continue

    # ... but will become senior to someone who was his junior
    if not any(aB < aX and bB > bX for (bX, aX) in zip(before, after)): continue

    # count the number of demotions
    d = sum(aX > bX for (bX, aX) in zip(before, after))

    printf("[d={d}, A: {bA}->{aA}; B: {bB}->{aB}; C: {bC}->{aC}; D: {bD}->{aD}; E: {bE}->{aE}]")
    b = dict(zip(before, "ABCDE"))
    a = dict(zip(after, "ABCDE"))
    r[d].add((b[4], a[4]))

# look for the fewest number of demotions
k = min(r.keys())
for (b4, a4) in r[k]:
  printf("{k} demotions: SoS = {b4}->{a4}")
```
  Solution: Chamberton is the current Secretary of Science. Edane will be the new Secretary of Science.
  
  There is only one scenario that has only 2 demotions:
  
  A: demoted from Deputy Premier to Head Whip
  B: demoted from Chancellor to Minister of Mails
  C: promoted from Secretary of Science to Chancellor
  D: promoted from Minster of Mails to Deputy Premier
  E: promoted from Head Whip to Secretary of Science
  
  We are told B is to be demoted, and someone else who was above B must also be demoted under B, so we cannot do better than 2 demotions.
  
  There are 2 further scenarios that satisfy the conditions that have 3 demotions.
  
  LikeLike
- John Crabtree 10:06 pm on 3 April 2020 Permalink | Reply
  
  Let the five positions in order of rank be P1, P2, P3, P4 and P5.
  
  C and E will be promoted, D can be promoted, and B has somebody senior to him, and so A is currently in P1.
  B can be demoted to P3, and so is currently in P2.
  A will move to P4 or P5 (lower than B), and so will be junior to C (P3 or above), and so D moves to P1, and so B moves to P3, and so C moves to P2.
  To minimise the number of demotions A moves to P5, and E moves from P5 to P4.
  The current positions are not in order, and so currently C is in P4, and D is in P3.
  
  Chamberton is the current Secretary of Science.
  Edane will become the Secretary of Science.
  
  LikeLike
Jim Randell 10:30 am on 31 March 2020 Permalink | Reply
Tags: by: Stephen Hogg ( 65 )
Teaser 2877: Four steadings and a numeral
From The Sunday Times, 12th November 2017 [link] [link]

Farmers Al, Bo, Cy and Di have different double-digit numbers of sheep kept in their respective steadings. Al has the fewest and his number of sheep is a certain fraction of Bo’s number of sheep. Also, Bo’s number of sheep is that same fraction of Cy’s number, and Cy’s number is that same fraction of Di’s.

If I told you the total of Bo’s number of sheep added to Cy’s, then you would be unable to work out all their numbers of sheep. Similarly, if instead I told you just Bo’s number of sheep, then you would be unable to work out all the other numbers.

What (in the order Al, Bo, Cy, Di) are their numbers of sheep?

[teaser2877]
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GeoffR and Jim Randell are discussing. Toggle Comments
- Jim Randell 10:31 am on 31 March 2020 Permalink | Reply
  This Python program looks for possible geometric sequences for A, B, C, D (the numbers of sheep). And then uses the [[ filter_unique() ]] function from the enigma.py library to find sets where (B + C) does not identify a single sequence, and where B does not identify a single sequence. The solution is in the intersection of these two sets.
  
  This Python program runs in 62ms.
  
  Run: [ @repl.it ]
```
from enigma import (subsets, irange, div, fraction, filter_unique, unpack, intersect, printf)

# collect results as (A, B, C, D)
rs = list()

# start with values for A and B
for (A, B) in subsets(irange(10, 97), size=2):
  # f = B / A
  # C = f * B = B^2 / A
  C = div(B * B, A)
  if C is None or C > 98: continue
  # D = f * C = BC / A
  D = div(B * C, A)
  if D is None or D > 99: continue

  printf("[A={A} B={B} C={C} D={D}, f={f[0]}/{f[1]}]", f=fraction(B, A))
  rs.append((A, B, C, D))

# B + C does not uniquely identify the solution
(_, s1) = filter_unique(rs, unpack(lambda A, B, C, D: B + C))
printf("[s1 = {s1}]")

# B does not uniquely identify the solution
(_, s2) = filter_unique(rs, unpack(lambda A, B, C, D: B))
printf("[s2 = {s2}]")

# so the solution is in both sets
for (A, B, C, D) in intersect((s1, s2)):
  printf("A={A} B={B} C={C} D={D}")
```
  Solution: The numbers of sheep are: Al=16, Bo=24, Cy=36, Di=54.
  
  Each term in the 3/2 times the previous term.
  
  There are 6 possible geometric sequences:
  
  [1] A=10, B=20, C=40, D=80; f=2
  [2] A=11, B=22, C=44, D=88; f=2
  [3] A=12, B=24, C=48, D=96; f=2
  [4] A=16, B=24, C=36, D=54; f=3/2
  [5] A=24, B=36, C=54, D=81; f=3/2
  [6] A=27, B=36, C=48, D=64; f=4/3
  
  [1], [4] have the same value for B + C (= 60).
  
  [3], [4] have the same value for B (=24), and [5], [6] have the same value for B (= 36).
  
  The solution occurs in both sets, so is sequence [4].
  
  LikeLike
- GeoffR 5:06 pm on 1 April 2020 Permalink | Reply
  The following programme gives the same six sequences for A,B,C and D.
```
for a in range(10,99):
  for b in range(a+1,99):
    for c in range(b+1,99):
      for d in range(c+1,99):
        # Ratios given are a/b = c/d and b/c = c/d
        if a*c == b*b and b*d == c*c:
          print(f"A={a}, B={b}, C={c}, D={d}")
 
# A=10, B=20, C=40, D=80
# A=11, B=22, C=44, D=88
# A=12, B=24, C=48, D=96
# A=16, B=24, C=36, D=54  << Answer - see Jim's logic above
# A=24, B=36, C=54, D=81
# A=27, B=36, C=48, D=64
```
  LikeLike
Jim Randell 4:46 pm on 27 March 2020 Permalink | Reply
Tags: by: Stephen Hogg ( 65 )
Teaser 3001: Tetragonal toy tiles
From The Sunday Times, 29th March 2020 [link] [link]

(see: [ @wikipedia ])
Thirteen toy tiles comprised a square, rectangles, rhombuses (diamonds on a playing card are rhombuses) and kites (as shown in the diagram). All of each different type were identical. A rhombus’s longer diagonal was a whole number of inches (equalling any diagonal of any other type). Its shorter diagonal was half this. Also, one side of a rectangle was slightly over one inch.A pattern I made, using every tile laid flat, had all the symmetries of a square. After laying the first tile, each subsequent tile touched at least one other previously placed tile. Ultimately, any contact points were only where a vertex of a tile touched a vertex of just one other tile; only rhombuses touched every other tile type.What, in inches, was a square’s diagonal?

[teaser3001]
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Brian Gladman, Jim Randell, and Robert Brown are discussing. Toggle Comments
- Jim Randell 6:29 pm on 27 March 2020 Permalink | Reply
  I came up with a pattern for 13 tiles that has the same symmetries as a square (I’ll make a diagram of it later), and that gave me a way to calculate the sides of the rectangle, in terms of the larger diagonal of the rhombus, x.
  
  Once these are calculated the value of x can be determined manually, or here is a quick program to do it:
  
  Run: [ @replit ]
```
from enigma import (sqrt, irange, inf, printf)

# multipliers for the sides of the rectangle
(ra, rb) = (sqrt(1, 8), sqrt(7, 8))

# consider the diagonal of the square piece
for x in irange(1, inf):
  # calculate the sides of the rectangle
  (a, b) = (ra * x, rb * x)
  if 1.0 < a < 1.1 or 1.0 < b < 1.1:
    printf("x={x} [a={a:.3f} b={b:.3f}]")
  elif a > 1.1:
    break
```
  Solution: The length of the square’s diagonal is 3 inches.
  
  I arranged 13 shapes (1 square, 4 rectangles, 4 rhombuses, 4 kites) into the following pattern:
  
  All the diagonals of all the shapes are equal (= x), except for the shorter diagonal of the rhombus (= x/2).
  
  In this arrangement the short side of the rectangle is the hypotenuse of a right-angled triangle where the other two sides have length x/4, so it has a length of x√(1/8), and so the longer side has a length of x√(7/8).
  
  Setting x = 3 gives dimensions of 1.061 × 2.806 inches. The smaller side being close to 1 + 1/16 inches. Which is “slightly over 1 inch” as required.
  
  The exact shape of the kite doesn’t matter (as long as both diagonals are x), it doesn’t affect the calculation for the rectangle. (In particular the kites could all be rotated through 180° to give a second arrangement).
  
  Placing the rhombuses the other way leaves a gap that cannot be filled by the required rectangle, and we don’t have enough shapes to fill the gap with multiple shapes.
  
  LikeLike
- Robert Brown 8:06 am on 28 March 2020 Permalink | Reply
  
  I did a scale drawing. My kite has the same aspect ratio as the one in the original text, which makes the large angle equal to that of the rhombus. I don’t think your program would have found my answer, which has the rectangle 1.03 inches high.
  
  LikeLike
  - Jim Randell 8:13 am on 28 March 2020 Permalink | Reply
    
    @Robert: My arrangement looks like this [ link ], so the exact shape of the kite doesn’t affect the calculations. But I didn’t look too hard for alternative patterns (although you would hope the setter would have made sure there was a unique answer to the puzzle).
    
    LikeLike
- Robert Brown 11:54 am on 28 March 2020 Permalink | Reply
  
  Interesting. Each rhombus has a thin & thick ‘corner’. My layout has the thin corners connected to the square, then the kites & rhombuses are all angled to fit round the square. I tried (but failed) to get the rectangles in the corner, to give it ‘all the symmetries of a square’ ! My rectangle is long & thin, with diagonal =9 inches. I see Brian has 3 inches, I wonder what his layout looks like . . .
  
  LikeLike
  - Jim Randell 12:36 pm on 28 March 2020 Permalink | Reply
    
    I tried my pattern with the rhombuses turned through 90°, but I found the gap between the remaining vertices was too large to fit any of the other shapes into.
    
    Looking at Brian’s attachment to the Google Sites page it looks like he has found the same layout I did (although I don’t think his kites are the right shape, but that doesn’t change the answer).
    
    LikeLike
    - Brian Gladman 1:15 pm on 28 March 2020 Permalink | Reply
      
      That is because the teaser doesn’t explicitly constrain the non-rhombus shapes to have equal diagonals (of course, all except the kite do). The longest rhombus diagonal can be any diagonal of any other shape and I chose to match the longest diagonals of the rhombus and the kite.
      
      LikeLike
      - Jim Randell 1:20 pm on 28 March 2020 Permalink | Reply
        
        @Brian: It does say that the longer diagonal of the rhombus should “equal any diagonal of any other type”, so I think the vertices of the kite must touch the sides of an x by x bounding box.
        
        LikeLike
        
        Brian Gladman 2:36 pm on 28 March 2020 Permalink | Reply
        
        Yes, you interpret it to mean “equal to the diagonals of the other types” (surely a simpler expression of this interpretation), whereas I interpret it to mean a choice between “any of the diagonals of any other type”. Fortunately this clear ambiguity (!) doesn’t have an impact on the answer.
        
        LikeLike
- Robert Brown 12:33 pm on 28 March 2020 Permalink | Reply
  
  So my alternative pattern lacks one of the required symmetries (it has clockwise & counter clockwise versions). We’ve had problems with words before . . . I guess Brian’s layout is similar to yours.
  
  LikeLike

Jim Randell 9:49 am on 24 March 2020 Permalink | Reply
Tags: by: Peter Harrison ( 6 )

Teaser 2500: [Rogue ball detection]

From The Sunday Times, 22nd August 2010 [link] [link]

A well-known puzzle asks:

“If among 12 snooker balls one is a different weight, how can the rogue ball be identified – together with deciding whether it is heavier or lighter – in three weighings on a balance?”

Recently I faced a very similar problem of finding a rogue ball among a consignment of 39 identical-looking balls – and deciding if it was heavier or lighter. I had at my disposal a two-pan balance.

How many weighings did I need?

This puzzle was originally published with no title.

[teaser2500]

Jim Randell 9:50 am on 24 March 2020 Permalink | Reply

We dealt with a similar puzzle in Enigma 1589 (also set by Peter Harrison). For that puzzle I wrote a program that searches for a minimal decision tree.

For an analytical solution the paper Weighing Coins: Divide and Conquer to Detect a Counterfeit by Mario Martelli and Gerald Gannon [ link ] provides an elegant analysis to determine the maximum number of coins that can be distinguished using a certain number of weighings for four related problems.

And this gives us the solution to this puzzle:

Solution: 4 weighings can be used to find the rogue object in a set of 39.

LikeLike

Jim Randell 9:44 pm on 24 March 2020 Permalink | Reply

For Enigma 1589 I wrote the following code to generate the decision trees for the four types of problem described in the linked paper.

It can be used to generate decision trees for either puzzle.

Run: [ @codepad ] [ @repl.it ]

from enigma import (irange, arg, printf)

# weigh 3^n coins in n weighings, suspect coin is lighter
def solve_lighter(n, coins, p=''):
  assert len(coins) == 3**n
  # 3 coins
  if n == 1:
    # weigh two of the coins
    printf("{p}{coins[0]} < {coins[1]} -> {coins[0]}L")
    printf("{p}{coins[0]} = {coins[1]} -> {coins[2]}L")
    printf("{p}{coins[0]} > {coins[1]} -> {coins[1]}L")
  else:
    # divide the coins into 3 piles
    k = len(coins) // 3
    (c1, c2, c3) = (coins[:k], coins[k:2 * k], coins[2 * k:])
    printf("{p}{c1} < {c2}")
    solve_lighter(n - 1, c1, p + '  ')
    printf("{p}{c1} = {c2}")
    solve_lighter(n - 1, c3, p + '  ')
    printf("{p}{c1} > {c2}")
    solve_lighter(n - 1, c2, p + '  ')


# weigh 3^n coins in n weighings
# suspect coin is in reds if heavier, in blacks if lighter
def solve_red_black(n, reds, blacks, red='H', black='L', p=''):
  assert len(reds) + len(blacks) == 3**n and len(reds) == len(blacks) + 1
  # 2 red, 1 black
  if n == 1:
    # weigh the two reds
    printf("{p}{reds[0]} < {reds[1]} -> {x[0]}{x[1]}", x=((reds[1], 'H') if red == 'H' else (reds[0], 'L')))
    printf("{p}{reds[0]} = {reds[1]} -> {blacks[0]}{black}")
    printf("{p}{reds[0]} > {reds[1]} -> {x[0]}{x[1]}", x=((reds[0], 'H') if red == 'H' else (reds[1], 'L')))
  else:
    # weigh (3^(n-1) + 1)/2 reds against (3^(n-1) - 1)/2 of the blacks
    k = (3**(n - 1) + 1) // 2
    (r1, r2, r3) = (reds[:k], reds[k:2 * k], reds[2 * k:])
    (b1, b2, b3) = (blacks[:k - 1], blacks[k - 1:2 * k - 2], blacks[2 * k - 2:])
    (left, right) = (r1 + b1, r2 + b2)
    printf("{p}{left} < {right}")
    solve_red_black(n - 1, r2, b1, red, black, p + '  ')
    printf("{p}{left} = {right}")
    solve_red_black(n - 1, b3, r3, black, red, p + '  ')
    printf("{p}{left} > {right}")
    solve_red_black(n - 1, r1, b2, red, black, p + '  ')


# weigh (3^n - 1) / 2 suspect coins in n weighings
# given a known good coin
def solve_test_coin(n, good, coins, p=''):
  assert len(coins) == ((3**n) - 1) // 2
  if n == 1:
    # weigh the coin against the good coin
    printf("{p}{good} < {coins[0]} -> {coins[0]}H")
    printf("{p}{good} = {coins[0]} -> #")
    printf("{p}{good} > {coins[0]} -> {coins[0]}L")
  else:
    k = (3**(n - 1) + 1) // 2
    (c1, c2, c3) = (coins[:k], coins[k:2 * k - 1], coins[2 * k - 1:])
    (left, right) = (c1, [good] + c2)
    printf("{p}{left} < {right}")
    solve_red_black(n - 1, c1, c2, 'L', 'H', p + '  ')
    printf("{p}{left} = {right}")
    solve_test_coin(n - 1, good, c3, p + '  ')
    printf("{p}{left} > {right}")
    solve_red_black(n - 1, c1, c2, 'H', 'L', p + '  ')


# weigh (3^n - 3) / 2 suspect coins in n weighings
def solve_general(n, coins, p=''):
  assert len(coins) == ((3**n) - 3) // 2
  # divide the coins into 3 piles
  k = len(coins) // 3
  (c1, c2, c3) = (coins[:k], coins[k:2 * k], coins[2 * k:])
  printf("{p}{c1} < {c2}")
  solve_red_black(n - 1, c1 + c3[:1], c2, 'L', 'H', p + '  ')  
  printf("{p}{c1} = {c2}")
  solve_test_coin(n - 1, c1[0], c3, p + '  ')
  printf("{p}{c1} > {c2}")
  solve_red_black(n - 1, c1 + c3[:1], c2, 'H', 'L', p + '  ')
  

# number of weighings and problem type
n = arg(3, 0, int, prompt="number of weighings")
p = arg(2, 1, int, prompt="problem number")
printf("[n={n}; p={p} (of 1-4)]")

if p == 1:
  # solve the general case
  k = (3**n - 3) // 2
  if k > 0:
    printf("---\n{n} weighings finds the suspect coin from {k} coins\n---")
    solve_general(n, list(irange(1, k)))

elif p == 2:
  # solve with a known good coin
  k = (3**n - 1) // 2
  printf("---\n{n} weighings finds the suspect coin from {k} coins, with an extra known good coin (0)\n---")
  solve_test_coin(n, 0, list(irange(1, k)))

elif p == 3:
  # solve red/black
  k = 3**n
  j = (k + 1) // 2
  printf("---\n{n} weighings finds the suspect coin from {k} coins, where {j} of the coins are red and {j1} of the coins are black\nif the suspect coin is red it is heavier, if it is black it is lighter\n---", j1=j - 1)
  solve_red_black(n, list(irange(1, j)), list(irange(j + 1, k)))

elif p == 4:
  # solve lighter
  k = 3**n
  printf("---\n{n} weighings finds the suspect lighter coin from {k} coins\n---")
  solve_lighter(n, list(irange(1, k)))

LikeLike

Jim Randell 4:25 pm on 20 March 2020 Permalink | Reply
Tags: by: Richard England ( 9 )
Teaser 3000: The three thousandth
From The Sunday Times, 22nd March 2020 [link] [link]

In this addition digits have been consistently replaced by letters, different letters representing different digits. But instead of an addition in base 10 in which the letters represent the digits 0 to 9 this is an addition in base 11, using X for the extra digit, in which the letters represent the digits 1 to X, with 0 unused.

Please submit the number (still in base 11) represented by TEASER.

Congratulations to The Sunday Times for publishing 3000 Teaser puzzles.

[teaser3000]
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Jim Randell, GeoffR, and Jim Randell are discussing. Toggle Comments
- Jim Randell 4:30 pm on 20 March 2020 Permalink | Reply
  We can use the [[ SubstitutedSum() ]] solver from the enigma.py library to solve this puzzle.
  
  The following run file executes in 126ms.
```
#! python3 -m enigma -rr

SubstitutedSum

--base="11"
--digits="1-10"

"THREE + THOUS + ANDTH = TEASER"
```
  Solution: TEASER = 12X523.
  
  There are two ways to assign values to the letters, as D and O are interchangeable:
  
  17322 + 17495 + X6817 = 12X523 / A=10 D=8 E=2 H=7 N=6 O=4 R=3 S=5 T=1 U=9
  17322 + 17895 + X6417 = 12X523 / A=10 D=4 E=2 H=7 N=6 O=8 R=3 S=5 T=1 U=9
  
  LikeLike
  - Jim Randell 12:08 pm on 1 July 2023 Permalink | Reply
    Here is a faster solution using the [[ SubstitutedExpression.split_sum ]] solver from the enigma.py library.
    
    The internal runtime of this run file is 495µs.
```
#! python3 -m enigma -rr

SubstitutedExpression.split_sum

--base="11"

"{THREE} + {THOUS} + {ANDTH} = {TEASER}"

--invalid="0,ADEHNORSTU"  # no 0 digits
--code="base_digits('0123456789X')"  # use X for digit 10
--answer="int2base({TEASER}, 11)"  # output solution in base 11
```
    LikeLike
- GeoffR 1:48 pm on 21 March 2020 Permalink | Reply
  The teaser uses the 10 different letters ([T,H,R,E,O,U,S,A,N,D] in this teaser), which can convieniently represent the digits (1..10), as zero is not required,
  
  The programme produces a single solution with the third digit as 10 (or X), alltough there are two sets of digits, with values of O and D interchangeable.
  
  The programme uses the standard method of adding digits in columns, with carry digits to adjacent columns for column digit sums exceeding 11.
```
% A Solution in MiniZinc
include "globals.mzn";

%    T H R E E
%    T H O U S
%    A N D T H
%  -----------
%  T E A S E R
%  -----------

% Using digits 1..10 in base 11
var 1..10:T; var 1..10:H; var 1..10:R; var 1..10:E;
var 1..10:O; var 1..10:U; var 1..10:S; var 1..10:A;
var 1..10:N; var 1..10:D;

constraint all_different ([T,H,R,E,O,U,S,A,N,D]);

var 0..2: carry1; var 0..2: carry2; var 0..2: carry3;
var 0..2: carry4; var 0..2: carry5;

% Adding up digits in columns, starting from right side
constraint (E + S + H) mod 11 == R 
/\ carry1 == (E + S + H) div 11;

constraint (E + U + T + carry1) mod 11 == E 
/\ carry2 == (E + U + T + carry1) div 11;

constraint (R + O + D + carry2) mod 11 == S 
/\ carry3 == (R + O + D + carry2) div 11;

constraint (H + H + N + carry3) mod 11 == A 
/\ carry4 = (H + H + N + carry3) div 11;

constraint (T + T + A + carry4) mod 11 == E 
/\ carry5 == (T + T + A + carry4) div 11;

constraint T == carry5;

solve satisfy;

output ["TEASER = " ++ show(T) ++ " " ++ show(E) ++ " " ++ show(A)
++ " " ++ show(S) ++ " " ++ show(E) ++ " " ++ show(R)  
++"\n" ++ "10 digits are [T,H,R,E,O,U,S,A,N,D] = " ++ show([T,H,R,E,O,U,S,A,N,D]) ];
```
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