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  • Unknown's avatar

    Jim Randell 7:19 am on 7 March 2019 Permalink | Reply
    Tags: by: Dr J A Anderson   

    Brain-Teaser 461: [Grovelball] 

    From The Sunday Times, 29th March 1970 [link]

    The finals of the Ruritanian grovelball championships have just taken place between the four leading teams, each team playing the other three.

    The Archers won the championships, scoring fourteen grovels to one against, while the Bakers scored eight and conceded five.

    The Cobblers, who scored four grovels, conceded the same number as the Donkey-drivers, who failed to score in any of their three games.

    In no game were fewer than three grovels scored, and there were only two games which resulted in the same score.

    What were the scores in the Archers v. Bakers and Archers v. Cobblers games?

    This puzzle was originally published with no title.

    [teaser461]

     
    • Jim Randell's avatar

      Jim Randell 7:22 am on 7 March 2019 Permalink | Reply

      We first note that C and D conceded the same number of goals, say x. And the sum of all the goals scored must be the same as the sum of all the goals conceded:

      14 + 8 + 4 + 0 = 1 + 5 + x + x
      2x = 20
      x = 10

      From there we can use the [[ Football() ]] helper class from the enigma.py library to consider this as a “substituted table” problem. We get a fairly short program, but not especially quick, but it finds the solution in 886ms.

      Run: [ @repl.it ]

      from enigma import Football, digit_map, ordered

# scoring system (all games are played)
football = Football(games="wdl")

# digits stand for themselves, A=10, E=14
d = digit_map(0, 14)

# the table (mostly empty)
(table, gf, ga) = (dict(played="3333", w="???0"), "E840", "15AA")

# consider match outcomes
for (ms, _) in football.substituted_table(table, d=d):

  # find possible scorelines from the goals for/against columns
  for ss in football.substituted_table_goals(gf, ga, ms, d=d):

    # additional restrictions:
    # "in no game were fewer than 3 goals scored"
    if any(sum(s) < 3 for s in ss.values()): continue

    # "only two games had the same score"
    # so there are 5 different scores involved
    if len(set(ordered(*s) for s in ss.values())) != 5: continue

    # output the matches
    football.output_matches(ms, ss, teams="ABCD")

      Solution: A vs B = 4-1; A vs C = 7-0.

      There is only one possible set of scorelines:

      A vs B = 4-1
      A vs C = 7-0
      A vs D = 3-0
      B vs C = 3-1
      B vs D = 4-0
      C vs D = 3-0

      So A has 3w, B has 2w+1d, C has 1w, D has 0w.

      Like

  • Unknown's avatar

    Jim Randell 8:16 am on 6 March 2019 Permalink | Reply
    Tags:   

    Teaser 2939: Betting to win 

    From The Sunday Times, 20th January 2019 [link] [link]

    Two teams, A and B, play each other.

    A bookmaker was giving odds of 8-5 on a win for team A, 6-5 on a win for team B and X-Y for a draw (odds of X-Y mean that if £Y is bet and the bet is successful then £(X + Y) is returned to the punter). I don’t remember the values of X and Y, but I know that they were whole numbers less than 20.

    Unusually, the bookmaker miscalculated! I found that I was able to make bets of whole numbers of pounds on all three results and guarantee a profit of precisely 1 pound.

    What were the odds for a draw, and how much did I bet on that result?

    [teaser2939]

     
    • Jim Randell's avatar

      Jim Randell 8:17 am on 6 March 2019 Permalink | Reply

      If the integer stakes are p for A to win, q for B to win, r for a draw, we have:

      (13/5)p = p + q + r + 1
      (11/5)q = p + q + r + 1
      ((X + Y)/Y)r = p + q + r + 1

      which, given X and Y, we can solve as:

      p = 55(X + Y) / (23X – 120Y)
      q = 65(X + Y) / (23X – 120Y)
      r = 143Y / (23X – 120Y)

      and we want these to be integers.

      X and Y are limited to be integers between 1 and 19, so we can easily check all possibilities.

      This Python program runs in 72ms.

      Run: [ @repl.it ]

      from itertools import product
from enigma import (irange, printf)

# consider possible values for X, Y
for (X, Y) in product(irange(1, 19), repeat=2):
  # calculate integer stakes:
  # p (for A to win)
  # q (for B to win)
  # r (for a draw)
  d = 23 * X - 120 * Y
  if not (d > 0): continue
  (p, x) = divmod(55 * (X + Y), d)
  if x != 0: continue
  (q, x) = divmod(65 * (X + Y), d)
  if x != 0: continue
  (r, x) = divmod(143 * Y, d)
  if x != 0: continue

  # output solution
  printf("X={X} Y={Y}, p={p} q={q} r={r}")

  # verify the winnings
  w = p + q + r + 1
  assert divmod(13 * p, 5) == (w, 0)
  assert divmod(11 * q, 5) == (w, 0)
  assert divmod((X + Y) * r, Y) == (w, 0)

      Solution: The odds for a draw were 11-2. The bet on a draw was £22.

      A variation: If the bets can be whole numbers of pennies, instead of whole numbers of pounds, then there is another solution where X-Y is 10-1 and the bets are £5.50 on A to win, £6.50 for B to win, £1.30 on a draw. The winnings are £14.30 for a total outlay of £13.30.

      Like

    • GeoffR's avatar

      GeoffR 8:22 am on 7 March 2019 Permalink | Reply 

      % A solution in MinZinc 
var 1..19: X; var 1..19: Y;

% Integer stakes are p for A to win, q for B to win, and r for a draw
% For bets of p, q and r, a return of (p + q + r + 1) makes a profit of £1
var 1..100: p; var 1..100: q; var 1..100: r;

constraint 13 * p = (p + q + r + 1) * 5;        % 8-5 odds for team A
constraint 11 * q = (p + q + r + 1) * 5;        % 6-5 odds for team B
constraint (X + Y )* r = (p + q + r + 1) * Y;   % X-Y odds for a draw

solve satisfy;

% X = 11; Y = 2; p = 55; q = 65; r = 22;
% ==========
% Finished in 195msec 
% Odds for a draw were 11/2 and the stake for a draw was £22

      Like

  • Unknown's avatar

    Jim Randell 11:16 am on 5 March 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 460: [Crates of Boppo] 

    From The Sunday Times, 22nd March 1970 [link]

    Boppo is made up in 1 lb cans at 8s, 2 lb cans at 15s and 4 lb cans at 27s. The contents of crates vary, but every crate contains 60 cans of mixed three sizes with a total value of £45.

    The foreman himself had packed two such crates. He told new worker Rob, “Just go along and nail down those two open crates in the shed, Rob. The lorry’s waiting.”

    Rob found the two crates standing beside the weighing platform, and for no particular reason he weighed them. One was heavier than the other. “Huh! Calls ‘isself a foreman” he said, as he removed some cans from the heavier crate. “That’s it. Both the same now. Lucky I weighed ’em”. So he nailed them down and the lorry went off with them.

    He met the foreman. “You made a mistake with those crates, chum”, he said, “but it’s OK now. Here’s three spare cans.”

    “!!” said the foreman.

    What were the weights of the two crates the foreman had packed?

    This puzzle was originally published with no title.

    [teaser460]

     
    • Jim Randell's avatar

      Jim Randell 11:19 am on 5 March 2019 Permalink | Reply

      Before decimalisation £1 was the same as 20 shillings, so each crate has a monetary value of 900 shillings.

      I solved this using a similar approach to Enigma 486 (and borrowed some of the code too).

      This Python 3 program runs in 108ms.

      from enigma import (defaultdict, irange, subsets, printf)

# express total <t> using denominations <ds>, min quantity 1
def express(t, ds, s=[]):
  if not ds:
    if t == 0:
      yield s
  else:
    (d, *ds) = ds
    for i in irange(1, t // d):
      yield from express(t - d * i, ds, s + [i])

# the weights and costs of the cans
weights = (1, 2, 4)
costs = (8, 15, 27)

# find ways to split the cost of a crate into cans
crates = defaultdict(list)
for ns in express(900, costs):
  # but each crate has 60 cans
  if not (sum(ns) == 60): continue
  # calculate the total weight of the crate
  t = sum(n * w for (n, w) in zip(ns, weights))
  printf("[{ns} -> weight = {t}]")
  crates[t].append(ns)

# possible weights of 3 cans to be removed
w3s = defaultdict(list)
for rs in subsets(weights, size=3, select='R'):
  w3s[sum(rs)].append(rs)

# find two crates with weights that differ by 3 cans
for (w1, w2) in subsets(sorted(crates.keys()), size=2):
  for rs in w3s[w2 - w1]:
    # check there are enough cans in the w2 crate
    if not any(all(not (r > n) for (r, n) in zip(rs, ns)) for ns in crates[w2]): continue
    # output solution
    printf("{w2} lb -> {w1} lb by removing {rs} x {weights} lb cans")

      Solution: The crates the foreman had packed weighed 122 lb and 126 lb.

      Rob reduced the weight of the 126 lb crate to 122 lb by removing 2× 1 lb cans and 1× 2 lb can.

      There are only 3 ways to pack the crates so they have 60 cans and are worth 900 shillings (900s) (where the is at least 1 of each type of can included):

      12× 1 lb @ 8s + 41× 2 lb @ 15s + 7× 4 lb @ 27s = 60 cans, 122 lb, 900s
      24× 1 lb @ 8s + 22× 2 lb @ 15s + 14× 4 lb @ 27s = 60 cans, 124 lb, 900s
      36× 1 lb @ 8s + 3× 2 lb @ 15s + 21× 4 lb @ 27s = 60 cans, 126 lb, 900s

      If we don’t require that all three weights of can are represented in the crate then we can make a crate with 60× 2 lb cans, which weighs 120 lb, and gives rise to multiple solutions.

      Like

    • GeoffR's avatar

      GeoffR 7:42 am on 6 March 2019 Permalink | Reply

      The only 3 ways to pack a crate are found in my MiniZinc programme as 122 lb, 124 lb and 126 lb

      The only way of removing 3 cans to make the crates equal is to remove 3 cans ( 2 of 1lb and 1 of 2lb = 4lb) to reduce the crate weighing 126 ilb to 122 lb.

      Therefore, the foreman had packed 122 lb and 126 lb into the two crates.

      var 1..60: n1; var 1..60: n2;  var 1..60: n3;   % Numbers of  three types of can
var 1..200: wt;   %total weight of all cans in 1 crate

constraint n1 + n2 + n3 == 60;  % Every crate contains 60 cans
constraint n1 * 8 + n2 * 15 + n3 * 27 == 900;   %sh £45 * 20 = 900 shillings for 1 crate
constraint wt = n1 * 1 + n2 * 2 + n3 * 4;

solve satisfy;

% n1 = 12;  n2 = 41;  n3 = 7;  wt = 122;
%----------
% n1 = 24;  n2 = 22;  n3 = 14;  wt = 124;
% ----------
% n1 = 36;  n2 = 3;  n3 = 21;  wt = 126;
%----------
%==========
%Finished in 199msec

      Like

  • Unknown's avatar

    Jim Randell 9:32 am on 4 March 2019 Permalink | Reply
    Tags: by: Arithmedicus,   

    Brain-Teaser 459: [Grandchildren] 

    From The Sunday Times, 8th March 1970 [link]

    Grandpa was as pleased as Punch when he heard that his large flock of grandchildren would be there at his birthday party, but he couldn’t remember how many there were. So he wrote to his youngest daughter, who knew he liked problems, and got back the answer:

    “There are just enough children to arrange them in pairs in such a way that the square of the age of the first of a pair added to thrice the square of the age of the second give the same total for every pair. The eldest is still a teenager. You remember that I have twins one year old.”

    How many grandchildren are there?

    This puzzle was originally published with no title.

    [teaser459]

     
    • Jim Randell's avatar

      Jim Randell 9:33 am on 4 March 2019 Permalink | Reply

      I think the wording of this puzzle could be improved. Saying there are “just enough” children sounds like we are looking for the minimum number of children to satisfy the conditions, given that there are two ages of 1 amongst the children, and one teenager. So we already have three values that must be used in the pairs. The smallest potential set would be two pairs, and this is achievable, giving an answer of 4 children.

      But it appears this is not the solution the setter had in mind. Instead it seems we want the largest possible number of children that can be formed into different pairs that give the same value for the function. If we were to allow duplicate pairs (i.e. pairs of children with the same ages in the same order), then we could increase the number of children to be as large as we want.

      Instead this program looks to group the ages of pairs of children into sets that give the same value for the function. We then need a set that has two 1’s and also value between 13 and 19. Then we look for the maximum number of different pairs of children possible, and this is the answer that was published.

      This Python program runs in 81ms.

      Run: [ @repl.it ]

      from enigma import (defaultdict, irange, subsets, printf)

ages = set(irange(1, 19))

# record the pairs by total
d = defaultdict(list)

# consider pairs of ages
for (a, b) in subsets(ages, size=2, select='M'):
  n = a**2 + 3 * b**2
  d[n].append((a, b))

# consider the values found
for k in sorted(d.keys()):
  ps = d[k]
  # look for pairs where 1 occurs twice
  if not (sum(p.count(1) for p in ps) == 2): continue
  # and the eldest is a teenager
  if not (max(max(p) for p in ps) > 12): continue
  # output solution
  printf("{k} -> {ps}")

      Solution: There are 12 grandchildren.

      It turns out there is only one possible total that gives a viable set of pairs. The pairs are:

      (1, 11) (8, 10) (11, 9) (16, 6) (17, 5) (19, 1)

      Each giving the value of 364 when the function f(a, b) = a^2 + 3b^2 is applied to them.

      So there two 1 year olds (the twins) and also two 11 year olds.

      If we had four children of ages 1, 1, 11, 19, we would have “just enough” to form them into 2 pairs – (1, 11) and (19, 1) – that give the same value under the function, and there are two 1’s and a teenager involved. So this would be the answer if we take a strict interpretation of the wording.

      Similarly, if we allow duplication among the pairs we could have 14 children, or 16, etc.

      Like

      • Jim Randell's avatar

        Jim Randell 10:07 am on 4 March 2019 Permalink | Reply

        A variation would be to look for the largest number of children of different ages (apart from the twins) that can form the pairs.

        In this case we would have to discard the (11, 9) pair to leave:

        (1, 11) (8, 10) (16, 6) (17, 5) (19, 1)

        So there would be 10 children.

        Like

  • Unknown's avatar

    Jim Randell 7:12 am on 3 March 2019 Permalink | Reply
    Tags: ,   

    Teaser 2940: Getting there in the end 

    From The Sunday Times, 27th January 2019 [link]

    A tractor is ploughing a furrow in a straight line. It starts from rest and accelerates at a constant rate, taking a two-digit number of minutes to reach its maximum speed of a two-digit number of metres per minute. It has, so far, covered a three-digit number of metres. It now maintains that maximum speed for a further single-digit number of minutes and covers a further two-digit number of metres. It then decelerates to rest at the same rate as the acceleration. I have thus far mentioned ten digits and all of them are different.

    What is the total distance covered?

    As stated this puzzle does not have a unique solution.

    This puzzle was not included in the book The Sunday Times Teasers Book 1 (2021).

    [teaser2940]

     
    • Jim Randell's avatar

      Jim Randell 7:12 am on 3 March 2019 Permalink | Reply

      The tractor starts at rest:

      u = 0

      It then accelerates, at constant rate a, in time t1, to velocity v:

      v = a.t1
      a = v/t1

      During this period it has covered a distance d1:

      d1 = (1/2)a(t1^2) = v.t1/2

      It continues at a constant velocity v, for time t2, and covers distance d2:

      d2 = v.t2

      It then decelerates to rest, at a rate of −a, covering distance d1 again.

      The required answer being:

      d1 + d2 + d1 = v(t1 + t2)

      We can consider the alphametic expressions where:

      t1 = AB; v = CD; d1 = EFG; t2 = H; d2 = IJ

      This can be easily solved by the [[ SubstitutedExpression() ]] solver from the enigma.py library.

      This run-file executes in 126ms.

      Run: [ @repl.it ]

      #! python3 -m enigma -rr

SubstitutedExpression

--answer="2 * EFG + IJ"

"div(AB * CD, 2) = EFG"
"CD * H = IJ"

      Solution: There are three possible total distances: 646m, 864m, 1316m.

      None of the solutions are particularly realistic. Each of them involve the tractor accelerating incredibly slowly for a very long time to reach an extremely modest top speed, and later take just as long to decelerate. Perhaps with a change of units the problem could made to apply to something that does take a long time to decelerate, like a train, oil tanker or spaceship.

      It would also have been easy to require just one of the possible answers, maybe by specifying the answer has 4 digits (1316m), or that all the digits are different (864m).

      The published solution is:

      Teaser 2940
      There were three possible correct answers.
      646, 864 or 1316m. We apologise for any confusion.

      Like

  • Unknown's avatar

    Jim Randell 8:54 am on 2 March 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 458: [Football tournament] 

    From The Sunday Times, 1st March 1970 [link]

    The table gives the results of an all-play-all soccer tournament is South America.

    If I tell you the number of scores of one goal by either side in the 3 games played by Brazil you can deduce the scores of the games between Argentine and Peru and Argentine and Ecuador.

    What are they?

    This puzzle was originally published with no title.

    [teaser458]

     
    • Jim Randell's avatar

      Jim Randell 8:55 am on 2 March 2019 Permalink | Reply

      The enigma.py library includes the [[ Football() ]] class to help in solving puzzles involving football tables, and the [[ filter_unique() ]] function to help in solving puzzles of the form “… if I told you this, then you could work out that…”. We can use both of these to solve this puzzle.

      This Python program runs in 98ms.

      Run: [ @repl.it ]

      from enigma import Football, filter_unique, unpack

# record possible outcomes
rs = list()

# scoring system
football = Football(games='wdl')

# the order of the teams in the table
(B, A, P, E) = (0, 1, 2, 3)

# digits in the table stand for themselves
d = dict((x, int(x)) for x in "01234789")

# the columns of the table
(table, gf, ga) = (dict(w="3100", d="0121", l="0112"), "9724", "3847")

# find possible match outcomes
for (ms, _) in football.substituted_table(table, d=d):

  # find possible scorelines using the goals for/against columns
  for ss in football.substituted_table_goals(gf, ga, ms, d=d):

    # record outcomes
    rs.append((ms, ss))

# if I told you...
# the number of scores of 1 goal by either side in the games played by B...
f = unpack(lambda ms, ss: sum(v.count(1) for (k, v) in ss.items() if B in k))

# you can deduce the scores in the AvP and AvE match
g = unpack(lambda ms, ss: (ss[(A, P)], ss[(A, E)]))

# find the unique solutions
(rs, _) = filter_unique(rs, f, g)

# output solutions
for (ms, ss) in rs:
  football.output_matches(ms, ss, "BAPE")

      Solution: A vs P = 1-1. A vs E = 5-3.

      The scores in all matches are uniquely determined. They are:

      B vs A = 4-1
      B vs P = 3-1
      B vs E = 2-1
      A vs P = 1-1
      A vs E = 5-3
      P vs E = 0-0

      Like

  • Unknown's avatar

    Jim Randell 12:31 pm on 1 March 2019 Permalink | Reply
    Tags:   

    Teaser 2941: Big deal 

    From The Sunday Times, 3rd February 2019 [link] [link]

    My wife and I play bridge with Richard and Linda. The 52 cards are shared equally among the four of us – the order of cards within each player’s hand is irrelevant but it does matter which player gets which cards. Recently, seated in our fixed order around the table, we were discussing the number of different combinations of cards possible and we calculated that it is more than the number of seconds since the “big bang”!

    We also play another similar game with them, using a special pack with fewer cards than in the standard pack – again with each player getting a quarter of the cards. Linda calculated the number of possible combinations of the cards for this game and she noticed that it was equal to the number of seconds in a whole number of days.

    How many cards are there in our special pack?

    [teaser2941]

     
    • Jim Randell's avatar

      Jim Randell 12:32 pm on 1 March 2019 Permalink | Reply

      If there are k players, and each player receives n cards in the deal, then there are n.k cards in total.

      The number of ways of dealing n cards to the first player from the total set of n.k cards are:

      C(n.k, n) = (n.k)! / [ n! (n(k − 1))! ]

      For the second player there are only n(k − 1) cards remaining, and dealing n cards from that we get:

      C(n(k − 1), n) = (n(k − 1))! / [ n! (n(k − 2))! ]

      For the third player:

      C(n(k − 2), n) = (n(k − 2))! / [ n! (n(k − 3))! ]

      and so on up to the final player

      C(n, n) = n! / [ n! 0! ] = 1

      Multiplying all these together we find that one of the terms in the denominator cancels with the numerator for the next player, giving a total number of possible deals of:

      deals(k, n) = (n.k)! / (n!)^k

      We can use this rather neat formula to find the required number of cards in the puzzle.

      This Python program runs in 74ms.

      from enigma import (irange, inf, factorial, arg, printf)

# number of players
k = arg(4, 0, int, prompt="number of players")

# multiple = number of seconds in a day (assumed constant)
m = arg(60 * 60 * 24, 1, int, prompt="multiple")

printf("[{k} players, m = {m}]")

# number of cards per player
for n in irange(1, inf):
  p = factorial(n * k) // pow(factorial(n), k)
  (d, r) = divmod(p, m)
  printf("{t} cards, {n} each, {p} deals, {d} days (+ {r}s)", t=n * k)
  if r == 0: break

      Solution: There are 28 cards in the special pack.

      Like

  • Unknown's avatar

    Jim Randell 2:05 pm on 28 February 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 457: [March-past] 

    From The Sunday Times, 15th February 1970 [link]

    The Duke of Teresa and Baron von Barin agreed to have a march-past of their respective men-at-arms to celebrate the wedding of the Duke’s son with Baron’s daughter. Each troop would march past in column of three (i.e. 3 men per file). The Duke has a few thousand men and the Baron about half as many.

    On the appointed day however, one of the Duke’s men could not go on parade and so the men were ordered into column of four, but 3 men were left over, whereupon they formed column of five and 4 men were left over, and this pattern of behaviour persisted as the Duke consecutively increased the number per file until all the men made an exact number of files.

    Meanwhile, the Baron has the same problem. One man reported sick and he consecutively increased the number per file and in column of four had 3 left over, column of five had 4 left over, etc., and eventually he reached a number per file that left no man spare.

    The Baron’s men proceeded to march past but the Duke was in a dilemma as each body of troops had a different number of men per file. He immediately solved this by ordering his men to form files with the same number of men as the Baron’s. This they did without any men being left over.

    How many men respectively had the Duke and the Baron on parade?

    This puzzle was originally published with no title.

    [teaser457]

     
    • Jim Randell's avatar

      Jim Randell 2:09 pm on 28 February 2019 Permalink | Reply

      The total number of men (including the indisposed man) must be able to form 3, 4, 5, … lines, if they are unable to form lines when one of them is missing, being short by one man.

      So, if we can form 3, 4, 5, … lines, the total number of men must be a multiple of 3, 4, 5, …, i.e. a multiple of lcm(3, 4, 5, …).

      This Python program looks for possible total numbers of men for the Duke D and the Baron B, and then finds the solution where the ratio D/B is closest to 2. I assumed the Duke had somewhere between 2,000 and 10,000 men. The program runs in 84ms.

      Run: [ @repl.it ]

      from itertools import count, combinations
from collections import defaultdict
from enigma import irange, mlcm, fdiv, Accumulator, printf

# suppose the number of men (n - 1) end dividing into k equal sized lines (k > 6)
# for all lowers numbers they are one short, so n must be a multiple of all lower numbers

# minimum/maximum numbers to consider for D
(m, M) = (2000, 10000)

# record lcms up to M by k
d = defaultdict(list)
for k in count(7):
  n = mlcm(*irange(2, k - 1))
  if n > M: break
  d[k] = n

# choose the result with D/B nearest to 2.0
r = Accumulator(fn=(lambda x, y: (x if abs(x - 2.0) < abs(y - 2.0) else y)))

# choose k's for D and B (kD < kB)
for (kD, kB) in combinations(sorted(d.keys()), 2):
  (mD, mB) = (d[kD], d[kB])
  # the Duke has "a few thousand men"
  for D in irange(mD, M, step=mD):
    if D < m: continue
    # (D - 1) is divisible by kD and kB
    if not ((D - 1) % kD == 0 and (D - 1) % kB == 0): continue
    # the Baron "about half as many"
    for B in irange(mB, M, step=mB):
      # (B - 1) is divisible by kB
      if (B - 1) % kB != 0: continue

      # output the solution (the numbers on parade are B-1, D-1)
      r.accumulate_data(fdiv(D, B), (D - 1, B - 1))
      printf("kD={kD} kB={kB}, mD={mD} mB={mB}, D={D} B={B} [D/B ~ {f:.3f}]", f=fdiv(D, B))

# output the solution
(D, B) = r.data
printf("parade: B={B}, D={D}")

      Solution: The Duke had 5159 men on parade. The Baron had 2519 men on parade.

      The ratio D/B is approximately 2.05.

      The Duke’s men are one man short of forming 3, 4, 5, 6 lines, but can form 7 lines (5159 = 737 × 7).

      The Baron’s men are one man short of forming 3, 4, 5, 6, 7, 8, 9, 10 lines, but can form 11 lines (2519 = 229 × 11).

      The Duke’s men can also form 11 lines (5159 = 469 × 11).

      There is a further solution where the Duke has 9779 men on parade (and the Baron still has 2519), but in this case the D/B ratio is 3.88.

      If we allow the Duke and the Baron more men then we can get the ratio closer to 2. There is a solution with D=60599 and B=30239, giving a ratio of 2.004, but it is hard to describe 60,600 men as “a few thousand”.

      Like

  • Unknown's avatar

    Jim Randell 1:55 pm on 27 February 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 456: [Chess competition] 

    From The Sunday Times, 8th February 1970 [link]

    The organiser of our lightning chess competition works on a strict timetable; he allows ten minutes a match, and makes no concession, in his planning, for draws or protracted games.

    Last year having five chess boards available for use, he arranged an American tournament (in which every competitor played one match against every other player). This year there would be exactly twice as many competitors as last year if one man hadn’t recently dropped out; and only one chessboard is available.

    He is therefore organising a knock-out competition: and he reports that his total time allocation will be precisely the same this year as it was last year.

    How many competitors are there this year?

    This puzzle was originally published with no title.

    [teaser456]

     
    • Jim Randell's avatar

      Jim Randell 1:56 pm on 27 February 2019 Permalink | Reply

      For n players there are T(n − 1) = n(n − 1)/2 matches in a tournament where every player plays every other player.

      So, with 5 simultaneous matches this would mean the number of allocated time slots would be:

      T = n(n − 1)/10

      In a knockout competition with m players there are (m − 1) games. (One player is eliminated in each game, until there is just one player left).

      So with only one match going on at a time, the total number of allocated time slots would be:

      T = m − 1

      And we are told that the number in the knockout tournament is m = 2n − 1.

      So, equating the times:

      m − 1 = n(n − 1)/10
      10(2n − 2) = n(n − 1)
      20(n − 1) = n(n − 1)
      n = 20 (as n ≠ 1)

      So last year there were 20 competitors. This year there are 39.

      Solution: This year there are 39 competitors.

      Liked by 1 person

  • Unknown's avatar

    Jim Randell 10:07 am on 26 February 2019 Permalink | Reply
    Tags: ,   

    Brain-Teaser 455: [Darts teams] 

    From The Sunday Times, 1st February 1970 [link]

    Every week we field a village darts team of 3 men, one from each of our pubs (Plough, Queen’s Head, and Roebuck). Altogether we can call on 14 players (whose names, luckily, start respectively with the first 14 letters of the alphabet): five of them frequent the Plough, six the Queen’s Head and seven the Roebuck, the apparent discrepancy is explained by the thirsts of Ernie, Fred, Len and Mark, all of whom are “two-pub men” and are thus eligible to represent either of their haunts.

    For over three years we have picked a different team each week and have just exhausted all 162 possible combinations. The last two teams were:

    Joe, Nigel, Charlie
    Charlie, Fred, Harry

    For the next seven weeks we are waiving the one-man-per-pub rule and have picked teams which have not so far played together. The are:

    Ernie, Len, Mark
    Ian, Fred, Alan
    Joe, Fred, George
    Len, Mark, Keith
    Fred, Keith, Nigel
    Ernie, Len, Nigel
    Ian, Joe, and one other to be picked from a hat on the night.

    Which darts players frequent the Roebuck?

    This puzzle was originally published with no title.

    [teaser455]

     
    • Jim Randell's avatar

      Jim Randell 10:07 am on 26 February 2019 Permalink | Reply

      I found two solutions for this puzzle, although with a slight change in the wording we could arrive at a unique solution.

      This Python program runs in 226ms.

      Run: [ @repl.it ]

      from itertools import product, permutations, combinations
from enigma import icount, unpack, printf

# all the players
players = "ABCDEFGHIJKLMN"

# players with dual allegiance
duals = "EFLM"

# the pubs
pubs = set("PQR")

# do the three sets form a team
def team(a, b, c):
  for (x, y, z) in permutations(pubs):
    if x in a and y in b and z in c: return True
  return False

# choose the "missing" pub for the duals
for s1 in product(pubs, repeat=4):
  (E, F, L, M) = (pubs.difference([x]) for x in s1)

  # ELM do not form a team
  if team(E, L, M): continue

  # choose (single) pubs for K, N
  for s2 in product(pubs, repeat=2):
    (K, N) = (set([x]) for x in s2)

    # KLM, FKN, ELN do not form teams
    if team(K, L, M) or team(F, K, N) or team(E, L, N): continue

    # CJN do form a team
    for s3 in permutations(pubs.difference(N)):
      (C, J) = (set([x]) for x in s3)

      # remaining assignments for given combinations, AGHI
      for s4 in product(pubs, repeat=4):
        (A, G, H, I) = (set([x]) for x in s4)

        # check the remaining combinations
        if not (team(C, F, H)) or team(A, F, I) or team(F, G, J): continue

        # which leaves B and D
        for s5 in product(pubs, repeat=2):
          (B, D) = (set([x]) for x in s5)

          table = (A, B, C, D, E, F, G, H, I, J, K, L, M, N)

          # P, Q, R should have teams of 5, 6, 7
          (P, Q, R) = (icount(table, (lambda x: p in x)) for p in 'PQR')
          if (P, Q, R) != (5, 6, 7): continue

          # count the total possible teams
          t = icount(combinations(table, 3), unpack(team))
          # there should be 162
          if t != 162: continue

          # find how many unused IJ? combinations there are
          ijs = list(p for (p, x) in zip(players, table) if p not in 'IJ' and not team(I, J, x))
          # there should be at least 2
          if len(ijs) < 2: continue

          # who plays for R
          Rs = list(p for (p, x) in zip(players, table) if 'R' in x)

          printf("Rs = {Rs}")
          printf("  A={A} B={B} C={C} D={D} E={E} F={F} G={G} H={H} I={I} J={J} K={K} L={L} M={M} N={N}")
          printf("  ijs={ijs}")
          printf()

      Solution: A, B, D, F, G, I, J are on the Roebuck team.

      This solution assumes that the final team member of the I+J+? team that is picked out of the hat can be any of the other players (i.e. I and J have never played together on a team before).

      The team allegiances in this case are:

      P = C, E, F, L, M [5]
      Q = E, H, K, L, M, N [6]
      R = A, B, D, F, G, I, J [7]

      However, I took the construction of final I+J+? team by picking a name out of a hat to require only that there should be more than one candidate to placed in the hat. If we have the following team allegiances:

      P = E, F, J, L, M [5]
      Q = E, H, K, L, M, N [6]
      R = A, B, C, D, F, G, I [7]

      Then this is also a solution to the puzzle. In this case the remaining possible partners for I+J+? are: A, B, C, D, F, G.

      Like

  • Unknown's avatar

    Jim Randell 8:14 am on 25 February 2019 Permalink | Reply
    Tags:   

    Teaser 2945: Infernal indices 

    From The Sunday Times, 3rd March 2019 [link] [link]

     

    The above is the size of the evil multitude in the last “Infernal indices” novel: “A deficit of daemons”.

    Spoiler Alert: At the end, the forces of good engage in the fewest simultaneous battles that prevent this evil horde splitting, wholly, into equal-sized armies for that number of battles. The entire evil horde is split into one army per battle, all equally populated, bar one which has a deficit of daemons, leading to discord and a telling advantage for the forces of good.

    How many battles were there?

    [teaser2945]

     
    • Jim Randell's avatar

      Jim Randell 8:15 am on 25 February 2019 Permalink | Reply

      Although the given number (let’s call it N) is large (it is 36,306 decimal digits), it is not too large for Python to cope with.

      We need to find the smallest whole number n that does not divide N. This is a much smaller number.

      The following Python program runs in 107ms.

      Run: [ @replit ]

      from enigma import (irange, inf, printf)

# construct the large number N
x = 6**6
N = (6**x) - (x**6)

# find the smallest number that does not divide it
for n in irange(2, inf):
  if N % n != 0:
    printf("n={n}")
    break

      Solution: There were 17 battles.

      (N mod 17) = 14, so the best the evil horde could manage would be 16 armies the same size and 1 army that had 3 fewer demons.

      Like

      • Jim Randell's avatar

        Jim Randell 4:46 pm on 2 March 2019 Permalink | Reply

        We can use the technique given in Enigma 1494 to compute the residues of large powers without the need to construct the huge numbers.

        The internal runtime for the following program is significantly less than the simple program given above (169µs vs. 961µs for the program given above). Although in practise both programs run instantaneously.

        from enigma import (irange, inf, printf)

# compute a^b mod n
def mpow(a, b, n):
  s = 1
  while b > 0:
    (b, r) = divmod(b, 2)
    if r > 0: s = (s * a) % n
    a = (a * a) % n
  return s

# N = 6^x - x^6
x = 6**6

# find the smallest number that does not divide it
for n in irange(2, inf):
  # find the residue N mod n
  if mpow(6, x, n) != mpow(x, 6, n):
    printf("n={n}")
    break

        Using the 3-argument version of the Python builtin [[ pow() ]] instead of [[ mpow() ]] gets the internal run time down to 40µs.

        For a manual solution we can use “the difference of two squares” method to express the number as:

        N = (6^36)(6^23310 − 1)(6^23310 + 1)

        From which it is easy to see that N will have a large number of factors of 2 and 3, and the fact that any power of 6 ends in a 6 means that the difference of two powers of 6 will end in a 0, so it will also be divisible by 5.

        So we can start by considering the primes: 7, 11, 13, 17, 19, 23 when looking for a suitable n.

        When evaluating mpow(6, x, n) the exponent can be reduced modulo (n − 1) (as n is prime) using Euler’s Theorem, and we can always evaluate the mpow(x, 6, n) in 3 iterations.

        In the end we don’t even need to consider all these primes in order to get the answer, so we can arrive at the solution with only a modest amount of calculation.

        Like

        • Jim Randell's avatar

          Jim Randell 12:14 pm on 3 March 2019 Permalink | Reply

          Here’s my complete manual solution:

          Writing:

          N = (6^36)(6^23310 − 1)(6^23310 + 1)

          We can start to examine divisors:

          For n = 2, it clearly divides the (6^36) term. Similarly for n = 3, 4, 6, 8, 9, 12, 16, ….

          For n = 5, it doesn’t divide the (6^36) term, so let’s consider the value of (6^23310 mod 5):

          5 is prime, so we can use Euler’s Theorem to reduce the exponent mod (p − 1):

          6^23310 mod 5 =
          6^(23310 mod (5 − 1)) mod 5 =
          6^2 mod 5 =
          36 mod 5 =
          1

          And if (6^23310 mod 5) = 1 then 5 divides the (6^23310 − 1) term.

          For n = 7, we consider the value of (6^23310 mod 7):

          6^23310 mod 7 =
          6^(23310 mod 6) mod 7 =
          6^0 mod 7 =
          1 mod 7 =
          1

          So 7 divides the (6^23310 − 1) term.

          For n = 10 we already know 2 divides the (6^36) term and 5 divides the (6^23310 − 1) term.

          For n = 11:

          6^23310 mod 11 =
          6^(23310 mod 10) mod 11 =
          6^0 mod 11 =
          1 mod 11 =
          1

          So 11 divides the (6^23310 − 1) term.

          For n = 13:

          6^23310 mod 13 =
          6^(23310 mod 12) mod 13 =
          6^6 mod 13 =
          12

          And 12 is equivalent to −1 mod 13, so 13 divides the (6^23310 + 1) term.

          For n = 14, we already have divisors of 2 and 7.

          For n = 15, we already have divisors of 3 and 5.

          for n = 17:

          6^23310 mod 17 =
          6^(23310 mod 16) mod 17 =
          6^14 mod 17 =
          (6^7 mod 17)^2 mod 17 =
          196 mod 17 =
          9

          9 is not equivalent to +1 or −1 mod 17, so 17 does not divide any of the terms, and gives our solution.

          Like

  • Unknown's avatar

    Jim Randell 7:39 am on 24 February 2019 Permalink | Reply
    Tags:   

    Teaser 2944: Happy families 

    From The Sunday Times, 24th February 2019 [link] [link]

    Teaser 2944

    Oliver lives with his parents, Mike and Nellie, at number 5. In each of numbers 1 to 4 lives a family, like his, with a mother, a father and one child. He tries listing the families in alphabetical order and produces the table above.

    However, apart from his own family, there is just one father, one mother and one child in the correct position. Neither Helen nor Beth lives at number 3 and neither Dave nor Ingrid lives at number 1. George’s house number is one less than Larry’s and Beth’s house number is one less than Carol’s. Apart from Oliver’s family, who are correctly positioned?

    [teaser2944]

     
    • Jim Randell's avatar

      Jim Randell 8:05 am on 24 February 2019 Permalink | Reply

      The wording of the puzzle is a little strange. I’m assuming the columns of the correct table are placed in order of house number (not “alphabetical” order), and the names of the fathers, mothers and children have been placed in the correct rows, just not necessarily in the correct columns. It seems like a reasonable assumption, and leads to a unique solution.

      I think I would have described the filling out of the original table like this:

      Oliver knows the names of the fathers, mothers and children, but is not sure who lives where, so he filled out each row in alphabetical order (from left to right), to give the table above. This correctly places his own family at house number 5. However …

      We can easily consider all the possibilities and eliminate those that do not satisfy the requirements.

      This Python program runs in 71ms.

      Run: [ @repl.it ]

      from itertools import permutations
from enigma import printf

# generate permutations that are only correct in 1 position
def generate(vs, k=1):
  for s in permutations(vs):
    if sum(x == v for (x, v) in zip(s, vs)) == k:
      yield s

# permute the Mums
for mums in generate('BEHK'):
  (m1, m2, m3, m4) = mums

  # neither H or B live at number 3
  if m3 == 'H' or m3 == 'B': continue

  # permute the kids
  for kids in generate('CFIL'):
    (k1, k2, k3, k4) = kids

    # I does not live at number 1
    if k1 == 'I': continue

    # B's house number is 1 less than C's
    if not (mums.index('B') + 1 == kids.index('C')): continue

    # permute the dads
    for dads in generate('ADGJ'):
      (d1, d2, d3, d4) = dads

      # D does not live at 1
      if d1 == 'D': continue

      # G's house number is 1 less than L's
      if not (dads.index('G') + 1 == kids.index('L')): continue

      printf("1 = {d1}+{m1}+{k1}; 2 = {d2}+{m2}+{k2}; 3 = {d3}+{m3}+{k3}; 4 = {d4}+{m4}+{k4}")

      Solution: George, Kate and Larry were also correctly placed.

      Like

  • Unknown's avatar

    Jim Randell 10:31 pm on 21 February 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 454: Nine holes 

    From The Sunday Times, 25th January 1970 [link]

    The card of the course at Triangle Park Golf Club gives the total length of the first 9 holes as 3,360 yards. Par for the individual holes (numbers 1 to 9 respectively) being 5, 4, 5, 4, 3, 4, 4, 3, 5 (total 37).

    Closer inspection of the card would show that the lengths of the holes (each a whole number of yards) are such that holes 1, 2 and 3 could each form a different side of the same right-angled triangle (Triangle A) and that other right-angled triangles could similarly be formed from lengths equal to those of holes 4, 5 and 6 (Triangle B); 7, 8 and 9 (Triangle C); 1, 4 and 7 (Triangle X); 2, 5 and 8 (Triangle Y) and 3, 6 and 9 (Triangle Z).

    Moreover, the total length of the three holes forming Triangle A, the total length of the three holes forming Triangle B, and the total length of the three holes forming Triangle C could similarly form the three sides of another right-angled triangle (Triangle ABC) while yet another right-angled triangle could similarly be constructed from the total lengths of the holes forming triangles X, Y and Z (Triangle XYZ).

    In triangle ABC the sides would be in the ratio 5:3:4.

    The length of each of the par 3 holes is between 150 and 250 yards (one being 168 yards); the par 4 holes are between 250 and 450 yards; and the par 5 holes are between 475 and 600 yards (one being 476 yards).

    What are the lengths of holes 2, 6 and 7?

    This puzzle is included in the book Sunday Times Brain Teasers (1974).

    [teaser454]

     
    • Jim Randell's avatar

      Jim Randell 10:33 pm on 21 February 2019 Permalink | Reply

      This Python program uses the [[ pythagorean_triples() ]] function from the enigma.py library to find appropriate triangles. (Originally written for Teaser 2910).

      Run: [ @replit ]

      from enigma import (defaultdict, pythagorean_triples, sq, printf)

# par for distance x
def par(x):
  if 149 < x < 251: return 3
  if 249 < x < 451: return 4
  if 474 < x < 601: return 5

# find triangles in the appropriate classes
ts = defaultdict(list)
for t in pythagorean_triples(600):
  k = tuple(par(x) for x in t)
  if None not in k:
    ts[k].append(t)

# select tuples from d[k] ordered by indices in ss
def select(d, k, ss):
  for x in d[k]:
    for s in ss:
      yield tuple(x[i] for i in s)

# does (x, y, z) form a right-angled triangle
def is_triangle(x, y, z):
  return sq(x) + sq(y) + sq(z) == 2 * sq(max(x, y, z))


# choose triangle A (5, 4, 5)
for (d1, d2, d3) in select(ts, (4, 5, 5), [(1, 0, 2), (2, 0, 1)]):

  # choose triangle B (4, 3, 4)
  for (d4, d5, d6) in select(ts, (3, 4, 4), [(1, 0, 2), (2, 0, 1)]):

    # choose triangle C (4, 3, 5)
    for (d7, d8, d9) in select(ts, (3, 4, 5), [(1, 0, 2)]):

      # check the total distance
      if not (sum((d1, d2, d3, d4, d5, d6, d7, d8, d9)) == 3360): continue      

      # check the other triangles
      (X, Y, Z) = ((d1, d4, d7), (d2, d5, d8), (d3, d6, d9))
      ABC = (d1 + d2 + d3, d4 + d5 + d6, d7 + d8 + d9)
      XYZ = (d1 + d4 + d7, d2 + d5 + d8, d3 + d6 + d9)
      if not all(is_triangle(*t) for t in (X, Y, Z, ABC, XYZ)): continue

      # check ABC is a 3, 4, 5 triangle
      if not (5 * min(ABC) == 3 * max(ABC)): continue

      # check the given pars
      if not (168 in (d5, d8) and 476 in (d1, d3, d9)): continue

      printf("d1={d1} d2={d2} d3={d3} / d4={d4} d5={d5} d6={d6} / d7={d7} d8={d8} d9={d9}")

      Solution: Hole 2 is 280 yards. Hole 6 is 357 yards. Hole 7 is 420 yards.

      There is only one solution:

      Hole 1 = 525 yards
      Hole 2 = 280 yards
      Hole 3 = 595 yards
      Hole 4 = 315 yards
      Hole 5 = 168 yards
      Hole 6 = 357 yards
      Hole 7 = 420 yards
      Hole 8 = 224 yards
      Hole 9 = 476 yards

      In fact if the triangle constraints for A, B, C, X, Y, Z, ABC, XYZ are satisfied, there is only a single solution even if the conditions for the total distance, ratios for ABC, and the specific distances for par 3 and 5 are ignored (lines 38, 47, 50).

      Like

      • Jim Randell's avatar

        Jim Randell 9:53 am on 23 February 2019 Permalink | Reply

        This puzzle is a good candidate for solving with a declarative set of constraints.

        Here is a MiniZinc model, that is run via the minizinc.py module. It solves the puzzle in 93ms.

        %#! python3 -m minizinc use_embed=1

% hole distances
{var("150..250", ["d5", "d8"])};  % par 3
{var("250..450", ["d2", "d4", "d6", "d7"])};  % par 4
{var("475..600", ["d1", "d3", "d9"])};  % par 5

% total distance
constraint d1 + d2 + d3 + d4 + d5 + d6 + d7 + d8 + d9 = 3360;

% right angled triangle constraints
predicate is_triangle(var int: a, var int: b, var int: c) =
  pow(a, 2) + pow(b, 2) + pow(c, 2) = pow(max([a, b, c]), 2) * 2;

constraint is_triangle(d1, d2, d3); % triangle A
constraint is_triangle(d4, d5, d6); % triangle B
constraint is_triangle(d7, d8, d9); % triangle C

constraint is_triangle(d1, d4, d7); % triangle X
constraint is_triangle(d2, d5, d8); % triangle Y
constraint is_triangle(d3, d6, d9); % triangle Z

constraint is_triangle(d1 + d2 + d3, d4 + d5 + d6, d7 + d8 + d9); % triangle ABC
constraint is_triangle(d1 + d4 + d7, d2 + d5 + d8, d3 + d6 + d9); % triangle XYZ

% ABC is a 3, 4, 5 triangle
constraint 5 * min([d1 + d2 + d3, d4 + d5 + d6, d7 + d8 + d9]) = 3 * max([d1 + d2 + d3, d4 + d5 + d6, d7 + d8 + d9]);

% some lengths are given
constraint d5 = 168 \/ d8 = 168;  % par 3
constraint d1 = 476 \/ d3 = 476 \/ d9 = 476;  % par 5

solve satisfy;

        Like

    • GeoffR's avatar

      GeoffR 11:07 am on 23 February 2019 Permalink | Reply 

      % A Solution in MiniZinc
include "globals.mzn";

% The nine holes are H1..H9 with Par distances as below:
var 475..600: H1; var 250..450: H2; var 475..600: H3; % Par 5, 4, 5

var 250..450: H4; var 150..250: H5; var 250..450: H6; % Par 4, 3, 4

var 250..450: H7; var 150..250: H8; var 475..600: H9; % Par 4, 3, 5

% Given hole distances
constraint H5 == 168 \/ H8 == 168;
constraint H1 == 476 \/ H3 == 476 \/ H9 == 476;

% Total distance for all holes
constraint H1 + H2 + H3 + H4 + H5 + H6 + H7 + H8 + H9 == 3360;

% Triangle A - H1 and H3 are both Par 5
constraint (H2 * H2 + H3 * H3 == H1 * H1) \/(H2 * H2 + H1 * H1 == H3 * H3);

% Triangle B - H4 and H6 are both Par 4
constraint( H5 * H5 + H6 * H6 == H4 * H4) \/ (H5 * H5 + H4 * H4 == H6 * H6);

% Triangle C - H9 is higher Par than H7 or H8
constraint (H7 * H7 + H8 * H8 == H9 * H9);

% % Triangle X - H1 is higher Par than H4 or H7
constraint H4 * H4 + H7 * H7 == H1 * H1;

% % Triangle Y - H2 is higher Par than H5 and H8
constraint (H8 * H8 + H5 * H5 = H2 * H2); 

% % Triangle Z - H3 and H9 are both Par 5
constraint (H6 * H6 + H9 * H9 == H3 * H3)\/ (H3 * H3 + H6 * H6 == H9 * H9);

% Triangle ABC sides
var 500..2000: A1;
var 500..2000: B1;
var 500..2000: C1;

constraint A1 = H1 + H2 + H3;
constraint B1 = H4 + H5 + H6;
constraint C1 = H7 + H8 + H9;

% Sides are in ratio 5:4:3
constraint A1 mod 5 == 0 \/ B1 mod  5 == 0 \/ C1 mod 5 == 0;
constraint A1 mod 4 == 0 \/ B1 mod  4 == 0 \/ C1 mod 4 == 0;
constraint A1 mod 3 == 0 \/ B1 mod  3 == 0 \/ C1 mod 3 == 0;

constraint (A1 * A1 + B1 * B1 == C1 * C1)
\/ (A1 * A1 + C1 * C1 == B1 * B1)
\/ (B1 * B1 + C1 * C1 == A1 * A1);

% Triangle XYZ sides
var 500..4000: X1;
var 500..4000: Y1;
var 500..4000: Z1;

constraint X1 = H1 + H4 + H7;
constraint Y1 = H2 + H5 + H8;
constraint Z1 = H3 + H6 + H9;

constraint (X1 * X1 + Y1 * Y1 == Z1 * Z1)
\/ (X1 * X1 + Z1 * Z1 == Y1 * Y1)
\/ (Y1 * Y1 + Z1 * Z1 == X1 * X1);

solve satisfy;

% H1 = 525; H2 = 280; H3 = 595;  Holes H1..H9 lengths (yd)
% H4 = 315; H5 = 168; H6 = 357;
% H7 = 420; H8 = 224; H9 = 476;
% Solution: H2 = 280, H6 = 357, H7 = 420
 
% A1 = 1400; B1 = 840; C1 = 1120;  Triangle ABC sides
% X1 = 1260; Y1 = 672; Z1 = 1428;  Triangle XYZ sides
% ----------
% ==========
% Finished in 237msec

      Like

  • Unknown's avatar

    Jim Randell 3:06 pm on 21 February 2019 Permalink | Reply
    Tags:   

    Teaser 2910: Living on the coast 

    From The Sunday Times, 1st July 2018 [link] [link]

    George and Martha are living on Pythag Island. Not surprisingly, it is in the shape of a right-angled triangle, and each side is an integral number of miles (under one hundred) long. Martha noticed that the area of the island (expressed in square miles) was equal to an exact multiple of its perimeter (expressed in miles). George was trying to work out that multiple, but gave up: “Please write down the appropriate digit” he said. “Impossible!” said Martha.

    What are the lengths of the coasts?

    [teaser2910]

     
    • Jim Randell's avatar

      Jim Randell 3:07 pm on 21 February 2019 Permalink | Reply

      Suppose the triangle has sides x, y, z where 0 < x < y < z < 100.

      We have:

      x² + y² = z²

      area = A = xy/2
      perimeter = P = x + y + z

      ratio = k = A/P = xy/2(x + y + z)

      Consider:

      (x + y + z)(x + y − z) = (x² + y² − z²) + 2xy = 2xy
      ⇒ xy = (x + y + z)(x + y − z)/2

      So:

      k = (x + y + z)(x + y − z)/4(x + y + z) = (x + y − z)/4

      This Python program uses the [[ pythagorean_triples() ]] function from the enigma.py library to find appropriate triangles. The routine itself uses the technique described at [ @Wikipedia ]. It runs in 71ms.

      Run: [ @repl.it ]

      from enigma import (pythagorean_triples, div, printf)

# consider all pythagorean triples with hypotenuse less than 100
for (x, y, z) in pythagorean_triples(99):
  # find ratio of area to perimeter, an integer with more than 1 digit
  k = div(x + y - z, 4)
  if k is None or k < 10: continue
  # output solution
  printf("x={x} y={y} z={z}, k={k}")

      Solution: The lengths of the coasts (in miles) are 65, 72, 97.

      Like

    • GeoffR's avatar

      GeoffR 1:26 pm on 22 February 2019 Permalink | Reply 

      % A Solution in MiniZinc
include "globals.mzn";

var 1..99:X; var 1..99:Y; var 1..99:Z;  % 3 sides of triangle
constraint X < Y /\ Y < Z;

var 1..5000: area;
var 3..297: perim;
var 2..50:ratio;

constraint X * X + Y * Y == Z * Z;
constraint area = (X * Y) div 2;
constraint perim = X + Y + Z;

constraint ratio = area/perim /\ ratio > 9;  % Single digit ratio is impossible (Martha)

solve satisfy;

% X = 65;  Y = 72; Z = 97;
% area = 2340; perim = 234; ratio = 10;
% ----------
% ==========
% Finished in 267msec

      Like

  • Unknown's avatar

    Jim Randell 8:23 am on 19 February 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 453: [Square field] 

    From The Sunday Times, 18th January 1970 [link]

    Old Hassan has been at it again! He has thought of a new way of parcelling out his square field between his three sons (Rashid and Riad the twins, and Ali).

    Calling the three into his tent on the eve of Ramadhan, he addressed them thus:

    “My sons, as I am now 74 years of age, I desire that you shall come into part of your inheritance. I have therefore divided my field into four triangular plots such that we each have an area proportional to our age.”

    “To avoid any friction regarding upkeep of fences I have also arranged that none of you shall have a common boundary line with either of your brothers. Further, knowing Rashid and Riad’s objection to being too strongly identified with each other, I have arranged for their plots to be of a different shape.”

    The twins are 11 years older than Ali.

    How old is Ali?

    This puzzle was originally published with no title.

    [teaser453]

     
    • Jim Randell's avatar

      Jim Randell 8:40 am on 19 February 2019 Permalink | Reply

      Let’s suppose that the field is a unit square, and that Ali’s age is a (a whole number).

      Then the total of all the ages is d = a + 2(a + 11) + 74 = 3a + 96, so the corresponding areas of the triangular plots are:

      Hassan = 74 / d
      Rashid, Riad = (a + 11) / d
      Ali = a / d

      Splitting two sides of the square at x, y ∈ (0, 1) we have:

      For Rashid and Riad:

      (a + 11) / d = x / 2
      (a + 11) / d = (1 − x)(1 − y) / 2

      and for Ali:

      a / d = y / 2

      So, given a value for a we have:

      d = 3a + 96
      x = 2(a + 11) / d
      y = 2a / d
      (1 − x)(1 − y) = x

      So we can consider whole number values for a that give a solution to these equations:

      from enigma import (Rational, irange, printf)

Q = Rational()

# choose a value for a
for a in irange(1, 50):
  # total age
  d = 3 * a + 96
  # calculate x and y
  (x, y) = (Q(2 * a + 22, d), Q(2 * a, d))
  # do they satisfy the third equation?
  if x == (1 - x) * (1 - y):
    printf("a={a} [x={x}, y={y}]")

      Solution: Ali is 24.

      A viable solution to the equations is:

      a = 24, x = 5/12, y = 2/7

      so the fields look like this:

      Hassan has 37/84 (≈ 44.1%) of the field. The twins have 5/24 (≈ 20.8%) each. And Ali has the remaining 1/7 (≈ 14.3%).

      There is a further solution to the equations at:

      a = −208/5, x = 17/8, y = 26/9

      but this doesn’t give a viable answer.

      Like

    • John Crabtree's avatar

      John Crabtree 4:24 pm on 16 April 2021 Permalink | Reply

      The four equations involving a, d, x and y can be reduced to:
      5a^2 + 88a -4992 = 0. Then a = -8.8 +/- 32.8

      Like

  • Unknown's avatar

    Jim Randell 8:25 am on 17 February 2019 Permalink | Reply
    Tags:   

    Teaser 2943: Keep your balance 

    From The Sunday Times, 17th February 2019 [link] [link]

    George and Martha have a set of a dozen balls, identical in appearance but each has been assigned a letter of the alphabet, A, B, …, K, L and each is made of a material of varying density so that their weights in pounds are 1, 2…. 11, 12 but in no particular order. They have a balance and the following weighings were conducted:

    (1) {A, C, I} vs {G, J, L}

    (2) {A, H, L} vs {G, I, K}

    (3) {B, I, J} vs {C, F, G}

    (4) {B, D, I} vs {E, G, H}

    (5) {D, F, L} vs {E, G, K}

    On all five occasions, there was perfect balance and the total of the threesome in each was a different prime number, that in (1) being the smallest and progressing to that in (5) which was the largest.

    In alphabetical order, what were the weights of each of the twelve balls?

    [teaser2943]

     
    • Jim Randell's avatar

      Jim Randell 8:38 am on 17 February 2019 Permalink | Reply

      We can solve this using the [[ SubstitutedExpression() ]] solver from the enigma.py library. Treating each symbol as standing for a different non-zero base 13 digit.

      This run-file executes in 332ms.

      Run: [ @replit ]

      #! python3 -m enigma -rr

SubstitutedExpression

--base="13"
--digits="1-12"

--template=""

# the weighings
"A + C + I == G + J + L"
"A + H + L == G + I + K"
"B + I + J == C + F + G"
"B + D + I == E + G + H"
"D + F + L == E + G + K"

# the total weights are prime
"is_prime(A + C + I)"
"is_prime(A + H + L)"
"is_prime(B + I + J)"
"is_prime(B + D + I)"
"is_prime(D + F + L)"

# the primes are in increasing order
"A + C + I < A + H + L"
"A + H + L < B + I + J"
"B + I + J < B + D + I"
"B + D + I < D + F + L"

      Solution: The weights of the balls are: A=4, B=8, C=5, D=9, E=12, F=11, G=1, H=6, I=2, J=7, K=10, L=3.

      Like

    • Jim Randell's avatar

      Jim Randell 10:19 am on 17 February 2019 Permalink | Reply

      There are six symbols in the first weighing, and together they must sum to twice the smallest prime, so the smallest possible value for the smallest prime is:

      divc(1 + 2 + 3 + 4 + 5 + 6, 2) = 11

      Similarly the largest possible value for the largest prime is:

      divf(7 + 8 + 9 + 10 + 11 + 12, 2) = 28

      So the list of possible primes is: 11, 13, 17, 19, 23.

      There are only 5 primes on the list, so the weighings give us 10 equations in 12 variables.

      Again using the [[ SubstitutedExpression() ]] solver, this run file executes in 131ms.

      Run: [ @replit ]

      #! python3 -m enigma -rr

SubstitutedExpression

--base="13"
--digits="1-12"
--template=""

# the weighings
"A + C + I == 11"
"G + J + L == 11"
"A + H + L == 13"
"G + I + K == 13"
"B + I + J == 17"
"C + F + G == 17"
"B + D + I == 19"
"E + G + H == 19"
"D + F + L == 23"
"E + G + K == 23"

      Like

      • Jim Randell's avatar

        Jim Randell 7:30 am on 5 March 2019 Permalink | Reply

        Here’s a manual solution:

        Given the 10 equations above and adding the equation:

        A + B + C + D + E + F + G + H + I + J + K + L = 78

        Gives us 11 equations in 12 variables.

        These can be solved to give the other values in terms of B:

        A = 44 − 5B
        C = 4B − 27
        D = 25 − 2B
        E = B + 4
        F = 27 − 2B
        G = 17 − 2B
        H = B − 2
        I = B − 6
        J = 23 − 2B
        K = B + 2
        L = 4B − 29

        And all the variables have to take on different values from 1 to 12.

        From I and E, we have:

        B ≥ 7
        B ≤ 8

        So B is 7 or 8.

        The formula for F gives:

        B = 7 → F = 13
        B = 8 → F = 11

        The first of these is not possible so B = 8, giving:

        A = 4; B = 8; C = 5; D = 9; E = 12; F = 11; G = 1; H = 6; I = 2; J = 7; K = 10; L = 3

        Like

        • Frits's avatar

          Frits 3:43 pm on 30 July 2020 Permalink | Reply

          @Jim

          “These can be solved to give the other values in terms of B:” –> easier said than done.

          Can your Enigma functions generate these 11 equations in terms of B?

          Like

          • Jim Randell's avatar

            Jim Randell 5:03 pm on 30 July 2020 Permalink | Reply

            You could use SymPy to reduce the set of equations down using a single parameter.

            But you can also choose a value for B and then you have 12 equations in 12 variables, which you can solve using the [[ Matrix.linear() ]] solver in enigma.py.

            Run: [ @replit ]

            from enigma import (Matrix, irange, join, printf)

eqs = [
  # A  B  C  D  E  F  G  H  I  J  K  L
  ( 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ), # B = ?
  ( 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0 ), # (A, C, I) = 11
  ( 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1 ), # (G, J, L) = 11
  ( 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1 ), # (A, H, L) = 13
  ( 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0 ), # (G, I, K) = 13
  ( 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0 ), # (B, I, J) = 17
  ( 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0 ), # (C, F, G) = 17
  ( 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0 ), # (B, D, I) = 19
  ( 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0 ), # (E, G, H) = 19
  ( 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1 ), # (D, F, L) = 23
  ( 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0 ), # (E, G, K) = 23
  ( 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ), # (A ... L) = 78
]

# the collection of weights
weights = set(irange(1, 12))

# choose a value for B
for B in weights:
  # evaluate the equations
  s = Matrix.linear(eqs, [B, 11, 11, 13, 13, 17, 17, 19, 19, 23, 23, 78])
  # the answers should be the values in weights
  if not (set(s) == weights): continue
  # output solution
  printf("{s}", s=join((join((w, '=', v)) for (w, v) in zip("ABCDEFGHIJKL", s)), sep=" "))

            Like

            • Frits's avatar

              Frits 8:27 pm on 30 July 2020 Permalink | Reply

              Thanks, I ran the following code at [ https://live.sympy.org/ ]:

              from sympy import linsolve, symbols, linear_eq_to_matrix
A,B,C,D,E,F,G,H,I,J,K,L = symbols("A,B,C,D,E,F,G,H,I,J,K,L", integer=True)
variables = [A,C,D,E,F,G,H,I,J,K,L,B]

def my_solver(known_vals):

    eqns = [A + C + I - 11,
G + J + L - 11,
A + H + L - 13,
G + I + K - 13,
B + I + J - 17,
C + F + G - 17,
B + D + I - 19,
E + G + H - 19,
D + F + L - 23,
E + G + K - 23,
A + B + C + D + E + F + G + H + I + J + K + L - 78]

    # add the known variables to equation list
    for x in known_vals.keys():
        eqns.append(x - (known_vals[x]))

    X, b = linear_eq_to_matrix(eqns, variables)
    solution = linsolve((X, b), variables)

    return solution


my_solver({})

              Output:

              {(44−5B, 4B−27, 25−2B, B+4, 27−2B, 17−2B, B−2, B−6, 23−2B, B+2, 4B−29, B)}

              Like

    • GeoffR's avatar

      GeoffR 3:38 pm on 19 February 2019 Permalink | Reply

      I assumed you did not want the answer shown?

      % A Solution in MiniZinc
include "globals.mzn";

% The dozen balls weigh between 1 and 12 pounds
var 1..12:A; var 1..12:B; var 1..12:C; var 1..12:D; var 1..12:E;
var 1..12:F; var 1..12:G; var 1..12:H; var 1..12:I; var 1..12:J;
var 1..12:K; var 1..12:L;

% All the balls A..K are of different weight
constraint all_different([A, B, C, D, E, F, G, H, I, J, K, L]);

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

% The weighing equations
constraint (A + C + I == G + J + L) 
/\ (A + H + L == G + I + K)
/\ (B + I + J == C + F + G) 
/\ (B + D + I == E + G + H)
/\ (D + F + L == E + G + K);

% the total weights in the equations are all prime numbers
constraint is_prime(A + C + I) /\ is_prime(A + H + L) /\ is_prime(B + I + J) 
/\ is_prime(B + D + I) /\ is_prime(D + F + L);

% The five primes are in increasing order
constraint (A + C + I) < (A + H + L) /\  (A + H + L) < (B + I + J)
/\ (B + I + J) < (B + D + I) /\  (B + D + I) < (D + F + L);

solve satisfy;

output [ " A = "++ show(A) ++ ", B = " ++ show(B) ++ ", C = " ++ show(C) ++ "\n"
++ " D = " ++ show(D) ++ ", E = " ++ show(E) ++ ", F = " ++ show(F) ++ "\n"
++ " G = " ++ show(G) ++ ", H = " ++ show(H) ++ ", I = " ++ show(I) ++ "\n"
++ " J = " ++ show(J) ++ ", K = " ++ show(K) ++ ", L = " ++ show(L) ];

      Like

  • Unknown's avatar

    Jim Randell 10:53 pm on 16 February 2019 Permalink | Reply
    Tags:   

    Brain-Teaser 452: [Coloured dice] 

    From The Sunday Times, 11th January 1970 [link]

    Three dice, numbered from 1 to 6, each had 2 blue, 2 red and 2 white faces. No number appeared in the same colour twice.

    When  Bill threw the dice they totalled 14, showing 2 white faces and 1 red. A second throw showed each die the same colour as before, but each had a different number totalling 15. On one die Bill noticed two similarly coloured sides added to 9.

    The telephone rang and Bill went to answer it, leaving little Tommy by himself, who then picked up one of the dice and painted a red face white and a white face red. He then threw the dice twice producing firstly 3 whites and then 3 reds, each throw showing one of the repainted sides.

    Tommy totalled the 3 whites to 8 and the 3 reds to 7. Unfortunately, sums were not his strong point and though he sometimes got them right, quite often he was 1 out, and sometimes even 2 out. However, he was certain that the 3 whites totalled more than the 3 reds.

    What numbers did Tommy repaint (1) red, and (2) white?

    This puzzle was originally published with no title.

    [teaser452]

     
    • Jim Randell's avatar

      Jim Randell 10:55 pm on 16 February 2019 Permalink | Reply

      This Python program runs in 200ms. (Internal runtime is 126ms).

      from enigma import (namedtuple, irange, partitions, subsets, cproduct, printf)

# represent a die
Die = namedtuple("Die", "B R W")

# collect possible dice
dice = set(Die(*s) for ns in partitions(irange(1, 6), 2) for s in subsets(ns, size=len, select='P'))

# can we make the numbers in <ns> from the opposite faces of <fs>
def make(ns, fs):
  ns = set(ns)
  (f1, f2, f3) = fs
  return any(ns.issubset([f1[i1] + f2[i2] + f3[i3], f1[1 - i1] + f2[1 - i2] + f3[1 - i3]]) for (i1, i2, i3) in subsets((0, 1), size=3, select='M'))

# change tuple <t> so index <i> is <v>
def change(t, i, v):
  t = list(t)
  t[i] = v
  return tuple(t)

# possible white, red values for Tommy (white > red)
ts = set((w, r) for w in irange(6, 10) for r in irange(5, w - 1))

# choose a set of three dice
for (d1, d2, d3) in subsets(dice, size=3):

  # "no number appeared in the same colour twice"
  if not all(len(set(x1 + x2 + x3)) == 6 for (x1, x2, x3) in zip(d1, d2, d3)): continue

  # "on one of the dice two similary coloured sides add up to 9"
  if not any(sum(x) == 9 for x in d1 + d2 + d3): continue

  # can we make a R, W, W combination that sums to 14 and 15
  if not any(make((14, 15), fs) for fs in [(d1.R, d2.W, d3.W), (d1.W, d2.R, d3.W), (d1.W, d2.W, d3.R)]): continue

  # Tommy picked one of the dice and swapped the colours of a red number and a white number, suppose he repaints t1
  for (t1, t2, t3) in [(d1, d2, d3), (d2, d1, d3), (d3, d1, d2)]:
    # choose red/white faces of t1 to repaint
    for (r, w) in subsets((0, 1), size=2, select='M'):
      # the repainted die
      t = Die(B=t1.B, R=change(t1.R, r, t1.W[w]), W=change(t1.W, w, t1.R[r]))

      # record possible 3 white/red scores (including the repainted faces)
      ws = set(sum(s) for s in cproduct([[t.W[w]], t2.W, t3.W]))
      rs = set(sum(s) for s in cproduct([[t.R[r]], t2.R, t3.R]))

      # do these create feasible values?
      vs = ts.intersection(cproduct([ws, rs]))
      if vs:
        printf("dice: {t1} {t2} {t3}")
        printf("  red = {r}, white = {w} -> {vs}", r=t.R[r], w=t.W[w])

      Solution: (1) Tommy repainted 3 to red, (2) and 4 to white.

      There is only one possible set of dice:

      die 1: blue = 1,5; red = 2,4; white = 3,6
      die 2: blue = 2,6; red = 1,3; white = 4,5
      die 3: blue = 3,4; red = 5,6; white = 1,2

      The throws for Bill, showing 2 white faces and 1 red:

      throw 1: die 1 = white 3; die 2 = white 5; die 3 = red 6; total = 14
      throw 2: die 1 = white 6; die 2 = white 4; die 3 = red 5; total = 15

      Both die 1 and die 2 have white faces that sum to 9.

      Tommy chooses die 1 and repaints 3 from white to red, and 4 from red to white, giving (with the repainted faces indicated in braces):

      die 1: blue = 1,5; red = 2,{3}; white = {4},6
      die 2: blue = 2,6; red = 1,3; white = 4,5
      die 3: blue = 3,4; red = 5,6; white = 1,2

      Tommy then throws 3 whites (totalling 6 to 10), and then 3 reds (totalling 5 to 9), the white total being more than the red total.

      throw 1: die 1 = {white 4}; die 2 = white 4; die 3 = white 2; total = 10
      throw 2: die 1 = {red 3}; die 2 = red 1; die 3 = red 5; total = 9

      So Tommy’s calculation of 8 for the reds and 7 for the whites were both out by 2.

      Like

  • Unknown's avatar

    Jim Randell 7:48 am on 14 February 2019 Permalink | Reply
    Tags: by: Philip Spencer   

    Brain-Teaser 451: [Soccer tournament] 

    From The Sunday Times, 4th January 1970 [link]

    The final weeks of the University soccer tournament proved unusually exciting this year. With only four games left, Arts, Biology and Classics had drawn clear and were battling for the honours. The position then was that Biology were top with 32 points, Arts second also with 32 points, and Classics third with 30 points.

    All three teams had scored 70 goals, but Classics had conceded 35 goals giving them an inferior goal average (i.e. ratio of goals for and against) to both Arts and Biology.

    The remaining games required Classics to play both Biology and Arts, who each also had to play one game against another team lower in the division.

    It turned out that after the four games, Arts had failed to score and, although having scored five of the total of ten goals scored, Classics worsened their goal average.

    Consequently, all three teams finished with the same number of points, but Arts won the league from Classics on goal average.

    What were the final goals for and against for each team?

    A prize of £3 was offered for this puzzle.

    This puzzle was originally published with no title.

    [teaser451]

     
    • Jim Randell's avatar

      Jim Randell 9:25 am on 14 February 2019 Permalink | Reply

      Here’s a programmed solution in Python that uses the [[ Football() ]] helper class from the enigma.py library. It runs in 108ms.

      Run: [ @replit ]

      # there are four remaining matches: AC, BC, AX, BX

from enigma import (Football, irange, cproduct, subsets, printf)

# the scoring system
football = Football(games='wdl', points=dict(w=2, d=1))

# possible scores (no team scored more than 5)
scores = dict()
scores['w'] = set((x, y) for x in irange(1, 5) for y in irange(0, x - 1))
scores['d'] = set((x, x) for x in irange(0, 5))
scores['l'] = set((y, x) for (x, y) in scores['w'])

# possible outcomes in the remaining 4 games
for (AC, BC, AX, BX) in football.games(repeat=4):

  # table for A, B, C in these four games
  A = football.table([AC, AX], [0, 0])
  B = football.table([BC, BX], [0, 0])
  C = football.table([AC, BC], [1, 1])

  # A and B got the same number of additional points, and C got 2 more
  if not (A.points == B.points == C.points - 2): continue

  # choose scores for the 4 matches
  for (sAC, sBC, sAX, sBX) in cproduct(scores[x] for x in (AC, BC, AX, BX)):

    # in total 10 goals are scored in the 4 matches
    if not (sum(x + y for (x, y) in (sAC, sBC, sAX, sBX)) == 10): continue

    # goals for/against A, B, C
    (fA, aA) = football.goals([sAC, sAX], [0, 0])
    (fB, aB) = football.goals([sBC, sBX], [0, 0])
    (fC, aC) = football.goals([sAC, sBC], [1, 1])

    # A failed to score, C scored 5 goals
    if not (fA == 0 and fC == 5): continue

    # C worsened their goal average: (70 + fC) / (35 + aC) < 70 / 35
    if not (fC < 2 * aC): continue

    # A, B started off with 70 goals for and a, b goals against
    # their goal average is better than 70/35, so b < a < 35
    for (b, a) in subsets(irange(1, 34), size=2):

      # and we ended with A having the best goal average, then C, then B
      if not ((70 + fA) * (35 + aC) > (70 + fC) * (a + aA)): continue
      if not ((70 + fC) * (b + aB) > (70 + fB) * (35 + aC)): continue

      # output solution
      printf("AC={AC}:{sAC} BC={BC}:{sBC} AX={AX}:{sAX} BX={BX}:{sBX} / a={a} b={b}")
      printf("A={A} f={fA} a={aA}", fA=70 + fA, aA=a + aA)
      printf("B={B} f={fB} a={aB}", fB=70 + fB, aB=b + aB)
      printf("C={C} f={fC} a={aC}", fC=70 + fC, aC=35 + aC)
      printf()

      Solution: A: for = 70, against = 35; B: for = 73, against = 37; C: for = 75, against = 38.

      There are two scenarios. Assuming 2 points for a win and 1 for a draw, we start with:

      A: points = 32; goals for = 70; goals against = 33; avg = 70/33 = 2.121
      B: points = 32; goals for = 70; goals against = 32; avg = 70/32 = 2.188
      C: points = 30; goals for = 70; goals against = 35; avg = 70/35 = 2.000

      Then the remaining four matches are played. The outcomes are one of:

      (1) A vs C = 0-0; B vs C = 3-5; A vs X = 0-2; B vs X = 0-0
      (2) A vs C = 0-2; B vs C = 3-3; A vs X = 0-0; B vs X = 0-2

      Each of these scenarios has 10 goals scored, none of them by A and 5 of them by C.

      The final table then becomes:

      A: points = 33; goals for = 70; goals against = 35; avg = 70/35 = 2.000
      B: points = 33; goals for = 73; goals against = 37; avg = 73/37 = 1.973
      C: points = 33; goals for = 75; goals against = 38; avg = 75/38 = 1.974

      Like

  • Unknown's avatar

    Jim Randell 11:54 am on 12 February 2019 Permalink | Reply
    Tags:   

    Teaser 2935: A palindrome 

    From The Sunday Times, 23rd December 2018

    → See: [ Teaser 2935: A palindrome at Enigmatic Code ]

    [teaser2935]

     
  • Unknown's avatar

    Jim Randell 11:52 am on 12 February 2019 Permalink | Reply
    Tags:   

    Teaser 2907: Combinatorial cards 

    From The Sunday Times, 10th June 2018

    → See: [ Teaser 2907: Combinatorial cards at Enigmatic Code ]

    [teaser2907]

     

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