From The Sunday Times, 5th November 1978 [link]
The Soccotia World Cup soccer squad consists of two goalkeepers and sufficient outfield players to ensure that each of these latter play in exactly two games, of the three scheduled. Each player is also allocated a consecutive number starting with one, including the goalkeepers. Each outfield player is equally capable of playing in defence and attack, but will not change from one role to another during the same game.
The manager selects the teams for each match such that the sum of the numbers of the outfield players remains constant for each game, and such that the sum of the numbers of the four defenders equals one half of the sum of the six attackers.
In each game Soccotia scored one goal, each of which was scored by an attacking player, three different players scoring one goal each.
In the first game, the goal resulted from a five man move involving four passes. The move was started by a defender, who was the only defender in the move. Each pass was made to a player whose number was a fixed integer multiple of the passer.
In the second game, the same four players were chosen to play in defence as in the first game, and the same defender started the five man/four pass move which led to the goal scored in this game. This time however the number of each player receiving the ball exceeded the number of the passer by a fixed integer.
In the third game the five man/four pass move leading to the goal followed the same pattern as that scored in the second game.
What were the numbers of the goalscorers in match order?
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There are now 900 Teaser puzzles available on the site, so I thought it appropriate to post Teaser 900.
[teaser900]
Frits and 
Hugo are discussing. 
GeoffR, and 
Tony Brooke-Taylor, and 
Pete Good, and 
S2T2 
Jim Randell 5:10 pm on 21 July 2023 Permalink |
This Python program runs in 65ms. (Internal run time is 4ms).
from enigma import (irange, subsets, gcd, is_duplicate, printf) # possible ages (no repeated digits) ages = list(n for n in irange(65, 99) if n % 11 > 0) # find pairs of ages with different digits that are coprime pairs = list((x, y) for (x, y) in subsets(ages, 2) if gcd(x, y) == 1 and not is_duplicate(x, y)) # extend the pairs to triples triples = list() for (x, y) in pairs: for z in ages: if z > y and gcd(x, z) == 1 and gcd(y, z) == 1 and not is_duplicate(x, y, z): triples.append((x, y, z)) # choose <k> more pairs from <ps> def choose(k, ps, ns, rs): if k == 0: yield rs else: # choose the next pair for (i, (a, b)) in enumerate(ps): # check numbers are all different if a in ns or b in ns: continue # and are pairwise coprime if not all(gcd(a, n) == 1 and gcd(b, n) == 1 for n in ns): continue # solve for remaining pairs yield from choose(k - 1, ps[i + 1:], ns + (a, b), rs + [(a, b)]) # solve the puzzle def solve(): # consider back rows in order (largest sum to smallest sum) for r1 in sorted(triples, key=sum, reverse=1): # find 4 more pairs to go with these yield from choose(4, pairs, r1, [r1]) # find the first solution for rs in solve(): printf("rows = {rs}") break # we only need one solutionSolution: The age of the eldest passenger on the back row is 95.
The ages in the rows are:
The sum of the ages in the back row being 254.
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Frits 1:52 pm on 24 July 2023 Permalink |
from math import gcd diffdgts = lambda *ns: len(set(s := ''.join(str(n) for n in ns))) == len(s) # check that element <a> is coprime with elements in <s> cp = lambda a, s: all(gcd(a, x) == 1 for x in s) # decompose <t> into <k> increasing numbers from <ns> with all different digits def decompose(t, k, ns, s=[]): if k == 1: if t in ns and t > s[-1] and diffdgts(*(s_ := s + [t])): yield s_ else: for (i, n) in enumerate(ns): if not diffdgts(*(s_ := s + [n])): continue yield from decompose(t - n, k - 1, ns[i + 1:], s_) # A B # C D # E F # G H # I J K sols = set() # check sum of ages from (90+80+70+6+5+3) to (80+70+60+0+1+3) for soa in range(254, 213, -1): # back row for I, J, K in decompose(soa, 3, [x for x in range(65, 99) if x % 11]): # check if I, J and K are coprime if not cp(I, [J]): continue if not cp(K, [I, J]): continue for A in range(65, 85): if not cp(A, sA := [I, J, K]): continue for B in range((A // 10 + 1) * 10, 99): if not cp(B, sB := sA + [A]) or not diffdgts(A, B): continue for C in range(A + 1, 86): if not cp(C, sC := sB + [B]): continue for D in range((C // 10 + 1) * 10, 99): if not cp(D, sD := sC + [C]) or not diffdgts(C, D): continue for E in range(C + 1, 87): if not cp(E, sE := sD + [D]): continue for F in range((E // 10 + 1) * 10, 99): if not cp(F, sF := sE + [E]) or not diffdgts(E, F): continue for G in range(E + 1, 88): if not cp(G, sG := sF + [F]): continue for H in range((G // 10 + 1) * 10, 99): if not cp(H, sG + [G]) or not diffdgts(G, H): continue # don't assume there is a unique answer sols.add(K) if sols: print(f"answer: {', '.join(str(s) for s in sols)} in years") break # we don't need to check a lower sum of agesLikeLike