From The Sunday Times, 11th June 1961 [link]
It is a lovely evening in August, and the final Test of the 1990 series has reached its climax. The rubber stands at two games each, the last ball of the last over of the match is now to come. Australia are batting and need six runs for victory. Whacker, their wicket-keeper, the last man in, has taken his stand. with his captain’s words ringing in his ears, “Six or bust!”.
Bumper, the England bowler, has just taken the previous two wickets in succession, With a dim opinion of Whacker’s batting, he feels sure that a fast straight one will not only give England the Ashes, but will give him his hat-trick and his seventh wicket of the match.
In a breathless hush he delivers a fast straight one. Striding forward like a Titan, Whacker mows …
The records of this match are scanty. The Australian total in two innings was 490. Except for one man run out in the first innings, their casualties fell to bowlers. The five England bowlers had averages of 14, 20, 25 (Bumper), 33, 43, each with a differing number of wickets.
How many wickets did each take and who won the Ashes?
This is another puzzle that I originally couldn’t find, but with a bit more searching of the archive I was able to unearth it.
This puzzle completes the archive of Teaser puzzles from 1961 (when the first numbered puzzles appeared).
This is the first of the numbered Teaser puzzles set by Charles Skeffington Quin (although it is credited to R. Skeffington Quinn), and was re-published as Brainteaser 1093 on 17th July 1983 to celebrate Mr Skeffington Quin’s 100th birthday.
[teaser15]
GeoffR and 
Linda, 
John Crabtree, and 
Frits, and 
S2T2 
Jim Randell 7:03 pm on 27 August 2021 Permalink |
This Python program looks for candidate solutions, but doesn’t show that the tickets can be assigned to tables in the required fashion. However there is only one possible answer, so this is not necessary. (However, it is possible to assign tickets in both cases – see [link]).
It runs in 234ms.
from enigma import (mgcd, mlcm, subsets, irange, intersect, divisors_pairs, tri, Primes, express, filter_unique, unpack, printf) # generate candidate solutions def generate(): # primes less than 150 primes = Primes(150) # look for gcds and lcms of 3 2-digit numbers (gcds, lcms) = (set(), set()) for ns in subsets(irange(10, 99), size=3): gcds.add(mgcd(*ns)) lcms.add(mlcm(*ns)) # consider number of tickets, n for n in intersect([gcds, lcms]): # total sum of tickets t = tri(n) # consider k tables, with q people per table for (k, q) in divisors_pairs(n, every=1): if k < 3 or q < 3: continue # consider 2 primes, for the ticket sums for (p1, p2) in subsets(primes, size=2): # express the total using 2 of the primes for (q1, q2) in express(t, (p1, p2), min_q=1): if q1 + q2 == k: yield (n, t, k, q, p1, p2, q1, q2) # look for candidate solutions, where p1 is ambiguous by p2 ss = filter_unique( generate(), unpack(lambda n, t, k, q, p1, p2, q1, q2: p2), unpack(lambda n, t, k, q, p1, p2, q1, q2: p1), ) for (n, t, k, q, p1, p2, q1, q2) in ss.non_unique: printf("n={n}: t={t}; {k} tables of {q} people; ({p1}, {p2}) -> ({q1}, {q2})")Solution: The larger of the two primes was 109.
There are 30 tickets:
And 30 is the only number that satisfies the conditions.
So the sum of the ticket numbers is T(30) = 465.
And the tickets can be arranged into 5 tables of 6 people, to give one of two prime sums for each table as follows:
In each case the larger prime is 109.
And there are many ways to achieve the same sums.
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GeoffR 9:30 am on 31 August 2021 Permalink |
I programmed this teaser in two stages, first re-using the few lines of Jim’s code to find the intersection of GCD/LCM values to give the number of guests.
The next stage was to examine possible table arrangements and arrangements of two prime numbers to make up the sum of tickets.
I also tested different prime number multipliers e.g 4/1 and 3/2 for the 5 table arrangement and 5/1 and 4/2 for the 6 table arrangement.
from itertools import permutations from enigma import mgcd, mlcm, subsets, irange, Primes # primes less than 150 primes = Primes(150) # look for 3 No. 2-digit numbers for gcd and lcm values gcds, lcms = set(), set() for ns in subsets(irange(10, 99), size=3): gcds.add(mgcd(*ns)) lcms.add(mlcm(*ns)) # Numbers of Guests (N) N = list(gcds.intersection(lcms))[0] print('No. of guests = ', N) # N = 30 # Ticket sum of numbers for N guests T = N * (N + 1)//2 print('Sum of all ticket numbers = ', T) # 30 people - possible table arrangements: # - 5 tables of 6 people or 6 tables of 5 people - two possibilities # - 2 tables of 15 people or 15 tables of 2 people - not viable # - 3 tables of 10 people or 10 tables of 3 people - not viable # try 5 tables of 6 people, using 2 primes to make sum of tickets # results were found for prime multipliers of 4 and 1 # no results from 2 and 3 prime multipliers were found print('Two possible prime values:') for p1, p2 in permutations(primes, 2): if 4 * p1 + p2 == T: print( (max(p1, p2), min(p1, p2)), end = " ") # (149, 79) (109, 89) (101, 61) (103, 53) (107, 37) (109, 29) (113, 13) # There is only maximum prime i.e. 109, where the other prime could not be known. # For maximum primes of 149, 101, 103, 107 and 113, the second prime is known. # try 6 tables of 5 people, using 2 primes make sum of tickets for p3, p4 in permutations(primes,2): if 5 * p3 + p4 == T: print(max(p3, p4), min(p3, p4)) # No solution # No. of guests = 30 # Sum of all ticket numbers = 465 # Two possible prime values: # (149, 79) (109, 89) (101, 61) (103, 53) (107, 37) (109, 29) (113, 13)This section of code find the two sets of three two digit numbers used by the setter for the HCF/LCM values in this teaser.
There were 33 numbers in the GCD set of numbers and 25,608 numbers in the LCM set of numbers, with a minimum value of 30 and a maximum value of 941,094. The setter used the minimum LCM value.
# The number of guests for dinner was the highest common # factor of three different two-figure numbers and the # lowest common multiple of three different two-figure numbers. # Q. What were these two sets of three digit numbers? from itertools import combinations from math import gcd, lcm for x in combinations(range(10, 100), 3): a, b, c = x if gcd(a, b, c) == 30: print(f"Three 2-digit values for HCF = {a}, {b}, {c}") for x in combinations(range(10, 100), 3): d, e, f = x if lcm(d, e, f) == 30: print(f"Three 2-digit values for LCM = {d}, {e}, {f}") # Three 2-digit values for HCF = 30, 60, 90 # Three 2-digit values for LCM = 10, 15, 30 # Therefore, a value of 30 was used for number of guests.LikeLike