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]
Frits and 
GeoffR and 
Linda, 
John Crabtree, and 
S2T2 
Jim Randell 10:13 am on 31 August 2021 Permalink |
There are two primes p and q and each square of the grid contains some product of the two primes, and the whole forms a multiplicative magic square.
Each square has a value: (p^m)(q^n) so the squares formed by the exponents (the “m-square” and the “n-square”) are both additive
magic squares.
If the PIN in the central square is (p^j)(q^k) then the magic constant of the multiplicative square is (p^3j)(q^3k) and the magic constants for the two additive magic squares are 3j and 3k.
This Python program runs in 59ms.
from enigma import (primes, irange, inf, subsets, div, seq_all_different, nsplit, ordered, printf) # generate multiplicative magic squares formed from two primes def solve(): # consider values for p and q for q in primes.irange(3, inf): for p in primes.irange(2, q - 1): # find 4 digit combinations of p and q ns = set() for k in irange(0, inf): qk = q**k if qk > 9999: break for j in irange(0, inf): n = (p**j) * qk if n > 9999: break if n > 3: ns.add(n) # choose E (the central number) for E in ns: # the magic constant is ... K = E**3 # choose A (the smallest corner) for A in ns: if not (A < E): continue # compute I I = div(K, A * E) if I is None or I not in ns: continue # choose B (less than D) for B in ns: # compute the remaining values C = div(K, A * B) if C is None or C < A or C not in ns: continue F = div(K, C * I) if F is None or F not in ns: continue D = div(K, E * F) if D is None or D < B or D not in ns: continue G = div(K, A * D) if G is None or G < A or C * E * G != K or G not in ns: continue H = div(K, B * E) if H is None or G * H * I != K or H not in ns: continue # return the magic square sq = (A, B, C, D, E, F, G, H, I) if len(set(sq)) == 9: yield sq for sq in solve(): # check no numbers have the same digits if not seq_all_different(ordered(*nsplit(x, 4)) for x in sq): continue # check none of the numbers is a factor of another if any(b % a == 0 for (a, b) in subsets(sorted(sq), size=2)): continue # output the square (A, B, C, D, E, F, G, H, I) = sq printf("[ {A} {B} {C} | {D} {E} {F} | {G} {H} {I} ]") break # we only need the first solutionSolution: The PIN in the middle is 2592.
And here is a complete grid:
The magic constant is: 2592^3 = 17414258688 = (2^15)(3^12).
The numbers in brackets show the powers of 2 and 3, which can be seen to form additive magic squares with magic constants of 15 and 12.
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Frits 6:15 pm on 16 September 2021 Permalink |
See also [https://brg.me.uk/?page_id=4071] for 2 different approaches.
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