From The Sunday Times, 25th May 1980 [link]
Large families often have problems, particularly when it comes to deciding who has first choice, etc.
My six brothers and I solved this particular one by devising our own dice game. Each of us in turn would roll a die four times and put the results together in that order to form a four-figure number. (So that, for example: 6 followed by 2, then 3 and 1, gives the number 6231). We would each do this and the one of us getting the highest four-figure number would be declared the winner.
One such game had some very strange results. Each of us got a different prime number, using the same four different digits. The average of the seven primes was a perfect square, to which Jon’s number was the nearest.
The difference between Ron’s and Don’s numbers, multiplied by one of the numbers from the die, gave the difference between Len’s and Ken’s numbers.
Omitting Ben’s score, the total could have been formed by throwing the die five times. And so could the total of the numbers thrown by Len, Ken, Ron and me.
What was my score, and who ate the biggest piece of apple pie?
The originally published version of this puzzle was flawed, so the version above was taken from the book Microteasers (1986), in which it was noted:
There is an interesting postscript to [this] puzzle. When it first appeared, the condition “each of us got a different prime number using the same four different digits” was given as “each of us got a different prime number; two digits never came up”. That allowed numbers like 5651, which was not the intention at all. The odd thing is that all the readers who sent in answers found the intended solution anyway. So perhaps, even if one relaxes the condition about all the primes using four different digits, the teaser still has a unique solution. The keener reader might like to adapt the program to take four digits in turn and see how many primes can be made from those four (but not necessarily using all four each time). Then you have to consider all possible sets of seven from those. We leave this much harder problem as an exercise.
However the original formulation of the puzzle does not give a unique solution.
[teaser931]
GeoffR and 
John Crabtree, 
Frits, 
Hugo, 
S2T2 
Jim Randell 7:47 am on 14 March 2023 Permalink |
(See also: Enigma 1198).
For a triangle with sides (a, b, c) (in order), we must have: a + b > c.
The triangle is then scaled up to give another similar triangle (f.a, f.b, f.c) (f > 1).
And two of the sides are the same, which must be: f.a = b, f.b = c.
Hence:
So we can choose a value for b, and then look for a, c sides that multiply to give b².
This Python program runs in 52ms. (Internal runtime is 288µs).
Run: [ @replit ]
from enigma import (irange, hypot, intf, divisors_pairs, div, sqrt, printf) # area of a triangle def area(a, b, c): return 0.5 * sqrt((a + b + c) * (a + b - c) * (c + a - b) * (b + c - a)) # dimensions of an A4 sheet (cm) (X, Y) = (21.0, 29.7) A4 = X * Y # choose a value for b for b in irange(1, intf(hypot(X, Y))): # calculate possible a, c values from divisors of b^2 for (a, c) in divisors_pairs(b, 2): if not (a < b < c and a + b > c): continue # calculate d d = div(b * c, a) if d is None: continue if area(a, b, c) + area(b, c, d) > A4: continue # output solution printf("({a}, {b}, {c}) -> ({b}, {c}, {d})")Solution: The smaller triangle has sides of length 8, 12, 18 cm.
And the larger triangle has sides of length 12, 18, 27 cm.
The dimensions of the second triangle are 1.5 times those of the first.
Here is a diagram of the two triangles, set on a sheet of A4 paper:
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