From The Sunday Times, 6th June 1971 [link]
There were six. Messrs. Butcher, Carpenter, Draper, Farmer, Grocer and Miller, who shared the fire-watching on Friday nights — three this week, three the next. By occupation they were (not necessarily respectively) a butcher, a carpenter, a draper, a farmer, a grocer and a miller.
Incidents were few and far between, until that Saturday morning when they found the “log” book signed by the Acting Deputy Assistant something or other, as follows: “All three present and fast asleep”.
Something had to be done about it: they decided to watch, to play whist, and to keep awake in the future. It was arranged thus: four did duty this week, next week two stood down and two others came in, and so on. Each did two turns in three weeks.
On the first Friday the carpenter, the draper, the farmer and the miller watched. Next week Mr Carpenter, Mr Draper, Mr Farmer and Mr Grocer played. On the third occasion Mr Butcher played against Mr Grocer, Mr Farmer against the butcher, and the miller against the draper. Each night the four cut for partners and kept them till morning.
If Mr Carpenter’s occupation is the same as the name of the one whose occupation is the same as the name of the one whose occupation is the same as the one whose occupation is the same as the name of the miller:
What is Mr Miller’s occupation?
As presented this puzzle has no solutions. In the comments I give a revised version that does have a unique answer.
This puzzle was originally published with no title.
[teaser521]
GeoffR and 
Frits, 
John Crabtree and 
S2T2 
Jim Randell 9:37 am on 7 July 2020 Permalink |
The triangle has sides of integer lengths (a, b, c) such that b = (a + c)/2.
We can use Heron’s Formula [ @wikipedia ] to equate the area of the triangle with the perimeter, which gives us some restrictions on the sides of the triangle.
However, we have encountered “equable” triangles before in Enigma 364, and generated a list of the 5 equable triangles, from which we can easily pick out the one that has one side that is an average of the other two.
I also wrote code to generate k-equable triangles (integer sided triangles where the area is k times the perimeter), and that code can be reused here:
Run: [ @repl.it ]
from enigma import irange, isqrt, divc, divf, printf # find all triangles with integer sides where area = k . perimeter def triangles(k=1): K = 4 * k * k for p in irange(1, isqrt(3 * K)): for q in irange(max(p, divc(K + 1, p)), divf(3 * K, p)): (r, z) = divmod(K * (p + q), p * q - K) if r < q: break if z == 0: yield (p + q, p + r, q + r) # consider 1-equable triangles for (a, b, c) in triangles(1): if b - a == c - b: printf("a={a} b={b} c={c}")Solution: The lengths of the sides of the plot are: 6m, 8m, 10m.
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GeoffR 10:54 am on 8 July 2020 Permalink |
for a in range (3, 25): for b in range (a+1, 25): for c in range(b+1, 25): # a < b < c and 2 * b = a + c if 2 * b != a + c: continue # Square of perimeter perim_sq = (a + b + c) ** 2 # Area squared formula is based in Heron's formula area_sq = 1/16 *( 4*a*a*b*b - (a*a + b*b - c*c)**2) if perim_sq == area_sq: print(f"Length of sides are {a}, {b} and {c}m") # Length of sides are 6, 8 and 10mA (5,12,13) triangle has the area equal to the perimeter numerically, but does not have one side as the average of the other two sides.
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