From The Sunday Times, 3rd January 1971 [link]
Joybells rang, fireworks went bang, and merry Bahreinis sang, celebrating not only their 500th Founding Festival, but also the coronation of the new monarch of Bahrein Teza.
“But”, said our special correspondent, intercepting the Lord High Solutioner as he was hurrying through the Palace gates, “I understand there has been a disturbance concerning rival claimants to the throne?”
“Yes, yes. A bit of bother. Long story”, replied the LHS. “Afraid I can’t stop now though. Banquet just starting. See you later. Wait! Here’s a copy of the royal genealogical tree. Tells you everything. Succession laws same as your own. Right?”, and he disappeared into the Palace.
Well, here’s a relevant extract:
CALCULUS – Son-in-law of NUMERA
DATA III – Daughter of LOGICUS XI
DATA – Mother of RIDELLA IV
DECIMUS – Husband of PUZELA
LOGICUS X – Father of NUMERA
LOGICUS XI – Father of RIDELLA IV
LOGICUS – Son of LOGICUS XI
MATHUS – Brother-in-law of PUZLUS V
NUMERA – Aunt of PUZLUS VI
PUZELA – Cousin of LOGICUS XI
PUZELA – Daughter-in-law of PUZELA
PUZLUS V – Father of RIDELLA
PUZLUS VI – Elder son of PUZLUS V
RIDELLA IV – Niece of PUZLUS VI
RIDELLA – Mother of SOLVUS II
RIDELLA – Wife of LOGICUS XI
SOLVUS II – Cousin of RIDELLA IV
So who is crowned and who is the claimant?
The following was published alongside this puzzle:
Brain-teased 500 times
“Tantalising but irresistible” sums up the reaction of thousands of readers to The Sunday Times Brain-teaser, which this week reaches its 500th example since numbering began on February 26, 1961.
Competitors, from emeritus professors of mathematics to 11-year-olds, with a growing proportion of women, submit their solutions from every county and from 56 oversea States, some never having missed a week. It is calculated that between 40,000/50,000 readers send a solution every year.
Set by Sunday Times readers, the problems are selected from about 200 a year, vetted for uniqueness of solutions by independent experts (with an occasional oversight) and are published in varying degrees of difficulty; entries then range weekly from around 4,000 to a handful from the master-minds.
— Anthony French
This puzzle was originally published with no title.
[teaser500]
Lise Andreasen and 
Steve Taylor are discussing. 
GeoffR and 
John Crabtree and 
S2T2 
Jim Randell 3:52 pm on 3 October 2019 Permalink |
We are to infer from the puzzle text that the lamb counts using base 4, with the standard symbols 0, 1, 2, 3.
So we need to find numbers in base 4 whose representation in base 10 is the same as the representation in base 4 but with the leftmost 2 digits removed.
This Python program runs in 221ms.
Run: [ @repl.it ]
from enigma import irange, inf, nsplit, nconcat, printf # base 4 digits for number n base4 = lambda n: nsplit(n, base=4) # consider k-digit base 4 numbers, read as base 10 numbers, they need to be (k + 2) digits for k in irange(1, inf): # smallest k-digit base 4 number read as a base 10 number if len(base4(10 ** (k - 1))) > k + 2: break # largest k-digit base 4 number read as a base 10 number if len(base4((10 ** k - 1) // 3)) < k + 2: continue # consider possible (k + 2) digit base 4 numbers for n in irange(4 ** (k + 1), 4 ** (k + 2) - 1): # represented in base 4 ds = base4(n) # does removing the first two digits give the base 10 representation? if nconcat(ds[2:]) == n: printf("prefix = {p} [k={k}: {ds} (base 4) = {n} (base 10)]", p=nconcat(ds[:2]), ds=nconcat(ds))Solution: The 2-digit prefix is 30.
The only viable numbers are:
The program considers base 4 numbers that are 5-, 6- and 7-digits long, and looks for corresponding 3-, 4- and 5- digit base 10 numbers.
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