A Holiday Brain Teaser
From The Sunday Times, 4th August 1957 [link]
The series below is peculiar in that if any two of the numbers are multiplied together and added to unity, the result is a perfect square:
1, 3, 8
Let A[4], be a fourth number in this series. There is a further series with the same characteristic:
8, B[2], 190, B[4]
in which not only is B[1], the same as A[3], but B[2] is also the same as A[4].
What is B[4]?
(Note: zero is excluded)
This is one of the occasional Holiday Brain Teasers published in The Sunday Times prior to the start of numbered Teasers in 1961. A prize of £5 was offered for a solution to the puzzle, and a further prize of £25 for finding A[5] or B[5].
[teaser-1957-08-04] [teaser-unnumbered]
Jim Randell 11:28 am on 17 October 2021 Permalink |
I am assuming we are looking for a sequence of increasing integers.
With computers we can easily calculate the next terms.
Even a simple program runs in less than a second:
But we can be a little bit cleverer, and this gives us a program that runs in just a few milliseconds:
Run: [ @replit ]
Solution: A[4] = 120. B[4] = 730236.
See OEIS A030063 [@oeis], which says that the sequence cannot be extended further (although there is a further term in rationals).
The following was published in the 18th August 1957 edition of The Sunday Times [link]:
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Jim Randell 12:10 pm on 17 October 2021 Permalink |
Euler found an infinite family of such 4-tuples:
(Which means that any pair of values from one quadruple can be used to seed another).
The A sequence of the puzzle is given by: a = 1; b = 3; r = 2.
And the B sequence is given by: a = 8; b = 120; r = 31.
In 1969 it was shown that the {1, 3, 8, 120} quadruple cannot be extended with another value, and in 2016 (a mere 59 years after the puzzle was set) it was proved that no Diophantine quintuple exists (see: [ @wikipedia ]). So the £25 prize was safe.
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Jim Randell 5:48 pm on 20 October 2021 Permalink |
And here is a solution using Pell’s equations (it uses the [[
pells()
]] routine written for Teaser 2994), which means fewer candidate numbers are considered when looking for solutions.LikeLike
GeoffR 3:09 pm on 17 October 2021 Permalink |
The best run-time I could get was 12.81 sec on an I9 processor. The Geocode solver took much longer. I had to use the answer to fix variable upper bounds.
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A Holiday Brain Teaser | PuzzlingInPython 7:48 pm on 17 October 2021 Permalink |
[…] From The Sunday Times, 4th August 1957 [link] […]
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GeoffR 10:38 pm on 17 October 2021 Permalink |
A Python solution ran in 258 msec.
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Frits 12:33 am on 18 October 2021 Permalink |
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GeoffR 9:17 am on 18 October 2021 Permalink |
@Frits:
Neat solution.
The combined use of
first()
in an iterator andinf
from the enigma.py library solves the problem I had in fixing the upper bound in a search for value B4.LikeLike
Frits 9:57 am on 18 October 2021 Permalink |
@GeoffR: the other way of achieving it is to break out of a while True loop for the first correct B4.
Also possible is using a wheel [2, 40, 152, 190] so that (i * i – 1) always is a multiple of 190.
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