Brain-Teaser 18: [Postal deliveries]
From The Sunday Times, 2nd July 1961 [link]
Mr Simpson, who lives at No. 1453 Long Street, is a keen mathematician, and so he was most interested when, [while delivering a letter], his postman mentioned a strange coincidence. If the numbers of [any] two houses to which he made consecutive deliveries were added together, the result came to the number of the next house to which he delivered a letter.
Mr Simpson asked him which houses he had visited, but the postman could only remember that some of them had single digits.
To which house did the postman deliver a letter immediately before delivering Mr Simpson’s letter?
I have changed the wording of this puzzle slightly for clarity.
This puzzle was originally published with no title.
[teaser18]
Hugh Casement, 
John Crabtree, and 
S2T2 
Jim Randell 10:26 am on 10 January 2021 Permalink |
I think the wording in this puzzle could have been made a bit clearer. (I have attempted to clarify the wording from the originally published text).
Evidently, if we write down the numbers of the houses that the postman delivered to, in the order he delivered, we are told that for every house he visited (after the first two houses), the number of that house is the sum of the numbers of the previous two houses delivered to.
So the numbers of the houses visited form a Fibonacci sequence. And we are told the sequence starts with two single digit numbers, and the number 1453 is in the sequence.
This Python program runs in 45ms.
from enigma import (irange, subsets, fib, printf) # collect solutions r = set() # choose a single digit number to start the sequence for (a, b) in subsets(irange(1, 9), size=2, select="P"): ns = list() for n in fib(a, b): if n > 1453: break ns.append(n) if n == 1453: r.add(ns[-2]) printf("[({a}, {b}) -> {ns}]") printf("answer = {r}")Solution: The postman had just come from house number 898.
There are in fact three possible starting pairs of house numbers:
but they all end up settling down to generating the same sequence of higher valued house numbers, which does eventually reach 1453:
And the number immediately preceding 1453 is 898.
As the ratio of consecutive terms in a Fibonacci sequence approaches the value of the Golden Ratio (𝛗 = (1 + √5) / 2), we can immediately calculate a good guess to the answer:
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Hugh Casement 2:02 pm on 11 January 2021 Permalink |
So for every third house visited, the postman crossed the road and back again.
That’s not the way our posties (male or female) work.
I think a better puzzle could have put Simpson in an even-numbered house,
all houses receiving post that day being on the one side of the street.
That was a good idea to divide by phi. Repeated division, each time rounding to the nearest integer, would get us most of the way to the start of the sequence, though it would be safer after a while to take S(n) = S(n+2) – S(n+1) .
Did we need to know that it starts with two single-digit terms? 1453/phi has a unique value!
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John Crabtree 4:26 pm on 11 January 2021 Permalink |
Good question, but I think that the answer is yes. Otherwise you can get shorter sequences which reach 1453, ie:
1453, 897, 556, 341, 215, 126, 89, 37, 52
1453, 899, 554, 345, 209, 136, 73, 63, 10
The ratio of successive terms oscillates about phi, and converges quite slowly.
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Hugh Casement 10:24 am on 12 January 2021 Permalink |
Thanks, John. The best approximations to phi are given by the original Fibonacci sequence.
So if we double each term we get numbers all on the same side of the street.
But perhaps the puzzle would then be too easy!
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