Spoiler for New Scientist Enigma 1777 “Around the houses” (Follow the link to see the puzzle.)
Kind of an odd puzzle and I wasn’t entirely sure how to interpret some of the wording.
But I think it’s pretty easy. First, if there are any houses 90 degrees apart then the number of houses must be a multiple of 4. Number of houses N=4n.
Second, if there are any digit reversal pairs 90 degrees apart then the house numbers differ by n: (10a+b)–(10b+a)=n. Then n=9a–9b. So n is a multiple of 9, and N=4n=36k.
Then there are only two possibilities, N=36 or 72. Examining these two — and I assume we don’t use leading zeros so 10 is not a reversal of 01, for instance, though in fact allowing that doesn’t permit any solution to the puzzle — we find:
For N=36, there are 2 90-degree pairs ((12, 21) and (23, 32)) and 1 180-degree pairs ((13, 31)).
For N=72, there are 5 90-degree pairs ((13, 31), (17, 71), (24, 42), (35, 53), and (46, 64)) and 2 180-degree pairs ((15, 51) and (26, 62)).
Now, does “the higher digit of the lowest-numbered house involved” refer to the larger digit or the digit in the higher place? If the latter then in either case it’s 1 and that’s the difference between the two counts for the N=36 case. If the former, then it’s 2 in the N=36 case, which isn’t the count difference, or 3 in the N=72 case, which is.
I think they mean the larger digit, in which case the answer is 72.
I suppose one should check the cases of N=18, 54, and 90, for which the counts would be 0 for 90 degrees apart (there are no houses 90 degrees apart) and nonzero for 180 degrees, but even if those cases turn up more solutions, I doubt if they’re what was intended… and besides, I’m not that interested!