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Listener 4125: Oyler’s Number Theory (or Mr and Mrs Doku and their Daughter Sue)

Posted by Dave Hennings on 11 March 2011

We have our regular quarterly numerical puzzle this week, and it’s Oyler’s annual outing. 18 different statements are listed, such as A. It is a prime number, F. It is a multiple of 5 and L. Its first digit is 3 times the units digit. Each entry is then clued by the letters of all the statements that apply to it. (In hindsight, this clueing idea somehow rang a bell, but I cannot find such a puzzle.)

I must say that I found this idea to be very entertaining. I’ve said before that one of the difficulties of a numerical puzzle is that, unlike a normal crossword, there are usually only one or two entry points to the solution. Not so here. It was interesting to see how little information was needed to identify some entries. No doubt this was partly due to the care taken in choosing the statements that were used. Of course, an entry was additionally clued by which statements did not apply to it.

My solution notes took the following form:

11ac JKLQ using JL 602 !Q 622 R 642 662 R 682 !Q

Possible entries were marked with a ! to indicate they didn’t meet one of the statements given, so !Q meant that it didn’t have a 3-digit prime factor. Conversely, they were just marked with a letter, eg R, to indicate that they met a statement that was not given for the entry. Underlining indicated that it was the solution; dashed underlining meant that it was one of several solutions. Some entries were fairly easy, like 11ac above; others took a little more effort, such as 12ac:

12ac BCE using E 13 !C 17 !C 21 !B 25 !B 29 !B 33 R 37 !C 41 !B 45 !B 49 !B 53 !C
57 61 !B 65 !B 69 !B 73 !C 77 R 81 !B 85 !B 89 !B 93 L 97 !C

It didn’t take long for me to suss that the nine digits in each row and column would be different, but I didn’t use that suspicion while solving the grid. That all went very well until I got to 16dn which was 4..5 and was divisible by 5 (F), had sum of first and last digits a perfect square (H), had a 3-digit prime factor and whose digit product was not 3 digits. I resolved this as 4215, but this resulted in 19ac being 922, and one of its statements was not R (in fact, because of the sudokuness of the grid, no entry was defined with an R).

A bit of checking of entries around that part of the grid, and I couldn’t find an error. This could only mean that a common feature of mathematical puzzles for me reared its ugly head: I would have to solve it again. My notes were reasonably clear, but I took the opportunity to write them out in a more structured and logical way given what I had discovered about the ease with which some, seemingly daunting, entries could be solved.

The second time through, and 16dn loomed large. On this occasion … exactly the same problem!! Examining the other squares in that row meant that 16dn had to be 48.5 or 49.5. I can’t say that it dawned straight away, but eventually I realised that the converse of statement P, digit product 3 digits, was not just digit product 2 digits, but also included digit product 4 digits! Was this a deliberate trap?

So that was the grid complete, and the ambiguities and central square could be resolved from the sudoku nature of the grid. Except that I had only two ambiguities! These were the second digit of 21dn (either 8293 or 8693) and the middle digit of 41ac (217 or 237). So should I read ambiguities in four cells as ambiguities with four entries? Surely not. I proceeded to the statement letters for the two 9-digit numbers in the central row and column: 139854276 (D, N and O) and 627953481 (just E). What word did this give: DONE or NODE. Well DONE was almost what QED signified, at least it did when I was at school: Quite Easily Done! It didn’t strike me as particularly good.

‘Only two ambiguities!’ Well, you probably know what came next. Yep, another run through my solution. Luckily I had attacked this fairly early on in the 12-day solving window, but I must admit I was getting a bit bored with it. I left it for a couple of days, and then checked my notes again. The same result! This was leading towards it being the first time since my all-correct streak began two years ago that I knew that I would not be able to fully resolve a solution. Actually, that’s wrong: in 4036 Base Jog by Brimstone (another numerical) I didn’t get the significance of 714 and 715, which are, apparently, Ruth-Aaron pairs.

I double-checked my entries. Yes, all nine rows and columns consisted of all of the nine digits. I noted that each of the 3×3 blocks that you would normally get in a sudoku grid were also all-different. Finally a quick check of the 9-digit entries and I’d be done. Oh My God! What’s this? 139854276 is a perfect square, and I didn’t spot that first time through, so it was D, I, N and O. How did I miss that!? And what’s more, 627953481 is a square as well, so it was E and I. Such sloppiness, but thankfully I check everything fairly thoroughly now, so the mistake was bound to get caught? And now the word to be entered under the grid was IODINE. I know the title of the puzzle was Elementary Number Theory, but I still thought the link was a bit tenuous. Still everything fitted (and I’d double-checked the letters for the two 9-letter words at least three times by now).

All in all, a really entertaining puzzle, and I for one would not be averse to a similar clueing method appearing again in a couple of years time. Except, knowing (but not having met) Oyler, the preamble will probably include the clues detail all the statements that apply to the associated entry, except each contains a misprint for one of the letters!!


One Response to “Listener 4125: Oyler’s Number Theory (or Mr and Mrs Doku and their Daughter Sue)”

  1. Oyler said

    Mmm Now that gives me an idea ….

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