Listener 4229: Cleopatra’s Needled by Ruslan (or Coming Apart at the Seams?)
Posted by Dave Hennings on 8 March 2013
I thought that I’d got used to the three-monthly mathematical appearing on the penultimate Saturday of the appropriate month, but there I was, completely surprised by its appearance when I turned to the Listener page. Not that the length of the preamble looked at all mathematical-oriented. It surely couldn’t have been a coincidence that the longest mathematical preamble in living memory followed on from one of the shortest.
Sadly, a note after the preamble revealed that this puzzle was being published posthumously since Ruslan had died last November. Although Ruslan has had only three previous Listeners, I remember them as being very enjoyable:
|3970||23/02/2008||Beating the Bookie|
|4151||20/08/2011||Number or Nummer|
Now that all the trauma of moving house was behind me, I had resolved to post my entry to St Albans as early as possible, preferably by the Thursday following publication. I had been feeling sorry for JEG having to trawl through the several hundred entries that he must get on Mondays and Tuesdays to weed out my entry for the previous week! Yet here I was starting Ruslan’s puzzle at 8am on the Saturday a week after publication.
It is always enjoyable when a mathematical puzzle uses a new clueing technique. Here we were told that the grid entry was the clue answer plus the sum of its individual digits. This sounded simple enough, but in the early stages of solving, it was amazing how often I got confused as to which number I was dealing with.
Despite all the information that we were given about the two answers (but one grid entry) at 19ac, this wasn’t where we could start. By simple trial, the answer to 9ac 2 + last digit of the grid entry (1) was 8, with its entry 16, and 18ac The grid entry reversed (2) was answer 45 and entry 54.
17ac The product of the two digits of 1 more than twice the grid entry (2) was between 11 and 49. Going through all the possibilities didn’t take long, eg 11 × 2 = 22 + 1 = 23 and 2 × 3 gives 6 as the answer and 12 as the entry, which does not equal 11. What worked was 22 × 2 = 44 + 1 = 45 and 4 × 5 = 20 as the answer and 22 as the entry. Obviously all other values up to 49 had to be checked to see if there was more than one possibility, but there wasn’t.
This enabled a great leap forward, with all the constituent parts being present for 10ac and then 20ac, 6dn, 14ac and 16dn. A few clues later, and I began to notice an awful lot of 2s appearing in the grid. No doubt that would be the most frequently occurring digit, which would form part of the endgame.
Less than an hour after solving 9ac, the grid was complete, apart from 19ac (102•), 13dn (64•0) and 1dn (1•1•). A quick examination of the possibilities for 19ac showed that the extra information given by the preamble was needed to confirm the two answers as 999 and 1017, both of which gave 1026 as the entry. The possible answers for 1dn were 1005, 1006, 1007, 1008 and 1009, which gave corresponding entries of 1011, 1013, 1015, 1017 and 1019, and 1019 was popped into the grid, being 1017 (19ac) – 8 (9ac) = 1009 + 1 + 9.
All the 2s in the grid formed a giant 4, so I replaced the 2s with 4s and then proceeded to add up all the entries. The grand total was 4444453! Hmmm! That seemed as though it should be 4444444, with this value appearing in row 5 and column 4. I rechecked my addition, but it came to the same total. Obviously an entry was 9 too big. Luckily, it didn’t take long for me to realise that it was 1dn that was wrong and should be 1010, ie 999 – 8 = 991 + 9 + 9 + 1. What a devious little trick from Ruslan.
The final bit of highlighting that was required was the 4s that were not part of the grand totals in row 5 and column 4. These formed the shape of a piece of thread going through the eye of the needle in row 5. My entry was in its envelope and in the post box before the 11:30am collection.
Of course, the reason why I had been posting my entry so near the deadline was that I wanted time to ensure that I had completed all parts of the puzzle satisfactorily. Not that it stopped me making stupid mistakes, like failing to highlight the different pentominoes in Oyler’s mathematical 2×2×2 last year.
So there I sat, having posted my entry, and I read the preamble again … don’t ask me why! “A digit that appears in the total also features in six other cells which must be highlighted to suggest what Cleopatra would do when needled by a clue.” Well it was obviously sewing, and I got that. But should the preamble be read as “must be highlighted, to suggest” or as “to be highlighted [in a way] to suggest”. I decided that it was the latter, with the highlighting in the form of a wispy bit of thread as in my grid above … and my entry was wrong. The question is: “How strict are we going to be marked this week?” I think, as in “What is a church/campanile?” from two years ago, I should be marked wrong. As it was, I was marked right then, and I hoped that I would be lucky this time too.
Whatever the outcome, a lovely last puzzle from Ruslan.