Listener No. 4438, Can’t You Do Division?: A Setter’s Blog by Oyler
Posted by Listen With Others on 12 March 2017
At last year’s Listener dinner in Windsor, I was talking to Shane and mentioned that the March numerical puzzle in The Magpie which was Factors by Amos should really have appeared in February as its inspiration was the Listener puzzle No. 1916 Can You Do Division? by Rhombus which had also appeared in that month and even better they should have held it over to Feb 2017.
Later on in the evening after the dinner and speeches I was talking to Roger in the bar and mentioned that February 2017 would be the 50th anniversary of Can You Do Division? and would they be interested in celebrating that anniversary. Ten years ago, Viking set the puzzle Factory which celebrated the 100th anniversary of the birth of Rhombus. Roger said he was quite happy to celebrate the puzzle’s 50th anniversary.
On the train journey back up north I started thinking about the puzzle. Not only had Amos and Viking used the idea but Tangent as well in his book Cryptic Crossnumber Puzzles. How could I put a different twist on it? After all the idea is beautifully simple as it’s just the number of factors that the entry has. Then inspiration struck — use the original grid but with different numbers. I knew I had a copy of the original at home and that I’d solved it but couldn’t recall the grid and I was desperate to get home to see it. To make matters worse our train got caught behind a cross country train coming out of Newcastle that dawdled and stopped at every station!
Once home I took down the relevant folder and found the puzzle. A 9×9 grid — nice I like those — but aargh what’s this? The barring pattern was strange to say the least. I looked for my solution and couldn’t find it. So the next day I re-solved the puzzle which took a couple of hours as I didn’t want to use the same numbers.
At this stage I must point out that I had no idea that my puzzle Square Time Sums was to be the May mathematical last year (Listener No. 4399).
The 9-digit entries just screamed out zero-less pandigital squares to me so I found my list and set to work. I had already looked at the problem of fitting them round the perimeter to obtain a Sudoku type grid and remembered that it had been a difficult challenge which I eventually abandoned! Perhaps a computer buff could write a program to see if this can be done and if so I’d love to have the results! So no Sudoku then, but a Latin Square should be possible. I decided to go down this route in that the Latin Square information might not need to be given. (Having the grid a Sudoku is a more difficult task for the setter but makes it easier for the solver in that there’s the extra criterion of the 3×3 blocks.) I wanted to avoid using the squares I’d used in previous puzzles and I managed to do that. I had wanted to include the three 3-digit squares that between them contain all of the digits from 1 to 9 inclusive along with 576 but that proved to be impossible thanks to the strange barring pattern Rhombus had used. This style of puzzle usually has a reasonable number of entries that have an odd number of factors. The reason for this is that they must be square numbers and so help to provide a way in. The following table gives the grid size, number of entries and those that were square numbers for the puzzles that I know of.
With the 9-digit entries in place, it was then a case of placing some other entries, like 576 which is the only 3-digit number to have 21 factors. Of course there were times when a really good entry resulted in a situation where a Latin Square was impossible and so I had to backtrack a few times. Strange as it may seem but I have just realised that two of the entries in my puzzle were also used by Rhombus and both appear in exactly the same places namely the entries for m and p. Furthermore, Encota has pointed out that the digits in the corner cells starting at the bottom left and working anti-clockwise give the original puzzle number 1916, which I hadn’t spotted. Eventually the puzzle was complete and I did the cold solve. It was during this process that I realised that the Latin Square condition would have to be given to solvers otherwise it would have taken forever to solve or there may have been multiple solutions.
I sent the puzzle to two other setters for their comments and both were favourable with one rather more so than the other. So I sent it to Roger to see what he thought and he reckoned it was fine but wouldn’t solve it until Shane had had a go. I submitted it to Shane with the note that the publication date should hopefully be February 2017. This is the way to beat the long queue that has built up for the mathematicals — set a puzzle that celebrates some anniversary!
How solvers solved this puzzle is entirely up to them. Some started with the square numbers that went round the perimeter. However, they would have to have had some perimeter cells entered as the squares can be reflected in the NW–SE diagonal.
Hopefully some invoked the spirit of 1967 and listened to some Pink Floyd, The Nice and Procol Harum whilst foraging away with just paper and pencil. Others, however, will have used the tools that are available today which is fair enough and which weren’t around in 1967; calculators, PCs and internet. So nowadays we are indeed very fortunate when it comes to solving the Listener mathematical puzzles and we shouldn’t forget that!
Some solvers mentioned Sudoku. Well it isn’t one as a Sudoku is a subset of Latin Squares. All Sudokus are Latin Squares but not all Latin Squares are Sudokus.
The original Rhombus puzzle was much harder although it did have more gimmes at the start with some of the square entries having to be a single prime raised to one less than the number of factors. It is still a tough challenge nowadays even with all our fancy tools.
Finally, one comment I received made my day — ‘50 years after submitting 1916!‘.
PS For those of you not ‘au fait’ with Scottish slang, ‘Gaesintaes’ translates as ‘Goes intos’ and is what Scottish pupils may reply to the question ‘ What sort of sum do we do if we want to share out 24 marbles between 3 people equally?’