Perfect Brain Teaser? Not quite!
By Oyler
I’m sure that I’m not the only setter who, whenever a numerical appears in The Listener or Magpie and has an interesting idea or something novel, has a go at setting their own version of it minus the dénouement. After all that’s how we get started. So, for me it was back in 1993 and Listener puzzles by Adam and Piccadilly that I copied. In those days it was only really the Listener that had these type of puzzles apart from mathematical journals. Nowadays there are more outlets and I’ve spent many a happy hour coming up with my versions of puzzles by the likes of Arden and Hedgehog. Arden’s Listener puzzle from last year, Versed Couplets, was one that appealed to me, and my version, involving subtraction instead of addition, appeared in CQ32.
When I was doing my teacher training way back in the dim and distant past I joined the Mathematical Association. One of their journals was Mathematics in School and at that time each issue had a section entitled Puzzles, Pastimes Problems by Canon D. B Eperson who had taught Alan Turing at Sherborne School. Some of the material involved you having to express numbers in the form 100a + 10 b + c, for example, and then do some algebraic manipulation.
If Perfect Brain Teaser had appeared with no pseudonym and you were asked to guess who set it, I’m confident that most of you would have plumped for my co-editor of CQ, namely Zag. After all he has form for this sort of thing where an answer must be modified in some way using the symmetrically opposite entry. So, it should come as no surprise that the inspiration for this puzzle was Zag. Solving this type of puzzle involves doing some algebra along the lines of that used in Eperson’s articles.
I’d never set a puzzle like this and I felt it was high time to rectify that and give it a go. So, for 2-digit numbers they can be represented ab and cd, say, with their values being 10a + b and 10c + d respectively.
For the across entries you get
| ans + cd | = | 10a + b | I |
| ans – ( a + b ) | = | 10c + d | II |
This yields
and
| 2 x ans | = | 11a + 10c + 2b + d( 1 – c ) | I + II |
Of course, you can swap ab for cd but you get the same type of solutions just with the letters swapped.
By using a similar argument for the 2-digit down entries we obtain
| 11a + 2b | = | 10c + d( 1 – c ) |
and
| 2 x ans | = | 9a + 10c + d( c + 1) |
Again, the ab and cd can be swapped.
This I was able to do with a spreadsheet. The algebra for the 3-digit entries is more complex so I contacted my son with the relevant equations and he ran a program to find the solutions.
Given Zag’s liking for small grids I opted for 4×4 with the symmetric barring pattern that gave eight 2-digit entries and four 3-digit entries. I set quite a few, some had different clues for each location, others had the same clue for symmetrically opposite locations, along with various combinations of modifications. Some of these efforts found their way into CQ.
All these puzzles had the solution pathway starting with the 2-digit entries and I wanted to try one that started with a 3-digit entry instead. I hit on the idea of trying to see if I could set the puzzle with only one clue. I wanted to have a unique starting number so something like prime, square etc was out. I opted for perfect but annoyingly there were two possible solutions – see AJ’s solution. As it turned out having only that one clue was impossible as that led to three possible solution grids. I noticed that only one of them could have a clue of a palindrome and that sorted things out.
So, it was not as neat as I intended but hopefully solvers had fun getting their brain around it.