Many years ago I contacted Tarquin Press and suggested that a book of cryptic cross number puzzles like those in The Listener would make a fine addition to their catalogue only to be told that they wouldn’t really be interested in that sort of thing. So imagine my surprise when browsing through their catalogue last year to find just such a title!! Unfortunately it was out of stock but good old Amazon had one so I had a copy in a few days. The book is written by Tangent and full of lovely cryptic puzzles – no stupid dénouements and such.

As I leafed through the book, puzzle 16 was Tangent’s take on one of the most famous cross number puzzles – Rhombus’s Can You Do Division? and was similar in style to the original as was Viking’s Factory. Puzzle 17 was called Odds and Evens in which all odd numbered entries were odd and all even numbered entries even apart from one which broke the rule, all entries had the same digit sum and given the digit sum for each row and column along with a few other bits of information the 5×5 grid could be filled.

This appealed to me greatly and I wondered if I could obtain a grid in which the parity of the clues was totally preserved. So I got out the 1cm squared paper ( being a maths teacher I have an almost unlimited supply!! ) and set to work. I wanted to use a larger grid so opted for 7 x 7 and played around for a while with the bars until eventually I had a suitable grid. Now for clues. At this stage I looked again at puzzles 16 and 17 which were on facing pages in the book and illumination – could I combine the two together in some way so that parity was preserved as well as some divisibility property?

I quite liked the idea of having the entries exactly divisible by their clue number and decided to use this but it could only work for the down entries so I needed to find something else for the across clues. I remembered a letter I’d received years ago from Piccadilly in which he described a puzzle that had all the entries as palindromes that were exactly divisible by their number of digits. A quick check revealed that I couldn’t use this as some of the odd across entries had an even number of digits. So I looked again at Tangent’s puzzle and decided to have the entries exactly divisible by their digit sum.

The setting process went well, rather too well, as the inevitable happened – an across entry that wasn’t exactly divisible. However I noticed that its remainder was equal to the digit sum of the divisor so decided to progress along those grounds and soon the puzzle was finished. I did the cold solve and found that I couldn’t remove the digit sum information that was given for every row and column which was a shame but you can’t have everything!!

I can heartily recommend Tangent’s book so get yourself a copy.

Posted on Monday, May 2nd, 2011 at 8:36 am under Setting the scene.
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