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Archive for March, 2011

Listener 4124, Travel Guide: A Setter’s Blog by Aedites

Posted by Listen With Others on 16 March 2011

There have been several Listener crosswords based on the London Underground, so I thought that it would be appropriate to use the Glasgow Underground, known locally as the “Subway” or the “Clockwork Orange”. The Subway opened in 1896 and was the third such system in the world (after London and Budapest). When I lived in Glasgow I used to use it to travel from Hillhead in the West End to the City Centre. Little had changed since 1896 and the original rolling stock was still in use; the only significant modernisation had been to replace the cable haulage with electric motors in the motor carriages. Subsequently there was a major upgrade in the late 1970s, and some of the stations were renamed. Fortunately I noticed this before starting the construction of the crossword.

Gioconda’s crossword Circumnavigation (Listener 3586) was based on 18 stations on the Circle Line which were included in answers clued by rows and columns and had to be omitted before entry. A similar approach was not practical since I wanted to include THE RIVER CLYDE and have the stations in approximately the correct geographical position, which would create a major problem in the bottom left hand corner. In addition some of the Glasgow station names would be quite awkward to clue. Thus I resolved to encode the station names as extra words in the row and column pair which intersected at the correct place. Fortunately all the station names could be anagrammed in this way. A similar technique was used in Aragon’s crossword Linear G____ (3655), where each jumbled station name had to be replaced by a single letter before the clue could be solved. Splitting the additional words across two clues to give a set of co-ordinates for a grid square is, as far as I know, a new idea.

It took about a day to construct the grid by hand, and an initial set of clues were constructed over the next three days. I then put the crossword to one side for a couple of months before reviewing the clues and sending it to two test solvers who made some helpful suggestions on further improvements to the clues.

This was the first time that I have used either misprints or extra words. At the time I did not realise that the editors’ preference was for the correct letters to spell the message. The original message was going to be CLOCKWORK ORANGE, but I decided that this would give Glaswegians an unfair advantage and would not be helpful to most other solvers. I therefore used UNDERGROUND CIRCLE instead, which required two additional entries in the construction of the grid. I thought that most solvers would be familiar with IBROX and PARTICK (football grounds) and GOVAN (former major ship-building centre), but the editors decided that a pointer to GLASGOW was needed, particularly for overseas solvers. Comments received suggest that about half the solvers did require the seven letters spelling GLASGOW in order to finish the crossword. The editors also changed my instruction to highlight THE RIVER CLYDE to drawing a straight line through it to avoid the ambiguity in the final E.

I am grateful to the many solvers who have sent positive feedback, indicating that they enjoyed the crossword. The only negative comment came from one solver who disliked the grid that I used and felt that this led to some poor surfaces in the clues.

Aedites.

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Listener 4125, Elementary Number Theory: A Setter’s Blog by Oyler

Posted by Listen With Others on 16 March 2011

On some occasions I’ve set puzzles just to make use of a title, for example The Times They are A Changin’? [Magpie Issue 58]. I’d wanted to set a puzzle with the title Elementary Number Theory for some time. After all it is one of the many areas that Euler made significant contributions to, as well as providing the title for numerous textbooks, the contents of which are anything but! I also wanted to make use of the word ‘elementary’ in a chemical sense as, although I’m a maths teacher at Madras College in St Andrews, my degrees are in chemistry.

The first ideas started to take shape at about the same time as Digimix appeared last year. I was using a resource based on one of Tony Gardiner’s excellent books, namely Maths Challenge I with my superb S1 class 1XP or, as I knew them, Windows!! Quite simply they were the best S1 class I’ve ever taught as they were incredibly receptive and insightful. The resource in question was Define a Number, in which there were 26 statements about numbers lettered A to Z. A – it is prime, B – it is perfect square, etc, and I’d get them to find me a number less than 100 say that satisfied certain statement letters or which statements did the number 42 satisfy or which number satisfied most or least of the statements as part of a class competition. It was great fun and maths teachers reading this should give it a go if they haven’t done so already!!

At this point I was reminded of Piccadilly’s Properties of Numbers puzzle and my own Euler’s Spoilers that used number definitions, and realised that I could use this resource as the basis for a puzzle. As the puzzle was based on number definitions I felt that I could at last make use of the title Elementary Number Theory as it described what the puzzle was about albeit in a fairly loose sense.

My first thoughts were to have two unclued 5-digit entries crossing in the middle of a 7×7 grid and, in order to find the statements that applied to them, solvers would have to decode two 3-digit entries to get the statement letters. But there was a twist: using the conventional A=1, B=2 would lead to nonsense, so instead the individual digits of those entries referred to the atomic numbers of the elements so H=1, B=5, C=6, N=7, O=8 and F=9. I set the puzzle and sent it off to The Magpie who quickly rejected it on the grounds that they hadn’t twigged the code and felt that solvers wouldn’t either. In addition they couldn’t suggest any way the puzzle could be saved.

I went back to the drawing board and came up with the idea of reversing the process, namely having the unclued entries fully checked apart from the middle square, and having solvers find the statement letters that applied to them and then rearrange the letters to give the name of an element.

I informed The Magpie that I had found a way to save the idea and would be in touch. There were a number of things that I had to consider – size of grid, what statements to use and which element. As I was going to send the puzzle to The Magpie and as this was at the same time as Digimix I decided to use the 1 – 9 digits theme and revisit two of my previous Magpie puzzles Quadratum and Quadratum II which were, as I never tire of saying, fully sudoku compliant and published well before The Times published their first sudoku. Quadratum had three 9-digit squares and Quadratum II had three sums of the type 391 + 284 = 675 arranged in the 3×3 blocks with a further hidden one using the centre squares of each 3×3 block. So I wondered if I could combine the two. I retrieved the list I’d used for Q and QII and set to work. I wanted to use two 9-digit squares crossing in the middle and have sums at the top left and bottom right along with the hidden variety. I pored over the lists of squares (there are only 30 of them) and selected two then went to the sums lists (there are 336 of these if you count A + B = C and B + A = C as different) to achieve a fit. This was of course done by hand and took about an hour. No doubt other setters would write a computer program to get the desired fit but I don’t do that.

That done I had 37 out of the 81 digits placed in the grid and all I had to do was to hope that the sudoku had a solution, which took about 20 minutes to get by hand. The next stage was to bar off the grid so that there were no unchecked cells apart from the middle one, and that each entry was unique and no more than 4 digits long. After all solvers would have to deal with a couple of 9-digit numbers and given the feedback I’ve received on puzzles from solvers bemoaning the fact that their calculators only have an 8-digit display I decided to be kind and have short entries. (Goodness only knows how they coped with Elap’s 11-digit triangular numbers which I personally felt was a step too far.)

The next stage was the prime factorisation of all the entries and this was achieved using my new calculator a Casio fx-85GT+ that cost about £7 from Amazon and has a 10 digit display and a prime factorisation function.

I had to think about the statements that I was going to use and which element. The element would have to have two letters the same, be fairly short and preferably have letters that weren’t too far on in the alphabet. A quick scour of a list of elements soon revealed Iodine as a candidate and as it had two Is, the statement I was now fixed as “It is a perfect square” – and what was this? Iodine has atomic number 53 which would be my age on my next birthday.

I had followed with interest the discussion on the Answer Bank about Digimix and the relevance of the number 73194 which was just a check sum. Some speculated that it was my birthday using the American date method. So I was either very young or very old. Here was a way of answering that speculation in the best way possible with a puzzle. At this point I decided that the puzzle would go to The Listener and not The Magpie as it is only Listener puzzles that are talked about on the Answer Bank as The Magpie has their own on site discussion forum.

I studied the prime factorisations of the entries as well as the entries themselves for some time and needed to have statements that would yield the correct letters for the 9-digit squares. Of course I couldn’t just add any old statements as I had to check that once I had found the ones for the 9-digit squares that any other statements added wouldn’t apply to the 9-digit squares.

Eventually I had 16 statements and writing the clues was very easy!! Then came the most important part of any numerical puzzle – the cold solve or the logical pathway through the puzzle. It is very easy to set a numerical puzzle (honest). You could start with a completed sudoku for instance then bar it off and write clues!! The important bit though is that it can be solved without recourse to exhaustive trial and error and computational power. I recognise that many solvers don’t want to spend hours sitting at their computers using spreadsheets and writing programs but prefer to sit in an armchair relaxing with their favourite tipple armed with a calculator and a few lists of numbers.

It was during the cold solve that I realised I would have to have more statements as there were too many loose ends, so I added one more and that seemed to rectify the situation. I started off again but realised I had made an assumption, namely that there were no repeated digits in any entry during my first cold solve. So I added statement R which said “it has a repeated digit” which of course would never appear in the clues. That done I finished up with 4 cells that had multiple possibilities and of course the unchecked centre cell. I had always intended that solvers would have to study their final grid in order to resolve the central cell but hadn’t thought about any others. I decided to leave things as they were in that there were enough completed rows and columns which satisfied the Latin criterion as well as completed 3×3 blocks which would hopefully force solvers into realising they had a sudoku. And there you have it!!

In retrospect I should perhaps have used 26 statements so that solvers may have thought that the word to be found was sudoku. But maybe that would have been just a tad too clever.

Xenon for next year perhaps?

PS. As Simon Long and Arden have pointed out, there was only one ambiguous cell apart from the centre one and that was the middle digit of 41 across. The other three could be found by logic and I should have found that if only I hadn’t been so incredibly thick.

Oyler.

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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!!

 

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Elementary number by Oyler, a perfect square

Posted by shirleycurran on 11 March 2011

Gloom! A numerical. However, Numpty number one explains what the rubric says. These are elementary rules – simple mathematics! All you have to do is apply each of those rules to all  of the answers to make sure that only the stipulated ones apply. That’s the way you eliminate the others – like that rule R, ‘It has a double digit’. None of them obeys rule R, but a lot of the possibilities do – so they are ruled out! Simple! Best to start with the ones with the most rules and the two-digit numbers, as those are likely to be the easiest to solve.

Well, I leave him to it and soon he has the north-west corner almost complete and an intriguing feature emerging. There are no zeros at all (we knew there couldn’t be any as leading zeros) and we already have most of the digits from 1 to 9 and EACH ONCE ONLY!

When the south-east corner is almost complete and the centre top, our suspicion becomes near certainty and even Numpty 2 is capable of participating. Too much of life is already consumed by crosswords, and I have promised myself never even to look at a S—-U for fear of being hooked. However, Oyler’s construction obliges me to break my promise. Completing this was almost a pleasure and I made a little bit of mathematical progress, learning what a ‘composite number with a three-digit prime factor’ and a ‘sum of four consecutive integers’ are.

With a complete grid, we have to apply those 18 rules to the two long unclued entries: 139854276 and 627953481. My little plastic Donald Duck calculator has problems with such long numbers but between us, we tease out DINO and EI, learning, with surprise, that the vertical unclued light is I, a perfect square.

DINO and EI produce only one convincing word, IODINE, symbol I (see above – a perfect square) and isn’t that exactly what Oyler has produced? A perfect square! That’s clever! What is more, Wikipedia tells us that the element Iodine was discovered just 200 years ago. That must be why that number 5 appeared in the centre of our ‘perfect square’ with a 3 to complete iodine’s atomic number just below it in the grid.

Easy for the experts, I am sure, (still, nothing is stopping them from having a go at those fiendish Magpie numericals is it?) but just right for many of us. What an encouraging first numerical puzzle of the year. I almost enjoyed it (almost!) Thank you, Oyler.

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Listener 4125: Elementary Number Theory by Oyler

Posted by erwinch on 11 March 2011

So, the first numerical Listener of the year and now a week earlier than previously.  This is Oyler’s tenth Listener and he certainly needs no introduction from me.  I am not going to trawl back through my files but I do not recall seeing clues of this form before.  We were told which of 18 statements (A to R) applied to each entry so the statements that did not apply were of equal use to the solver.  There was one statement (R: It has a repeated digit) that did not apply to any of the clued entries.  The expectation was that it would apply to the two unclued entries: 23ac and 5dn, both of 9 digits.
 
There appeared to be numerous starting points but I started with 20ac to which only one statement applied: I: It is a perfect square (3).  There are 22 3-digit perfect squares and all but one could be eliminated by considering the 17 statements that did not apply – statement R was particularly useful here.  20ac = 196
 
On to 20dn HK (1?) where we told that the sum of the first and last digits were a perfect square (H), so 20dn = 10, 13 or 18, and that it is the sum of four consecutive integers (K) The general formula for the sum of four consecutive integers is 4n + 6 where n is the first integer therefore they are all even numbers (from 10 step 4): 20dn = 10 or 18
 
32dn A B E H M P (3) had the greatest number of statements that applied:
A: It is a prime number.
B: Its digit product is odd – all digits are odd.
E: It has a remainder of 1 when divided by 4 – the number ends with 1, 3, 5, 7 or 9 but not 5 (A) nor 1 (P) here.
H: The sum of its first and last digits is a perfect square (ie 1, 4, 9 or 16 but not 1 or 4 here).
M: Its digit sum is a prime number (ie 11, 13, 17, 19 or 23 here).
P: Its digit product is a three digit number (all digits are >2 here).
Again R was very useful since we knew that it had no repeat digits: 32dn = 937
 
39dn IJ (2): it is a perfect square (I) and its digits are all even (J): 39dn = 64
 
38dn BCFG (2) and all digits are odd (B):
C: It is the product of two distinct primes that are less than 100.
F: It is a multiple of 5.  It equals 5 × 3, 7 or 19.
G: It is a triangular number.  38dn = 15
 
32ac BCDEL = 9?  We only needed to consider one statement here:
L: Its first digit is 3 times the units digit.  32ac = 93
D: It has a remainder of 2 when divided by 7.
 
37ac HQ = ?31  We only needed to consider H since R did not apply: 37ac = 831
Q: It is a composite number with a three-digit prime factor.  831 = 3 × 277
 
41ac CEH = ??7  From H it equals 2?7 or 9?7 and from E it equals 217, 237, 257, 297, 917, 937 or 957.  But it is not prime (A) and not all digits are odd (B), which leaves 217, 237 or 297.  Also, its digit product is not a three-digit number (P) leaving 217 or 237.  From C: 217 = 7 × 31 and 237 = 3 × 79 so 41ac = 217 or 237
 
42ac EP = 54?  From P and R it is > 545 and from E it equals 549: 42ac = 549
 
26ac EFMPQ (3)  From F and Q it equals ??5 and from M and P, the first two digits are both even and greater than 2.  Possibilities are 465, 645, 485, 845, 685 or 865.  These all comply with E but not M or Q: 26ac = 685 or 865
 
36ac EFIM (2)  36ac = 25 is the only possible fit.
 
14ac AB (2) and 1dn CF (3)  1dn must end with 5 so 14ac = 53 or 59.  15dn CL (3) = 3?1 or 9?3 but H does not apply so 15dn = 9?3 and 14ac = 59
 
18ac CEGM (2)  There are ten two-digit triangular numbers (G) but only 21 and 45 comply with E while F does not apply so 18ac = 21
 
9dn EHLQ = ??1  From L and H it equals 3?1 and from Q it equals 3 × 107 or 3 × 127: 9dn = 321 or 381
 
13ac KM = ?2 or ?8  From K it equals 22, 42, 62, 82, 18, 38, 58, 78 or 98.  But M excludes all but 38, 58 or 98: 9dn = 381  But 38 = 2 × 19 and 58 = 2 × 29 yet C does not apply so 13ac = 98
 
21ac DFPQ (3) must end with 5 so 22dn HNOP = 5?4 to comply with H.  O: It is a multiple of 12 so 22dn = 564 (504 does not comply with P).  N: It has at least 12 factors (including itself and 1)
 
39ac AM = 61 or 67 but 34dn P = ?19 or ?79 but must equal ?79 to comply with P: 39ac = 67  33ac CJK (2) is even (J) so 34dn = 279, 479, 679 or 879 but A, C and Q do not apply (479 is prime, 679 = 7 × 97 and 879 = 3 × 293) so 34dn = 279
 
The grid now looked like this with one column complete:
 
 
 
We could see that the completed column 9 contained the numbers 1 to 9 once only and that there were no repeated numbers in any row, column or major diagonal so far.  As a rule it is futile to make any assumptions with these puzzles but might this be the property that the completed grid was to display?  There was a 9 in column 2 so would 94 fit for 35ac CKM?  94 = 2 × 47 = 22 + 23 + 24 + 25 and 9 + 4 = 13 so indeed it did while it did not comply with any remaining statement.  However, I was very wary so continued by placing further entries in the top right corner of cells.  I had a moment of doubt when two sixes appeared on the main SW/NE diagonal but the grid was soon complete with all rows and columns complying with the assumption.
 
Factorising the two unclued entries:
 
23ac 139854276 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 73 × 73
 
The statements that apply are DINO
 
5dn 627953481 = 3 × 3 × 8353 × 8353
 
The statements that apply are EI
 
DINO + EI is an anagram of Iodine, the word to be written below the grid.
 
So, the puzzle was complete.  However, there was a little extra hidden in the grid.  Ninas are rare in the Listener since the grid is usually milked for all its thematic material.  It is bad enough trying to find things in the grid even when you are looking for them but I think that the last we saw was Pacific Ocean appearing unnoticed by most solvers, including myself, in News Quiz by Stan back in November.  A clueing device seen a couple of times over the past year is, for example, at No.10 to denote NEON or NE, well, at No.53 we have Iodine.  53 appears seven times in the grid, including diagonals and reversals, but only one is of note and the cells should really have been required highlighting – in an appropriate shade of course:
 
 
 
This also gives us the elemental symbol, albeit in a somewhat squat font, and to argue that highlighting any other 53 could also be judged correct is absurd – our highlighting is central and is found in the two entries that supplied the Iodine.
 
To sum up, this did rather live up to its title in being elementary, in all senses, but I found it a most enjoyable and absorbing puzzle, certainly DEF H! (see puzzle number)
 

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