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Archive for December, 2010

Three-Square by Elap (Macey’s Pizza Parlour!)

Posted by shirleycurran on 17 December 2010

This rather strange grid was accompanied by the announcement that next year the numerical puzzles will be set for the penultimate week of February, May, August, and November, to avoid their falling on UK bank holidays. I heard a fine rejoinder. We should persuade the Cameron-Clegg powers that be to declare every weekend a bank holiday – that way, perhaps we would be rid forever of these beastly things.

We had over a foot of snow and more falling steadily so this brute was not detracting from the lawn-mowing or weeding time, but for the entire weekend, the air was, to put it decently, somewhat blue, as dead-end followed fruitless path. Mr Math worked his way through the entire thing (sorry Elap, but nothing will persuade me to use kind words for it) with a silly little plastic calculator a pencil and a mountain of paper. (OK, the choice is ours – we abandon numericals or buy decent solving equipment – but should a cross WORD – my caps. require Excel and a state of the art calculator?)

‘A primitive triangle is one in which the lengths of the sides are relatively prime’. Very kind of Elap to tell us that but what, exactly, does it mean?. I got it wrong and thought they all had to be different primes – but not so, two of the sides could share a factor, but the only common factor for the three had to be 1. Give me ‘Stripey horse (5)’ any day! I know what a scalene triangle is now and a Heronian triangle, as that seems to be another name for them, but I’d rather have the usual menu of tsetses, asti and meris.

Well, we got there, and that last sprint towards the finish line was almost (repeat, almost) fun, as entry after entry confirmed that we were right. And we sat and gazed at it. What were we supposed to do with those purple patches in the corner? We soon realized that none of the individual columns or rows provided the figures for sides of Pythagorean triplets, primitive or otherwise. We attempted to factorise the numbers to produce a pattern, we added, multiplied, subtracted, divided and cursed. We even considered cutting holes in the grid or cutting it up and reassembling it to see whether we could make three squares, an origami wren, or a Rubik cube, but to no avail this week.

READ THE PREAMBLE! I’ve been reiterating it all year long and some of last year too. That sentence ‘Solvers must insert appropriate digits in the triangular groups of shaded cells to ensure all rows and columns are thematically consistent’,  has to be giving us a hint, otherwise, any numbers we insert will be sheer guess work. The theme seems to be ‘triangles’. I feed one row into Google and seem to have hit lucky. Up comes the phone number of Macey’s Pizza Parlour in Twin Forks, Idaho. No, seriously, I see a string of vaguely familiar numbers … 1,3,6, 10, … Had it been any other row (I tried!) the answer ‘Triangular numbers’ would not have come up!

So, by an astonishing stroke of good luck, we hit on the theme but there was still hard labour to perform (with my little plastic calculator and a formula) to complete those 9 and 11-digit numbers where the first three digits, in two cases, were missing.

I have to admire Elap, all the same. How long must he have spent working through ludicrously large triangular numbers to produce the few that would give him a5 and b4, for example, – two surface areas of scalene triangles, running vertically? Despite being a complete mathematical numpty, I can recognise the remarkable feat of construction that has gone into the compilation of this grid. It must be even more complex than making nine and eleven-letter words intersect in rows and columns (as there are only ten available digits and twenty-six letters). This sort of numerical genius is out in that deep snow – way beyond my comfort area. So I should stop being churlish about being a hopeless mathematician and appreciate this numerical gem! As always, I do hope we’ll be honoured with a setter’s blog.

Warm thanks, Elap, (but please, please Mr Cameron, decree 52 bank holiday weekends next year!)

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Listener 4114: Three-Square by Elap

Posted by erwinch on 17 December 2010

Elap has provided us with a numerical puzzle every year since 2003 – let us hope that this is a tradition that will long continue.  I first had to do a bit of revision for I had forgotten the meaning of the term relatively prime, if indeed I ever knew it.  I rather fancied to pronounce primitive as prime-itive in this context but am reliably informed that anyone doing so would be open to ridicule.  Of the four sets of clues, the Threes reminded me of a notable sequence.  Which of the following is the odd one out?
The last is the only one that isn’t prime (= 17 × 19607843) and there are no more primes in the sequence until 333333333333333331 at 18 digits.
It was the Threes which provided the opening here for me since P2 and j2 intersected (the p’s and q’s in italics are the values derived from the Pythagoras theorem):
333333 = E3 * P2 * j2
Factorising 333333 = 3 × 3 × 7 × 11 × 13 × 37
So {E3, P2, j2} = {117, 77, 37}, {143, 37, 63}, {693, 37, 13} or {777, 33, 13}
But 333 = B3 + N2 + j3 so j2 = 13
Looking at the second Right-angled clue:
E3, E3 – M3, V2 * X2 + Q3
All possible values of p and q for E3 = 693 or 777 (the odd term so p^2 – q^2) were looked at.  So for E3 = 693 (Sq rt > 26) I started with p = 27 and then 28 looking for integer values of q until M3 in 2pq was too big.  27^2 – 6^2 = 693 and 2pq = 324 but M3 = 369 whereas we knew that the final digit was 3.  I found just one fit with M3 = ??3: p = 29, q = 8: E3 = 777, M3 = 313, P2 = 33, N2 = 13, V2 * X2 + Q3 = 905, B3 = 18? or 19?
From 3333333 = P2 * X6 + W3 we find that X6 falls within the range 100980 to 101007 so X2 = 10
Since V2 * X2 + Q3 = 905 then Q3 = ??5, R2 = ?5, S3 = 5?? and g4 =???5
From X3 * X3 = R2 + g4, X3 = 100, X4 = 1009 and g4 = 99?5
From h2 * h2 = S3 + i2 we see that h2 falls within the range 23 to 26, e3 = 12? and D2 = 27
And from e3 * e3 = F5 + Q3, where F5 = ????1 and Q3 = ??5, we see that e3 = 124 or 126
Now looking at the first Right-angled clue:
D2, s3, q3
We know that D2 = 27 (p > 5) so s3 is even = 3?? and q3 = 3??
Just one fit was found: p = 14, q = 13: s3 = 364, q3 = 365, {W3, X6} = {894, 100983} or {564, 100993}
Then the third Right-angled clue:
G2, Q2, k2
G2 = 4? or 6? and k2 = ?3
Since we had k2 = ?3 (p^2 + q^2) the possible values of p and q were limited and we soon had two fits: {G2, Q2, k2} = {45, 28, 53} or {48, 55, 73} and e3 = 124
If G2 = 45 then e3 * e3 – Q3 (285) = F5 (15091) but F5 would equal ?45 ?1 so G2 = 48, Q2 = 55, k2 = 73, F5 = 14821, Q3 = 555, V2 = 35, R2 = 55, g4 = 9945  [Up until this point I had thought that Elap had kindly presented the first two terms in Right-angled clues in the order that could be written p^2 – q^2 then 2pq]
The grid now looked like this:
And so it went: from W2 * W2 = b4 + u2 – we knew that W2 = 89 or 56 and b4 = ?878 or ?978, where 56^2 (3136) could not fit, so W2 = 89, W3 = 894, X6 = 100983, b4 = 7878, u2 =43 and B3 = 18?  But I shall not go any further with the working here.  A full solution should be available on the Listener site but I doubt that it will tally with my route much beyond the first few steps if at all.  I also suspect that it will expose my route as being clumsy and inept.
Of the 22 clues, all but the second Scalene (16th solved) were left until the very end.  The final clue solved was the third Scalene:
G3 + L2, N3 + z2, y3, b4
This had boiled down to:
7878 * 7878 = (504 + x/2 + y/2) * (x/2 + y/2 – 1) * (301 – x/2 + y/2) * (204 + x/2 – y/2) where x was the second digit of z2 and y the third of y3.  I started off using a calculator with x = 0 and y = 0 to 8 step 2 (both would be odd or both even) but got fed up with the calculation and so used BBC Basic.  Not for the first time with the numerical puzzles, I should have stuck with the calculator since x = 1 and y = 5.
So, all the clues were solved and we had five complete rows and five complete columns – what was the common link between these large numbers?  They couldn’t be relatively prime since more than one was even so I first thought that they might be perfect cubes since that would link the threes, squares and three sides of the triangles – to my mind at least – but no.  I then tried triangular numbers and this fitted.
The general formula for the nth triangular number, ½n(n+1), was useful in completing the grid, the positive root of the quadratic equation telling us roughly where to look, but it was troublesome in the two cases where we did not know the first digit to have to repeat the calculations – I started with 9 as the first digit and went down.  Here is my final grid:
As ever with Elap, this was all tremendous fun and I eagerly await his next.  However, of the four sets of clues, I couldn’t see what was thematic about the Squares other than the link to the title and possibly Pythagoras.  Here is a final thought: if you wanted to play snooker with 96972777028 red balls, the 440392nd and largest triangular number in the grid, then you would need a frame with sides over 23km long! 

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Yo-ho-ho …

Posted by Dave Hennings on 17 December 2010

I would like to wish everyone who posts, comments, or just visits here a very happy Christmas. As well as my co-bloggers, especially Shirley and Erwin, I would like to thank all the setters and the editors for the enjoyment, happiness and panic that has gripped me at various times during the fifty-one puzzles that we’ve had. May 2011 bring as many superb Listener puzzles as have been presented to us this year.


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Listener 4114: Three-Square by Elap (or Gridlock, Gridlock, Gridlock)

Posted by Dave Hennings on 17 December 2010

It’s probably because I don’t get much practice at them, but I approach every mathematical with a certain amount of dread. Actually I approach every Listener with a certain amount of dread! I suppose one of the big differences is that mathematicals will probably only have one way in, whereas you can start a crossword pretty much anywhere.

So here was Elap’s latest, and a somewhat daunting grid: 9×9 but with three extra squares poking out from each side. Clues were in four categories: Threes, Squares, Right-angled and Scalene. We were also given a couple of ‘solvers may like to know’ hints. The first was that the sides of a right-angled trianle can be expressed as p2-q2, 2pq and p2+q2 where one of p is even, the other odd. The other was that, if the sides of a triangle are a, b and c, and s is half the perimeter, the area is √s(s-a)(s-b)(s-c). These hints actually made it seem even more daunting! If only Rhombus had obliged in a similar fashion!

I found the entry point fairly quickly. Numbering the clues 1-22, clue 4 was 333333 = E3 * P2 * j2. 333333 factorises as 32 7 11 13 37, and the second digit of j2 is the first of P2. From the factors, j2 can only be 11, 13, 33 or 37, but 11 and 13 require E3 to be too big. From clue 1, j3 must be between 100 and 222, thus limiting j2 to 22. From 13, s3 * s3 = j6 + v3, so max(s3 * s3) = 139999 + 999 (140998) and max(s3) = 375. The last digit of P2 is the first digit of s3, so P2 cannot be 37 but is 33. This also gives E3 = 777 and j2 = 13.

Next came clue 15: E32 + (E3 – M3)2 must be square and the first bit of trial and error; I think this is called ‘by inspection’ or ‘by examination’, but you know what I mean. I’ve also developed my own shorthand for making notes of my route through these minefields. In this case it’s ‘try 7772 + [674 reducing by 10]2 for square’. Before too long I get E3 – M3 = 464, M3 = 313. Of course it’s necessary to try all the possibilities in case there is more than one.

I won’t go into all the details of my solving process, but clue 14 resolved D2 (27), s3 (364) and q3 (365); clue 5 gave X2 as 10 and clue 15 gave Q3 ending 55.

I find that there’s always a point at which you know that everything’s going to be downhill from there. It’s very satisfying when a 4- or 5-digit number fits with two or three digits already in place. All in all, I think it took about three hours to solve the clues and fill the grid with the exception of the six pretty pink squares in three of the corners.

So that was the downhill bit. What followed was an uphill struggle of gargantuan proportions. I tried dividing by 333, 3333, 33333, 333333, 3333333 (and even 3 and 33), but, apart from all the acrosses being divisible by 3, the downs were a real mish-mash. I didn’t expect that to work, so I wasn’t disappointed. What I was really expecting was that the thematic link between all the rows and columns would be primeness in some way. I had been a bit surprised that it had been unnecessary to use the relatively prime aspect of the triangles during the solving process, and that worried me a bit. The rows and columns were not prime, and nor were they in groups of three, with each being relatively prime to its neighbours … too many even numbers for that. To my surprise relatively prime is in Chambers (under prime), but there was nothing in its definition that gave me any clue.

At some point I checked that the sides of the triangles were indeed relatively prime, and the first six were, but the seventh had sides of 505, 204, 305, so two of the sides had 5 as a common factor. My heart plummeted! Hadn’t I just finished the puzzle correctly? I wondered if I had made a mistake near the very end; it was unlikely to be near the beginning as everything had come out so neatly in the wash. I carried on wondering for about a day, with no light dawning about the thematic connection, and decided that there was only one thing to do … check everything from the beginning! B*gg*r!

Needless to say, I didn’t find a mistake … but it took nearly as long to check all the logic as it had the first time! I was beginning to panic. As far as I am aware, I am all-correct this year. True, I may have made a silly mistake, but I haven’t checked any of my entries against the official printed solutions. I know that I have solved all the final steps, at least in theory, but who knows what I may have put in the grid, or not highlighted, or …! Hence the panic.

My mind focussed on the title: Three-Square. Well I’d already proved that not all numbers were divisible by 3 and none of them was prime. So did 9 come into it in some way? I checked Chambers. It was there: having an equilateral triangle as cross-section. Well, zippy-doo! Those of you who didn’t make a dog’s dinner out of the final step of this puzzle must be wondering why I couldn’t see what was staring me in the face … I just wish I could’ve told you.

Finally, it was the Tuesday evening before the day my entry had to be posted. Up to that point I had made use of a nice site called The Number Empire which was perfect for checking factorisations. Who knows why, but I thought I’d try WolframAlpha. I’d used it for some trivial stuff before (something like ‘what is the 76th day after 19 February’?!). It describes itself as a computational knowledge engine and is incredible, not just for mathematics, but for a whole host of other stuff as well. For example, key in ‘London’ and it’ll give you its population, local time, current weather and a couple of maps.

So I just keyed in the number from the central row, 55544611650 to see what would pop out. I was told, among other things, that:

  • it’s 110011101110101101111000001101000010 in binary
  • has factors 2×3×52×11×79×101×4219
  • is an even number, and
  • is the 333,300th triangular number!

I’d come across triangular numbers in an earlier Listener mathematical (back in 2006 with 3866, Numbers Game by Polymath), but they obviously didn’t make much of an impact. I tentatively keyed in the number in the central column, 78783658878:

  • it is 1001001010111110111110011101101111110 in binary
  • has prime factors 2×3×19×1741×396947
  • is also an even number and
  • is the 396,947th triangular number.

I found the formula for triangular numbers: n(n+1)/2 and tried the first row as 442189150, doubled it and took its square root to give 29738.4986 and change. I asked WolframAlpha ‘what is the 29738th triangular number’ and it told me: 442189191!

Ten minutes later, and I had a grid with eighteen of these damned numbers all interlocking perfectly. This was as near to magic as I have seen in a Listener mathematical. Absolutely amazing, and the cause of much relief to yours truly. So many thanks Elap for a masterpiece of engineering (a bit like Pieman’s Liberty Bell last week, but with numbers). If we meet at next year’s Listener Setters’ Dinner in Sheffield and I shake your hand, you’ll know that, despite the smile on my face, somewhat darker thoughts will be going through my head :wink:!

Oh, and the relatively primeness of the three sides of the triangles? I am told (by the man himself) that it means that ‘there is no common factor for all three sides – ie the triangle is not a magnification of a smaller one which has integral sides’. I should have know that a Listener mathematical wouldn’t have a little mistake like that!

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Listener 4113: Liberty Bell by Pieman

Posted by erwinch on 10 December 2010

This was only Pieman’s fourth Listener as a solo effort, he paired with Mr Magoo in Collusion, but he has gained a formidable reputation with this small output.  He went undercover for 10 by 8 last year, which was one of the most difficult Listeners in recent memory.  So, anticipation was high when we saw a large grid, with many bars missing, accompanied by a shortish but tricky looking preamble.  I thought it odd that we were told to expect no indication of obsolete or other restricted usages in clues.
I was able to make steady progress but struggled to understand the clue for ciborium (42dn) until realising that Pieman had exploited the current vogue for removing apostrophes, even from long established usages to be found in the OED such as mind your P’s and Q’s (1779) and the three R’s (1825):
Is welcoming neighbour aboard with vessel (8) ciborium – BOR in (I + I) in CUM
Pieman liked the device so much he used it a second time, in 27dn:
Irritation as flushed from bad anaesthetics (10) tetchiness – (A)N(A)ESTHETICS (anag)
Hands up all those who first looked for an anagram of (A)NAE(S)THETICS – there is an unsatisfactory one.
I have yet to see us in this context but perhaps Pieman thought he would get the two in early for I can see this becoming as instantly hackneyed as a bit of a kip for at.  Also, it is patently absurd for we are never going to see the likes of the following sentence: embarrassed is spelled with two rs and two ss.  I dislike seeing people getting Trussed up over such trivial matters.
Talking of hackneyed, when did you last see rhino in a crossword clue (63dn) that was not to be read as money?
I thought the anagram entries a clever way to delineate certain areas of the grid and liked the way that bars could be deleted by joining letters or words: E|V|I|TATE became evitate and MAY|FAIR Mayfair, etc.  Eleven formerly unchecked letters could be arranged into the four-word quotation:
I have a dream … (Martin Luther King, 1963)
Finally, changes affecting the anagrammed entries only left a cryptic representation of a second quotation:
Let freedom ring (Samuel Francis Smith, 1831)
If you look at the full text of the I Have a Dream Speech you will see that King also used the phrase Let Freedom Ring repeatedly.  Both quotations appeared in the grid as follows:
The preamble told us to leave only real words and names when adjusting the red bars so are the isolated single letters (G,R and B) real words?  I would say that they certainly are.  Chambers lists all the letters of the alphabet as nouns and they appear in the very first books that we read: A is for Apple, B is for Bear …  Besides, we had two clues where the plurals of single letters were to be considered valid.  Thematically we might have freed the three ‘imprisoned’ letters and had they been MLK or SFS then I would have had to think again.
Following from Xanthippe last week, I felt that there were no real PDM’s here.  The red bars could readily be seen to be forming letters although there was something funny going on in the SW corner.  So, great fun but perhaps on the easy side for Pieman.  I have to say that I did not notice the lack of obsolete or other restricted usage signals in clues.

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