## 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 p^{2}-q^{2}, 2pq and p^{2}+q^{2} 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 3^{2} 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: E3^{2} + (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 777^{2} + [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 2×3×5
^{2}×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 4421891*50*, 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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