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Posts Tagged ‘Oyler’

Listener No 4438: Can’t You Do Division? by Oyler

Posted by Dave Hennings on 10 Mar 2017

What a fabulous start to the preamble this week, harking back to the days of mathematical puzzles by Rhombus. I have to admit that they always seemed far too mathematical, and I tended to give them a wide berth. I wondered if this would be more gentle.

As most of us probably did, I dug out my copy of the Listener in question and read its preamble, although I’m not sure that’s what they were called way back then:

“An integer x has y distinct divisors, including 1 and x, and this is represented by the relationship x(y).”

That was it! No wonder I gave them short shrift. However, if your shrift isn’t so short, the puzzle is at the end of this blog, and I hope you enjoy it.

I was a bit worried by the title of Oyler’s offering. Rather than Can You Do Division? — II, he called it Can’t You Do Division? It was almost as though he was looking down his nose at me and challenging me to succeed!

With a crossword, there are normally many ways into the puzzle. The trick with a mathematical is finding the one way in, as I think I’ve said before. [Yes, you have. Ed.] This week, I started off by expanding the information provided by the preamble. If a number has 2 factors, it is prime. If it has 3, it is a square. If it has 4, it is the product of two primes or a cube. If it has 5, it is a fourth power. Despite this last note, I failed to realise that any number with an odd number of factors is a square. (I know this because Oyler has just sent me his setter’s blog for the puzzle!) I’m not sure this would have helped particularly since, for example, m (21) was either 22×36 or 26×32. The first was too big, 2916, but the second was just right, 576.

Before that, I got S (5) which had to begin 1, 3, 7, 9 (t (2) is prime), so had to be 16 (24) rather than 81 (34). After that, K (7) was 64 and then p (35). (4+1)×(6+1) = 35, so we had a fourth power times a sixth power which was either 16×729 or 81×64. This first was too big (11664), so it was 5184. After that, I homed in on q, u, J, G, N, c, s, etc, etc. There were a couple of minor hold-ups but all in all, a reasonably quick solve just short of three hours.

I must admit that without a calculator I’d have been at a loss. The Listener web site mathematical appendix does have a list of factors for all numbers up to 1000, and that proved extremely useful, but I did need to use Wolfram to check the primeness of Q 14657 and C 428137!

Plus, and you’ll note that it is missing from Rhombus’s puzzle, every row and column contained all the digits 1–9. This certainly sped up the last few entries for me; indeed, some of the clues didn’t actually need solving. Of course, one of my last tasks was to solve them just to make sure. Even the squares in the perimeter didn’t need much work since the unchecked cells all came out in the pandigital wash.

For those of you interested, the solution to Rhombus’s puzzle (which I shall post here next week) began with the following explanation:

“If N = πxθyφz… where π, θ, φ are primes and x, y, z … are positive integers, then the number δ(N) of divisors of N is given by δ(N) = (x+1)(y+1)(z+1)…”

So special thanks to Oyler and/or the editors for making the puzzle slightly more understandable for the likes of me. The explanation in the preamble certainly helped make it a fairly straightforward solve.
Rhombus No 1916

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Can’t You Do Division by Oyler

Posted by shirleycurran on 10 Mar 2017

001The dreaded numerical! We download it and find Oyler. Well he’s the leader of the numbers brigade isn’t he? Co-editor of the new Crossnumbers Quarterly (here’s a plug – why not torture yourself an extra eight times or more every three months!) There isn’t much point my checking his right to admission to the Listener Tipplers’ Club is there since he has previously justified that with a few doubles and triples! I am fairly confident, too, that I’ll find that elusive HARE lurking somewhere in his final grid, though clearly it can’t be in a straight line (81185) as the preamble immediately tells us that each row and column contains all of the digits from 1 to 9 inclusive – a kind of sudoku without the division into 9 little mini grids and with the bonus that ‘The four entries around the perimeter are perfect squares to be deduced’.

While the other Numpty is muttering abusive words about numerical things not being crossWORDS, and beginning to fill scraps of paper with putative solutions, I run off an internet list of perfect squares with the digits 1 to 9 inclusive. I am sure there are gifted solvers who did this puzzle entirely with pencil and paper – even the list and number of factors of each solution – but I confess to using internet help and it was still far from easy. I hand over to the more numerical Numpty to explain how he proceeded.

We have 6 primes in the grid (total factors 2) and one squared prime (total factors 3). Useful, but the easiest way in seemed to involve K (form probably a**6), S (form probably a**4) and p (form probably a**6 x b**4), where the total numbers of factors implied entries 64, 16 or 81, and 5184. Happily this in turn suggested e to be 152843769. On the way! Limited options now for both B and b give the possible perimeters, and then it was a matter of inspection often for candidate numbers (such has u, xx6, of the form a x b**6).  One difficulty was that a search of an apparent  6-digit prime table didn’t actually contain the needed answer, for C, 428137!  A ‘number of factors’ test of the options (only 6 at the time) gave the required answer. However, despite its being a numerical, I enjoyed this one, which was attackable without hours of gazing forlornly into space, and didn’t (seem to) contain shoals of red herrings:  my thanks to Oyler (from the semi-numerical Numpty)!

I hope Oyler will comment on how it could be solved without quite a lot of labour, programming or the very useful internet tools?

oylers-alphanumerical-hares-001Later, when he sent his setter’s blog, Oyler commented “You missed the fact that m had to be 576 as it is the only 3-digit number to have 21 factors and was the first one I put in. Any number with an odd number of factors must be a perfect square”. Now I (the numberphobe Numpty) ask myself, why did I spend all those years learning every word of Shakespeare’s Hamlet and writing excited essays on Yeats’ and Keats’ poetry, Austen’s novels and Scott Fitzgerald’s The Great Gatsby when I could have been learning fascinating facts like that!

And those alphanumerical golden hares. Of course they were there – two of them!

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Square Time Sums by Oyler

Posted by shirleycurran on 10 Jun 2016

OylerPerfect SquaresI know that the editors are keen to publish ‘easy numericals’. For this Numpty, that is an oxymoron. However, if anyone is able to produce a numerical Listener puzzle to order, it has to be Oyler. We approached this ‘Square Time Sums’ with more than the usual trepidation (at least, the other Numpty did) as we were travelling between a son’s wedding and granddaughter’s christening with nothing but pencil and paper. However, I did manage to print out tables of the first 1000 prime numbers and square numbers of two, three and four digits and purchase a £1 calculator which later gave my three-year old grandson immense glee when he found out that you can change a whole row of nines to zeros by simply adding one. I suppose numbers can be fun.

Does Oyler qualify for the Listener Setters’ Toping Club? As I scan his clues and preamble, I realize that he is confirming his seat of honour as a founder member with not just doubles but triples – and even worse, four of them! How on earth did he stay sober enough to compile this! Cheers! Bottoms up.

The other Numpty grumbled and muttered for a few minutes and then announced  “I know I can compute the possible triples in my old favourite GWBASIC if I can convert my French laptop keyboard to an English one to find the odd characters like <, $ and % and there are at most about 20,000 triplets. The 1-9 appearing once only requirement might cut this down a lot!” Soon  afterwards he happily remarked “Well, there are only seven possible triples'”- and shortly after that he slotted in 1, 9 and 15 across and 2 down and fixed JKL and XYZ

ABC = 36, 729, 5184

JKL = 16, 784, 5329

PQR = 81, 576, 3249

XYZ = 25, 841, 7329

A steady solve followed, with the usual backtrack to find where the fat-fingering on the tiny £1 calculator had led to an error but a couple of hours later the grid was full and a careful check confirmed that it all worked.

What was left to do? Two complete rows and two complete columns “that are related to the theme” had to be highlighted. It didn’t take much nous to see that only four of the rows and columns contained all the digits from 1 to 9 but what was special about those? The answer astonished me – they are all perfect squares! How did Oyler manage that! Thanks to Oyler.

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Elementary number by Oyler, a perfect square

Posted by shirleycurran on 11 Mar 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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