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Posts Tagged ‘Oyler’

Listener No 4529: St Hubert’s College by Oyler

Posted by Dave Hennings on 7 December 2018

It had been nearly two years since Oyler’s last Listener, Can’t You Do Division? similar to an old Rhombus puzzle from the 60’s. I couldn’t recall an earlier Listener with this sort of theme, although there was an EV puzzle, also by Oyler, t20 based on a cricket match. (Sadly, I failed on that one, but luckily it wasn’t my blogging week.)

Here, bizarrely, we had a bit of time travel at St Hubert’s with a Master creating a way of remembering the security code for his wife’s bank account. Now this obviously put the date he wrote it sometime post-PIN days, which I would guess is in the mid 1960’s. With the added knowledge that the college had a time travel department, all sorts of things could be going on.

Anyway, as is usual with a mathematical puzzle, I had to start again when I had options for 15dn, the number of years since the college was founded, a square and multiple of 1ac. With the clue for 1ac The number of college graduates nominated for Nobel prizes in the 20th century, a factor of 14dn, I assumed that the Master was creating the code post-20c. but the only one that seemed to fit was 1995.

Of course, with the time-travelly bit, anything could be going on, and I eventually assumed that someone had sent Whitaker’s Almanac, or some such, back from the 21st century. A bit of a red herring with the time travel, I thought.

Still, it’s always nice to have a different mathematical and my reworking didn’t take too long. As expected a fun puzzle and not too tricky, so thanks, Oyler.

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St Hubert’s College by Oyler

Posted by shirleycurran on 7 December 2018

I dread those three-monthly numerical crosswords and downloaded this one by the king of the numerical crossword world with immense trepidation, then saw all those perfectly normal clues and heaved a sigh of relief. Years ago, one of the Listener editors told me that solvers are not very fond of this combination of text and numerical crosswords. I have no quarrel with it and there was a fine description of a truly scumbag college to delight us together with all those figures to work out. Yes, it took us until after midnight as I (using the Internet) landed on a list of five-digit primes that didn’t include the one we needed, and we left it rather late in our solve to calculate the three potential ‘bank security codes’ at 14d, one of which would confirm that the master’s wife didn’t produce 19 children, that the college had 19 Nobel nominees in its ranks and 37 post-grads.

Of course, it was Pheidippides, the speedy snail, who produced the whoop of joy at about midnight, when all our speculation and calculation was confirmed and I have no issue with him, but must severely castigate Oyler for publicising such a disgraceful seat of drunkenness and debauchery. Yes, of course I checked the alcohol content of the clues and was appalled when we learned that 699 bottles – almost certainly of the finest quality – shall we guess about 50 pounds a bottle? – were consumed at a single dinner. That’s 35,000 pounds. Nearly three times the Master’s annual entertainment allowance. No wonder his bursar spent over two and a half years in prison for tax evasion. They were probably in drunken cahoots (but cheers, anyway, Oyler!) Somebody has to fund all that boozing and we guessed they must be charging hefty fees from all those overseas students (yes, we did wonder for a while, whether they comprised part of the undergraduate/graduate body or had to be counted separately).

Fibonacci, the cat, seems to be the most effective of the whole bunch of them, but 55 mice! (Some were probably rats). With 54 non-academic staff, and probably even more academic staff, the staff-student ratio of this vermin-ridden place is totally skewed and the domestic staff, who should be shifting the snails, mice etc. are probably slewed with all that wine – certainly not doing their job – but spending their day with stopwatches timing snails round the quad. I ask you!

Or is that how the Master spends his day? The other Numpty declared that he should be fired for gross moral turpitude. 39 years old and he had already fathered two or three bastard offspring before baby-snatching a wife twenty years his junior, a kind of Lolita, just about as randy as he is with her own sprog or sprogs in tow. And what do they do? Produce ten more ‘in or out of wedlock’ (seducing pretty students?)

The whole set up is shady. With all those ‘graduates’ from a body of merely 349 students, something is fishy. There must have been about 140 graduates a year over the 361 years of the college’s existence so something is going on. Are they in the business of awarding shady postal doctorates for a fee? Shame, Oyler! It won’t do.


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

Posted by Dave Hennings on 10 March 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 March 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 June 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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