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Posts Tagged ‘Rhombus’

PQRST by Yorick

Posted by shirleycurran on 15 Sep 2017

The preamble told us there were three different types of clues; normal ones which would be involved in clashes, ‘most’ other ones that would have a misprint in the definition part and the remaining ones where the grid entry (given by wordplay) would have a wrong letter. That last one is an original device (and it led me to a truly Numptyish grid stare at the end, as I was stupidly attempting to make those letters spell the ‘further item’. Read the preamble! It clearly said ‘the correct letters can form the name of a further item and, of course, that RHOMBUS was a clanging penny drop moment.

First, of course, I had to confirm that Yorick retains his place in the Listener Setters’ Toping Society and, of course, he does with ‘Experts at making safe highballs and methaqualone (9, two words)’ “They are BOMBS”, said the other Numpty and he checked that ‘methaqualone’ in the Big Red Book is QUAD, so producing BOMBSQUAD. This was followed by ‘Regular drinks forgotten as America leads (6)’ We’re heading for the USA tomorrow and our USUALS there seem to be Starbucks’ coffees but ‘usuals’ are regular drinks in the pub in BRB and there is that forgotten old Germanic word ALS for ‘as’ so we had found one of the clues where the wordplay gave USAALS but the definition gave something different. So with highballs and the usuals, cheers, Yorick.

We didn’t find solving easy at all as this was very subtle cluing and a number of the words were new to us: MANDIRA, CLAES, UMBONAL, RALLIDAE, STEWCANS, FADIER and LATERIGRADE, for example, and I didn’t know that a jargon was another name for a ZIRCON, ‘Page one may be jargon, changing atmospheric treatment’s start to literary opposite extreme (6)’ No, this was  definitely not one of the relatively easy crosswords to encourage Listener newcomers. We had AGUIZE in place already and the Z?RCON, so this was likely to be ZIRCON but we had to work backwards. We needed an L corrected misprint (as B?ANKS was already in place) so we were probably looking for a PALE ‘zircon’ and, sure enough, the BRB told me that jargoon or jargon was just that. Of course, we had to work out the exact wordplay of every clue to check that one of those ‘wrong’ letters wasn’t there so we teased out AIR-CON with the A changing to the other alphabetic extreme, Z.

With just a break for dinner and rather a long struggle getting the computer to print the entire puzzle, which had not only arrived late but also somewhat truncated, it was late evening before we had a full grid and the message DRAW FIVE SHAPES THROUGH NAMES AND BLANKS. I needed a clue as my extra item seemed to be ASMDDAO. A little voice said “Read the preamble” and, of  course there it was. The CORRECT letters of the wrong wordplay entry. BSUORHM immediately gave me RHOMBUS and the light dawned.


Mad March(esa) MARA

relevant title) then had to think about that RHOMBUS . My first attempt was too small (I wonder how strict the marker will be!) and the other Numpty drew one for me that would fulfil the requirement (equal and parallel sides with an area equivalent to twenty cells. It just fitted nicely into the space left by the other shapes. Many thanks to Yorick for the challenge; there was so much in this impressive compilation.

The HARE? – ‘March'(ESA) was a big hint but it was a MARA this week mischievously hiding behind that rhombus – a  Mad March Mara.

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L1916: Can You Do Division? by Rhombus

Posted by Encota on 10 Mar 2017

I’d like to claim that it’s taken me 50 years to solve this one, following some music-induced haze that had resulted from listening to the 1967 releases of Sgt. Pepper, Pink Floyd’s The Piper At The Gates Of Dawn (aside: how surreal is that Chapter VII of The Wind In The Willows?) and The Velvet Underground & Nico but I can confess that Oyler kindly shared this puzzle with me when his ’50 years on’ L4438 Can’t You Do Division? came out in February 2017 – and of course I couldn’t resist the challenge.

So here is the puzzle…


I started by creating a small pencil-drawn grid with 10 to 90 along the top and 0 to 9 down the other, so I could begin to see how the number of factors distributed.


This allowed me to solve (for example) clue b, where the only two appropriately-sized numbers with 10 factors are 48 and 80 and, presuming that clue C doesn’t begin with a zero, 48 it is.

I wasn’t familiar with any of the mathematics behind factorising, so was childishly pleased with my very rusty skills when I managed to prove that, if p, q and r are prime, then

p^a*q^b*r^c has (a+1)(b+1)(c+1) factors

[Here I am using ^ to mean ‘raised to the power of’, e.g. 2^6 = 2*2*2*2*2*2 = 64 ]

Aside: this is actually a lot easier to prove than it may at first seem to non-mathematicians.  If you know how many factors p^a has, then multiplying it by q^b gives you all the original factors and then a new same-sized set when the original factors are all multiplied by q, then another same-sized set when they are all multiplied by q-squared etc all the way up to q^b.  So you have an additional b times the original number of factors (count them!), i.e. (b+1) times that number in total.  This can be extended to as many primes multiplied together as you like, resulting in (a+1)(b+1)(c+1)(d+1)… factors – or to fewer, to prove the original p^a must have (a+1) factors.

Once you know this, it then becomes straightforward to see the relatively few options that are available for some of the answers.  The ultimate example is clue p (down).  Here the clue says there are 35 factors.  The only ways to factorise 35 as a product of two integers are 1*35 or 5*7.  For the first the answer would have to be prime^34.  But even 2^34 is way too big, so the only other way is Prime1^4*Prime2^6, and the only one of those anywhere close to being between 1000 and 9999 is 3^4*2^6 = 4 * 1296  = 5184.

Clue N with 5 factors had to be a prime’s fourth power, but 3^4 is 81 and 7^4 is way bigger than 999, so 5^4 and 625 it is.  I think I knew being forced to learn all squares up to 25-squared at school would come in useful one day 😉

And Clue a had to be a nine digit prime to the tenth power.  Now anything to the tenth power gets big very quickly – clearly 10^10 is too long with eleven digits (1 followed by 10 zeroes) and 5^10 is less than a thousandth of that so won’t have enough digits, so Clue a can only be 7^10.  Pencil and paper readily reveals this to be 282475249.  Some of you will have set your own rules for what help you allow yourself here – likely to be somewhere on the [I did it all in my head > pencil & paper > abacus > calculator > spreadsheets > Internet lists > Wolfram > programming] scale.  I think I would have struggled without a pencil and paper…

Similarly Clue A is a prime^12 and the only one remotely the right size is 5^12 (which must end in 5).

Clue e is a prime^6, so with nine digits can only be 23^6, 29^6 or 31^6.  A bit of jigsawing with Clue A allows it to be identified as 29^6, i.e. 594823321.

Other answers where the number of factors was either prime or a product of two primes went in fairly quickly using the above, resulting in most of the central SW to NE entries being solvable.  LOI was, if I recall right, clue T, which had to be a multiple of 3^12, i.e. a multiple of 531,441.  As it ended in a 1 and needed to be around 1700 to result in a nine-digit number beginning in 9, it was soon clear that it could only be 531,441*1721, again given other jigsawing requirements.

So I finish my checking listening to the dulcet tones of Vivian Stanshall & co. on the Bonzo’s Gorilla and Captain Beefheart’s Safe As Milk, slightly worried by the quantity of 1967-released material in my music collection (given I was around five at the time).


Tim / Encota

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