One for the listener-a-day desk calendar

Hi and welcome back to a slightly lateish edition of George vs the Listener Crossword.  The US will be going on to daylight savings time next week which will probably also throw out my regular schedule somewhat, but hey, it brings up the number of hits to the site when you check back several times to see if your preferred site for bad jokes based around crosswords is up.

Crossnumber time!  It’s Oyler, which means the pendulum will swing from very easy to very hard.  A walk through previous Oyler Listeners shows that last year’s Digimix was very hard, and I had to resort to on-line aids to get through it. Oyler checked in on that one, so hi Oyler if you’re looking in again.  Before that came Pentomino Factory, which was right in my realm of mathematical knowledge and was complete over a single lunch in a bar with no calculator.  Before that was Odd One Out where I had the ignomy of two completely blank grids, and one grid with only a few entries.  So in my completely statistically proven theory of the Oyler Pendulum, this should be an easy one, and there’ll be a hard one next year.

Eighteen statements, A-R and the clues are all of the statements that match up.  Interesting set of statements – I started off by circling a few that I thought would get me started – all the cells are 2,3 and 4 digits except for the two long unclued ones, so O is useful, there aren’t that many multiples of 12 (bugger, there aren’t that many Os).  L should make a lot of them a giveaway (and there’s a few Ls) limiting them to 3-1, 6-2, 9-3.  H is also a nice one to work with – the digits would have to add to 4, 9 or 16.  And good old F is multiples of 5, so they have to end in 0 or 5.

My first stab at this one was in breaks during a lab I was proctoring, I had an old calculator and was starting to jot in digits here and there.  I think 36 ac being 25 was the first obvious one, then 38 down being 15, and with 39 down being 64 (the only two-digit perfect square where all digits are even), I jotted down that the Florida corner was looking awfully like a sudoku.  Add to that 18 across being 21, making the first digit of 9 down a 3, and my right column looked like 3-15-4–9 with 7 being one of the two digits that saitsfied the last number in 39 across, and I’m convinced that is what we’re looking for.  Not to mention I hadn’t found any 0’s yet and R (repeated digits) isn’t used.

Except that I had a mistake – I thought 13 across had to be 18, but that wasn’t going to fit with the 31 I’d written in for 10 across.  Here was the grid I had that I discarded.

My original (wrong) grid for Listener crossword 4125, Elementary Number Theory by Oyler

Time to start over

I know it’s not the kosher way to do this, but let’s follow the gut instinct.  I printed off a new grid and went back to work, now adding in the constraint that the answers had to fit in with the rules of a sudoku.  90 minutes and a calculator battery later, I’d found the two mistakes I’d made originally (3 down was 86 and 13 across was 98), and I had a grid.

Yes, I know it’s kind of silly, bit I went through all the letters looking for which ones my unclued entries fit.  I didn’t quite understand N at the start, though when I got E and I for the first number and just N for the second number, I figure it’s combinations of factors, not just prime factors.  That really threw me with 7 down, and I didn’t want to  write 476 in at first, but it was the only number that satisfied everything else apart from N

My grid for Listener Crossword 4125, Elementary Number Theory by Olyer

I wonder if anyone who doesn't like sudoku puzzles write NEIN as the answer?

So with INE and N, then we have NINE as our answer.  That was a really really nifty piece of grid construction, but once you cotton on to what is going on, it falls into place.  You nearly had me stumped, Oyler, but I’m going to continue this as proof of my Oyler’s pendulum theorem, which means I should come up a complete failure next time around.

Victory to George!  2011 tally: George 6, Listener 2.   Current streak, George 3.

Well buggerty bugger bugger boo.

Guess who went and transcribed two numbers on his calculating device again.  Just took a look at Listen with Others and I’ve missed that 139854276 is a perfect square, divisible by 12 and has a remainder of two when divided by 7.  The number I got from transcribing had only N as a property.

Victory to Oyler and the Listener Crossword!  2011 tally:  Listener 3, George 5.  Current streak, Listener 1.

I’m getting worse at this!

Feel free to leave comments below, and see you next week for something a little different.

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5 Responses

  1. Sorry, George, but your arithmetlc has let you down – 476 does haxe 12 factors (don’t fprget 1 and 476 !) Also 23ac satishies DIN&O.

  2. Oh no. I thought you’d get this one. The atomic number of Iodine is my age by the way!

  3. In poor defense to Peter M, I saw 476 as having four factors – 2, 2, 7, 17. I was mistaking factors for prime factors.

    And in poor defense to Oyler – thanks for looking in! No worries, I just messed up that last bit. I have amazing number dyslexia when I try, and trying to get nine numbers in a row into a calculator is a struggle. And I’ll look for a xenon-themed crossnumber next year! I used to work in the same room where xenon tetrafluoride was first synthesised, there was still a large crack in the wall marking where a vat of fluorinesulfonic acid broke.

  4. […] for me the most challenging of the numerical setters.  Let’s recount… last year was Elementary Number Theory, where a bad slip near the end made my grid incorrect.  Before that was Digimix, where I was […]

  5. […] Oyler gave us one of my favorites, the cube-net cutting puzzle 2X2X2 (which I got), before that was Elementary Number Theory, which I messed up near the end, Digimix, which I messed up near the middle, Pentomino Factory, […]

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