Friday, December 19, 2008
Euclid's calculator-button mashing
We're on to the last numerical puzzle of the year, and it's something I haven't seen before, a circular numerical puzzle. Aedites has given us one circular puzzle this year (the first circular I had ever solved), and after my boast earlier in the year that I can usually do numerical puzzles, I have only finished one of the three (and I didn't get Big Holes either).
First thing, I thought something had gone wrong with my printer when it spewed page after page and most of them looked blank. When I print off the Listener, I usually limit it to two pages (I print on American letter size paper, so it's a bit shorter than A4), but since numericals typically have fewer clues, I didn't bother. And here come the pages!
All entries are 5-digit integers who only have single prime factors 2-79, no factors repeated. Intriguing... trying to do them from memory, I write down the primes from 2-79 and completely miss 41. I'm off to a great start...
Each clue is the common factors, if there are none, there's no clue. Intriguing... so next step is to write out the list of definite factors for each letter and each number. This may not give all the factors (in fact in 24 and C it appears there's no factors in common with anything), but it gives some juicy starting points. I know some have six factors...
Y = f a r t n l (heh, fart)
17 = f t w n k u (heh, ftw)
A bit of poking and prodding on the calculator shows that there has to be 2, 3, 5 present to have six prime factors and still be a five-digit integer. So several pieces of good news... f, t and n have to be 2, 3, and 5. Anything with these three factors is going to end in 0 (multiple of 30), so I can confidently write in a 0 at the end of anything that has f, n, t as factors. Which is only T and Y (and 17-21 that cross Y), and 30. Three whole numbers in the grid!
Now 19 doesn't have f as a factor - so f has to be 3 (otherwise it couldn't end in 0). One of n and t is 2, the other is 5.
So let's see - the five digits of Y...
first digit crosses 1-4 which have n but not t
second digit crosses 5-8 which has neither n nor t
third digit crosses 9-12 which has t but not n
fourth digit crosses 13-16 which has neither n nor t
fifth digit is 0
So look for combinations of those last three factors that give a number that goes...
even - odd - five - odd - zero
five - odd - even - odd - zero
30 x 7, 13, 19 gives 51870, but that won't work as 30 x 11, 23, 29 is out of range for 17
30 x 11, 13, 19 gives 81510 for Y, so that makes 17 30 x 7, 17, 23 = 82110 and we have our first two full numbers in!
This means n is 2 and t is 5, so any entry with t in it ends in 5. l a and r are 11, 13, 19 and w, k u are 7, 17 and 19.
Z now looks like -50-5 and has factors l,a,r (don't know what they are but know what they come to as a product), t and k, so 13585,w means w = 7 and the entry is 95095.
And we are off! Most of these were filled by getting the known factors and looking for multiples that worked with the numbers that I knew. I only used the "cross-checking" of making sure the remaining factors didn't cross a few times, notably in getting C (factors of 29, 31, 37).
This was a single-sitting solve, rare for me. Didn't check the time, but it clocked in at a few hours of hunting and pecking. I didn't use a spreadsheet, but I used one of those dinky old calculators that will repeat the last function if you press equals. I'm glad it printed on four sheets, I actually needed to spill over to a fifth to get all of my notes in. Here's how things looked at the end...
What looked like a daunting prospect turned out to be a pretty nice exercise in logic and patience - I had fun with this one. And a victory to George! With just a few crosswords left in the year I've stuck my neck out again, one more completion guarantees me at least a draw.
Current tally: George 24, Listener 22. Current streak: George 3.
If you're doing the holiday thing, have a safe and drunk one! I'll be in Houston next Friday, I'll have internet so I should be able to do the blog, but I may not have any images up, taking the scanner on the plane sounds like a bad idea.