A cross cross crossnumber

Welcome back to George vs the Listener Crossword – after a pretty spectacular fall from grace last week (I was utterly convinced I was right, goes to show what I know!), we’re in the quadrannual world of the cross-number.  IOA is kind of a new setter to me – I am VERY slowly working through the Times Listener Crossword Book, and though I don’t have it close at hand, I’m pretty sure IOA’s last Listener (3789, Spotless) is one of the ones I was playing with last time.  I haven’t done an IOA Listener in the conventional sense before, so hi IOA if you’re looking in.

I’m rather fond of the numerical ones, though I’ve been kind of hit and miss on them lately, especially when there’s triangular numbers or a lot of adding up to do.  It doesn’t appear (yet) that there’s a lot of adding, but there’s a rather nifty device of the clue letters being a part of the puzzle.

I solved this in two sessions, one of them very long (about 4 hours, with four pages crammed full of notes), and the second a lot shorter, about an hour.  Let’s encapsulate the process…

  • A,R and V have to be 1, 5 and 13.  So nothing else can be 1.  The answers to A R and V (let’s call them ANS(A), ANS(R), ANS(V)) are a 3-digit and a 4-digit number, and all of them end in the start of another answer.  This means V can’t be 5, as (VXX+A)VXXX will end in 0
  • Wow that was a big logic step to find out that one letter can’t be one number
  • There’s only 4 2-digit answers and 2 5-digit answers.  So unless we’re putting more than one number in a square, K=H means H > 10 and K is one of 4, 12, 17, 23
  • One of the Rs is OKKK and has to be a 3- or 4- digit number.  so 23 is too big for K.  If K is 17, O has to be 2, if it’s 12 O has to be 2 or 3, if it’s 4, there’s lots of possible values for O
  • The other R is HHH + XXXXX and has to be 4-digits long since H>10.
  • ANS(W) is (WW-XXXXX)O.  The maximum value for W is 26, so XXXXX has to be less than 676.  X has to be 2 or 3
  • Hey – that means the OKKK one must be the 3-digit number!  That means K = 4.  O has to be 15 or less
  • One of the ANS(V) is (VXX+A)VXXX, the other is OOOOU – O – U.  Looking at the second one, that means O has to be less than 8 to make it a 4-digit number.  Since I know the limits of values for V, A and X, then the first one has to be 136, 378, 594, 5512 or 5928.  This means the other has to be a 4-digit number starting with 1, 3 or 5, or a 3-digit number starting with 5.  We can thus eliminate O=8 and 7 and 6 which have no possible combinations
  • So back to OKKK = we’re now limited to 128, 192,  meaning the other R is a 4-digit number starting with 1.  The only values of H which work for that are 10, 11.  But K = H and 4 down can’t end in 0.  H is 11!!!!  This is the longest I can remember for a numerical puzzle to get to the first number.  It also begins the cascade, as that means X has to be 2, and one of the R entries is 1363.  Since X is 2, O has to be 3 and the other is 192.  Those won’t fit at 1 across and down, so R is not 1.  The conditions for V eliminate V being 1, so V must be 13, R is 5 and A is 1.
  • I do myself absolutely no favors at this point by entering a wrong number in for the down part of R (1754, not sure how I came up with that).

This starts the cascade – and I have U = 7, W = 26, D = 9, Y = 19- meaning Q should be 16  and T should be  14 and then my numbers don’t fit.  This was the near the end of the marathon solving session.  I am thinking at this point that I’ve done something wrong mathematically, so it’s time to turn to the number dyslexic’s friend- excel!

In writing my excel spreadsheet I see what is wrong – it was in the third number I wrote down!  This meant that the 5-digit number I found (17256 for Y)  did’nt go at 19 but at 10.

Session 2 was with the spreadsheet in place, and having someone else do the math and not write down wrong numbers put me on easy street and with a complete grid!  I’m guessing we’ve got to turn the numbers in the grid into a message based on the number.  SHADE EVENS BUT RUB OUT ALL OF ODDS ERGO X.  Once I got the “SHADE EVENS” part I saw what was going on – the only even numbers in the grid were on the diagonals.

I’m not going to put the final final grid in here – here’s my glorious mess while I was working out the numbers…

Solution for Listener 4138, The X Factor by IOA

So take out the odd numbers and shade a big X through the evens and there you go

So there we go – a really neat piece of construction from IOA and even though it cost me last week, I’m going to go ahead and claim a victory here.

Victory to George!  2011 tally:  George 16, Listener 5.  Current streak:  George 1.

If you’re in the Asheville area this weekend, you can come see me in Keywords.  Otherwise, feel free to leave comments below and check back in next week where we will settle the spell with Nutmeg.

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