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Posts Tagged ‘by Oyler’

Alma Mater by Oyler

Posted by shirleycurran on 20 December 2013

Oyler's Alma Mater 001“Alma Mater”, said the other Numpty. ” Quite a coïncidence as 2013 is the 600th anniversary of my first Alma Mater, St Andrew’s University.  Well, down to work. Let’s work out the frequency of those letters. Hmmm! Strange, there are only 8 letters used. D, T, E, A, W, R, S, N!”

I hadn’t yet retreated out of firing range (as I habitually do until the later stages of the numerical Listeners – which can be several days of bearish growling later) and a minute’s fiddling with those letters provoked some muttering: “More than a coïncidence here. St Andrew’s! 1413 to 2013! This Saturday is also St Andrew’s day. Well done, Oyler! Preamble understood, perhaps? Do you suppose we put those two dates into 1ac and 31ac? That seems more than likely, and 1d can only be 1xx, so 1ac is 1413 and, 31ac is 2013. Progress! And the thematic shape must be a saltire, roughly corners to corners, passing centres of squares containing 1,4,1,3 and 2,0,1,3.” (Three minutes into a numerical Listener and the theme sussed – that has to be our record!)

Arithmetic might now be needed. A list of prime numbers is opened and there are only 21 from 11 to 97. Eight have to add up to 600 to thematically commemorate that anniversary. Some clues are numbers like perhaps 73^71. Awkward. I doubt that we have the calculator or device to work that out, but since Oyler is apparently a St Andrews man it’s probably not necessary to do a lot of maths, just find a  shortcut to get the last digit …..

To sum to 600, these 8 2-digit primes  are likely to be the forms x1, x1, x3, x3, x7, x7, x9 and x9, so the 8  tens parts need to add to 580. No 1x is possible, as  the maximum  for the 7 others is 559. Neither is 2x or 3x. That leaves only 13 primes to choose from ranging from 41 to 97. Looking at 18ac, 600+D-W-2E, this can only start with 6, 5, 4 or 3 so A can only be 71 or 79. Guess 71? From 23ac,  S, and 15d, AS, needing a common last digit, S must end in 1, so is 41 or 61. So is Oyler being kind, with S,T,A,N,D,R,E,W in increasing order, as 41, 67, 71, 73, 79, 83, 89, 97? Looks good! Digits interlock! Even the splendid WANT+REDS, DREW+STAN and SWAT+NERD fall into place.

What about 10ac, N^A or 73^71, and its friends, 2d, 3d and 20d, the giant numbers? No problem really, as  73*73 ends in 9. So 73^4 ends in 1, as does 73^68. So 10ac ends in last digit of 73*73*73 or 7. The others can be handled similarly. What an ingenious and pleasant puzzle, no computer needed, only fingers! I admit that the literary Numpty spotted the saltire-defining  squares first, though.

It had to be he saltire didn’t it – especially considering the shape of the grid (unlike that Swiss flag we had a few years back, that used one of the very rare square flag shapes). It was obviously a matter of searching along the diagonals, and, sure enough, there on the 1st, 3rd, 7th and 10th were the culprits. Saltire it is! Thank you Oyler for our speediest and most enjoyable numerical Listener ever!

Posted in Solving Blogs | Tagged: , , , | 3 Comments »

Digimix by Oyler

Posted by shirleycurran on 18 June 2010

Oh no, it’s the dreaded numerical again! I can hear whoops of joy but not so here. The slough of despond is even deeper when I hunt for the clues. Call those clues Rover? Bad dog! There are just rows of hieroglyphics. Perhaps I forgot to print the clues. But no; what you see is what you get. I hand over to the mathematical part of the numpty team.

The scope of the problem seems large at first: 5 digits and 6 digits for P and Q alone,  10^11 pairs! There is a drastic restriction to be applied, however, that 1-9 appear exactly once on both sides, in P_Q and X_Y_Z. This would seem to make a solution computable in a finite time.

How to use this, and where to begin, though? Line 1 of the clues is a good guess at where to start, and there we see h, 5h and if we are gazing hopefully, 6h again in Line 10 of the clues. The 5h immediately limits h to the form 1ab, to keep 5h as a 3-digit number as defined within the preamble.  Indeed, a can only be 2,3,4,5 or 6 for the same reason (1 is used, 6 x 17b too big,..) and b only 3,5,7 or 9 (no 0 allowed in 5h).  The 17 possible h combinations (h=123, .., 167) viewed with  5h and 6h reduces the choice of h to 127. One done (but I admit a false start after failing to multiply by 6 in my head correctly)!

Trying to avoid really starting work , we note  in Line 5 of the clues 5G appears, being of form 2a, and  in Line 9 of the clues 7J^2 appears.  These give G=25, 27 or 29  and J=11 (both 2 digits). But what now? Time to buckle down to it.

The ’1-9 once only’ rule now has to be applied to the clues. Looking at clue 1, this is trying to find a solution for P and Q as integers where P^2 + Q^2 = X_Y_Z, ie from (using h) 635127489 to 635127984.

I imported a free web version of BASIC (see for thinBasic) which offers ‘long integers’, in order  to do this by programmatic means. It is not too slow to do, as P can only be 1234-9865 in value and we have a smallish range too for X_Y_Z. So I just checked P^2 and  Q^2 where for each P in the range 1234-9865, Q had to lie between INT(SQR(635127489-P^2)) and INT(SQR(635127984-P^2))+1 for appearance of 1-9 once and once only in STR$(P^2)+STR$(Q^2) and STR$(P^2+Q^2). This is a loop over about 20000 P-Q pairs, reasonable to try, and quickly yielded P=9168, Q=23475 with X_Y_Z=635127849 so m=948 and G=29.

How to do this otherwise I fail to see! The need to use ‘large’ integers beyond 16 bits is beyond simple calculators, never mind the tedium involved, and the need to check strings of text (the 1-9 digits restriction) would seem to me tricky with Excel…

Much the same program can then be modified and applied to Line 4 (a good candidate, where P=145x and Q=9T, reducing the checks to be made a good deal). This yields x=27, T=3163, f=25  and thus s=313 in Line 1) and so on ….

The longest computation was for Line 8, 294 seconds and 3 M loops.  Others were typically 1-10 seconds.

Only two errors remained to make, failing to work out t=337 correctly in a mental addition and not noticing one of the missing entries when adding them up. The advantage of a non-numeric collaborator, who irritatingly but wisely suggested checking the results, was immediately apparent

About 5 hours of work, including much fumbling with a slightly different BASIC version which had a tendency to go silent if I missed out important syntax elements! I am pretty sure that the real mathematicians  know some drastic shortcuts, but it was an enjoyable and approachable puzzle.

What is left for the really numpty part of the team to do? Check all those calculations and copy it into a clean grid.  73194 – that has to mean something. Is it going to be an alphanumeric rendering of OYLER or do I turn it upside down and find a world-shattering message? Disappointment! I feed it into Google and discover that most references that come up are coupons for Pizzas in Oklahoma City. All that work for a pizza! (Or did we go wrong somewhere?)

Posted in Solving Blogs | Tagged: , | 4 Comments »


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