## 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 http://www.thinbasic.com 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?)

## erwinch said

In the solution notes on the Listener site,

4088is titledZero-less Pandigital Curiosities by Oylerand the final statement is that73194has no significance other than a check. So, there appears to have been some last minute changes for one reason or another.## Alastair Cuthbertson said

We decided to change the title as Googling Zero-less Pandigital Curiosities gives a link to the solutions. Digimix is shorter and much catchier! Oyler

## linxit said

I also used a computer to solve this, but in my case right from the beginning. I saw instantly that it would yield to computational brute force, so I wrote a short Perl script to generate all possible solutions, then went shopping as it would take at least an hour to run.

When I got back I transferred the output into a spreadsheet and was able to fill the grid in about 15 minutes, with very little thinking required. A mistake must have crept in somewhere though, as I got the total at the end wrong, even though my grid is 100% correct. I’ve also double-checked the addition, and there was no mistake there either. I had 72294, which is 900 out, so I must have transposed a couple of digits somewhere but I can’t see where. Just carelessness – I’m always making silly mistakes like that.

## Jake said

Hi Guys. I used this: http://codepad.org/DuNNQYrh.

I went a bit off track by having h dn as 177 although I had the other part M dn as 313. Most of what I had was correct in the end. Pandigitals 1-9 took into Excel and had a formulae of 78 values for P and Q – break them up into what’s needed and it all became easy -Once you know whats going on!