Listener 4088: Digimix by Oyler
Posted by erwinch on 18 June 2010
Oyler’s first Listener since, for me, the unforgettable Pentomino Factory last February. However, Digimix looked conventional and in appearance perhaps reminiscent of a Rhombus puzzle. I only had time for a brief blog on this since I was shortly off on holiday walking in the Mountains of Mourne, which I can highly recommend.
For the blog, I have numbered the clues 1 to 11 and clue 1 immediately caught my eye with X = 5h and Y = h. First of all though, I looked at the limits for P (1345 to 9876) and Q (12345 to 29876). I then spotted 7J^2 (clue 9) with only one fit: J = 11, 7J^2 = 847, 6J^2 (clue 5) = 726
Returning to h: with no zeros in the grid 5h must end in 5 and h must be an odd number in the range 123 to 197. There was also 6h (clue 10) and only one fit was found: h = 127, 5h = 635, m’ = 489 in some order, 6h = 762 and h’ (clue 2) = 721
Looking again at clue 10, D = ???1 and X_Y_Z = ??????762 so E = 2???9
Clue 5: G = 24, 28 or 29 but G = 29 is the only fit, 5G = 145, m’ = 489 or 849
We now had two possible values for X_Y_Z in clue 1: 635127489 or 635127849. It would have been all too easy to determine the corresponding values for P and Q using BBC Basic but that is not how these puzzles are designed to be solved. Applying the limits of P we find that Q must be in the range 23186 to 25165. Since X_Y_Z ends in 9, P & Q squared end in 4 & 5 and P & Q end in 2 or 8 & 5 or rather 8 & 5 since 2 is already accounted for. So, the limits for Q are now:
23415 to 24975 step 30
23418 to 24978 step 30
In fact there were only 32 eligible values to check, 16 in each group, and only one fit was found: Q = 23475, P = 9168. Factorising Q (3fs): 23475 = 3 × 5 × 5 × 313 so f = 25, s = 313 and m’ = 849. I did then use BBC Basic to check that this was indeed a unique solution to clue 1.
In clue 4: T = 3??? so can range from 3111 to 3999 and with X = 3Jf = 825 set the limits for Q (9T): 28134 to 28674. P = 5Gx = 145x so must end in 5 and only one fit was found: 5Gx = 3915, x = 27, 9T = 28467, T = 3163, 2p = 314, p = 157 and 5p (clue 7) = 785
The grid now looked like this:
And so it went, an absorbing meander through the clues until just clue 11 with its three unclued parts remained. Since P = 8796 and Z = 297, Q = 2???1 with the unknown parts = 345 in some order. This left six values for Q to check and Q = 23541 and X_Y_Z = 631548297 were soon found.
Finally, we had to sum the missing entries (later amended to asterisked items by the Listener site’s Stop Press) and enter their total in the box beneath the grid. I found this total to equal 73194, giving the completed grid as follows:
Some Rhombus puzzles contained a simple check like this but in recent years we have come to expect a strong thematic element to all the numerical puzzles so this seemed disappointing as an ending. 73194 is too big to be a possible value for Q, having ten digits when squared. I thought that it might factorise to give us a word or possibly initials but it didn’t:
73194 = 2 × 3 × 11 × 1109 gives only J (11) or A (22)
There again, all the digits are different and could leave 2568 in some order as a value for P but the preamble would certainly have said so. Nevertheless, I checked the twenty-four possible values for P and found the following:
5862 squared plus 73194 squared equals 5391724680, a truly pan-digimix dénouement!
I believe that Oyler originally intended this 10-digit number to be the total entered in the box but that it was rejected, which is a shame since I would have found it so much more satisfactory as an ending and also it would have been a good indication of a correct solution. We were told that there were no zeros in the grid, which I found essential to solving the puzzle, so is the box then part of the grid? Of course it is easy to say with hindsight but surely we would not have been confused had the final sentence of the preamble read something along the lines of:
To show that they have correctly deduced the asterisked parts, solvers must sum them and then derive a more fully thematic total to be entered in the box beneath the grid.
The inference is that the test solvers were indeed confused here. However, I must not dwell on the ending, which did not really spoil a puzzle that I found hugely entertaining and masterfully constructed – thank you Oyler.
As a postscript, here is a picture of me in The Mournes last Thursday with the Ben Crom Reservoir in the background – I wish I was still there.