Listen With Others

Listener 4190: 2×2×2 by Oyler

Posted by erwinch on 8 June 2012

This was definitely a puzzle of two halves and I especially enjoyed the novelty of the second half – the construction of one large cube out of eight smaller ones.  The numbers however turned out to be rather straightforward although it was not immediately apparent where to start.  Eventually, I settled on 27dn:

A^(H + A) (3) so A must be greater than 1

This gave three possible fits:

{A, H} = {2, 5}, {2, 6} or {3, 2}  Only A or H can equal 2

Now considering 32dn:

ANNA (2)

So A = 2, N = 3 or 4 and H = 5 or 6

34dn (M + A)^C (2) = (M + 2)^C and C must be greater than 1

So M = 1, C or N = 3 or 4 and H or O = 5 or 6

3dn CON (2) = 12O and cannot end in zero so O = 6 and H = 5

21dn HANOCH + C + O – ACH = 3570 (N = 3, C = 4) or 3579 (N = 4, C = 3) and also cannot end in zero so N = 4 and C = 3

This left 15 down clues to solve by direct substitution.  I dutifully used just a calculator and mental arithmetic but this was not a favourite part of the puzzle and had a check been required then I would have used BBC Basic here.

Most of the across clues were now also solved and it did not take too long to find the across assignments.

9ac NOAH = 36
Factorisation gave only one possible combination:
36 = 6 × 3 × 2 × 1 so C or M = 4 or 5

37ac (M + O)(OO)^N = 448
448 must be divisible by OO so O = 1 or 2
If O = 1 then 37ac is too small so O = 2
448 must be divisible by (M + O) {6 or 7} so M = 5, C = 4 and N = 3

35ac MA + N = 5A + 3 = 33 so A = 6 leaving H = 1

I found it rather laborious checking the 17 across clues that remained but the grid was then complete:

Finding the eight nets of a cube was not too troublesome:

The pair of twos in row 5 gave us an ambiguity that could be resolved since all the nets were to be differently-shaped.

For the final stage I considered making some blocks or buying sugar cubes to solve the problem as a three dimensional jigsaw but I found that I was able to do it on paper and here is my working:

I first classified the eight cubes as three pairs of opposite faces intending to match them with those labelled A to H on the larger cube:

a. 1/2, 3/5, 4/6
b. 1/4, 2/3, 5/6
c. 1/6, 2/3, 4/5
d. 1/3, 2/5, 4/6
e. 1/3, 2/4, 5/6
f.  1/2, 3/6, 4/5
g. 1/6, 2/4, 3/5
h. 1/4, 2/5, 3/6

(The three pairs for conventional dice are 1/6, 2/5, 3/4 so these were all non-conventional as was stated in the preamble.)

The 24 pairs could be sorted as follows:

1/2, 1/3, 1/4, 1/6 missing 1/5
2/1, 2/3, 2/4, 2/5 missing 2/6
3/1, 3/2, 3/5, 3/6 missing 3/4
4/1, 4/2, 4/5, 4/6 missing 4/3
5/2, 5/3, 5/4, 5/6 missing 5/1
6/1, 6/3, 6/4, 6/5 missing 6/2

It seemed clear that the missing pairs would be those of the larger cube: 1/5, 2/6, 3/4.  So we could place 1 on the right face and 6 on the bottom.  This left 3 and 4 for front and back.  I arbitrarily placed 4 on the front and then considered cube A.  The visible faces on A are 1, 2 and 4 so these could not be opposite pairs.  The relative position of the numbers was also important so it was necessary to check that the visible numbers would be visible when the larger cube was assembled.  Two of our eight cubes fitted for A.  In fact there were two fits for each of the cubes labelled A to H:

a. 1/2, 3/5, 4/6 cube D or F
b. 1/4, 2/3, 5/6 cube D or F
c. 1/6, 2/3, 4/5 cube A or G
d. 1/3, 2/5, 4/6 cube A or G
e. 1/3, 2/4, 5/6 cube C or E
f.  1/2, 3/6, 4/5 cube C or E
g. 1/6, 2/4, 3/5 cube B or H
h. 1/4, 2/5, 3/6 cube B or H

I then considered the vertical pairs A/E, B/F etc and found a match for the internal faces (shown in green) in two ways:

a. 1/2, 3/5, 4/6 cube D
b. 1/4, 2/3, 5/6 cube F
c. 1/6, 2/3, 4/5 cube G
d. 1/3, 2/5, 4/6 cube A
e. 1/3, 2/4, 5/6 cube E
f.  1/2, 3/6, 4/5 cube C
g. 1/6, 2/4, 3/5 cube H
h. 1/4, 2/5, 3/6 cube B

or

a. 1/2, 3/5, 4/6 cube F
b. 1/4, 2/3, 5/6 cube D
c. 1/6, 2/3, 4/5 cube A
d. 1/3, 2/5, 4/6 cube G
e. 1/3, 2/4, 5/6 cube C
f.  1/2, 3/6, 4/5 cube E
g. 1/6, 2/4, 3/5 cube B
h. 1/4, 2/5, 3/6 cube H

This confirmed 4 as the front face of the larger cube or did it?  Of course I had to check that it would not work if 3 was on the front:

g. 1/6, 2/4, 3/5 cube A?
h. 1/4, 2/5, 3/6 cube A?

On cube g we could show 1 on the right and 2 on top but 5 not 3 would then be at the front.  Similarly for cube h, 6 would be at the front:

It was the same for all eight, we could only show two of the required three visible faces on the larger cube at any one time.  So, the net below the grid was confirmed as having 4 at the front:

To conclude, this was another highly entertaining puzzle from Oyler.  The numbers in particular were on the easy side as these things go but this was none the worse for that – shades of Pentomino Factory perhaps.  My only regret is that it is another thirteen weeks or so until our next numbersfest.