Ed -

Begin by generating all five digit numbers whose digits are from {1, 2, 3, 4}.

Discard the primes and those with factors other than three or four primes less
than 100. Discard cubes and multiples of cubes and those with four factors not
all distinct. The following short list is exhaustive.

Conside C M M
31433 = 43 x 43 x 17 (3) Not valid since 33 cannot be on the square block
31423 = 67 x 67 x 7 (3) So C is 7 and M is 67

Consider D E H
11431 = 71 x 161 = 7 x 23 (3) Not valid since D must exceed 7
12341 = 43 x 287 = 7 x 41 (3) ditto
13114 = 83 x 158 = 2 x 79 (3) ditto
14413 = 71 x 203 = 7 x 29 (3) ditto
21243 = 97 x 219 = 3 x 73 (3) ditto
42413 = 83 x 511 = 7 x 73 (3) ditto
44233 = 89 x 497 = 7 x 71 (3) ditto
24211 = 71 x 341 = 11 x 31 (3) Not valid because H must be less than 53
31331 = 97 x 323 = 17 x 19 (3) ditto
42143 = 67 x 629 = 17 x 37 (3) ditto
12331 = 59 x 209 = 11 x 19 (3) ditto
22243 = 59 x 377 = 13 x 29 (3) ditto
41123 = 59 x 697 = 17 x 41 (3) ditto
44321 = 47 x 943 = 23 x 41 (3) Here, if E is 41 we haven't enough options for F, G and H
22231 = 47 x 473 = 11 x 43 (3) similarly
12121 = 31 x 391 = 17 x 23 (3) similarly
13243 = 41 x 323 = 17 x 19 (3) So D is 17, E is 19 and H is 41

Consider F F L & G G J
11323 = 67 x 169 = 13 x 13 (3) not valid since D is in any case 17 (and F must exceed D)
31211 = 59 x 529 = 23 x 23 (3)
41323 = 43 x 961 = 31 x 31 (3) So F is 23, G is 31, J is 43 and L is 59

For A D G H Find
43214 = 41 x 1054 = 31 x 34 = 2 x 17 (4) establishing that A is 2

Now turn your attention to A B K N & A B J P & A B P Q
24242 = 31 x 782 = 23 x 34 = 2 x 17 (4) not valid because B is either 3 or 5
31234 = 97 x 322 = 7 x 46 = 2 x 23 (4) ditto
32422 = 43 x 754 = 13 x 58 = 2 x 29 (4) ditto
43142 = 53 x 814 = 11 x 74 = 2 x 37 (4) ditto
11242 = 73 x 154 = 11 x 14 = 2 x 7 (4) ditto
12122 = 29 x 418 = 19 x 22 = 2 x 11 (4) ditto
13442 = 47 x 286 = 13 x 22 = 2 x 11 (4) ditto
21242 = 43 x 494 = 19 x 26 = 2 x 13 (4) ditto
14421 = 23 x 627 = 19 x 33 = 3 x 11 (4) ditto, and anyway A is 2
33231 = 53 x 627 = 19 x 33 = 3 x 11 (4) ditto
33411 = 43 x 777 = 7 x 111 = 3 x 37 (4) ditto
34221 = 61 x 561 = 17 x 33 = 3 x 11 (4) ditto
42441 = 47 x 903 = 7 x 129 = 3 x 43 (4) ditto
44421 = 67 x 663 = 17 x 39 = 3 x 13 (4) ditto
12243 = 53 x 231 = 11 x 21 = 3 x 7 (4) ditto
12441 = 29 x 429 = 13 x 33 = 3 x 11 (4) ditto
24123 = 43 x 561 = 17 x 33 = 3 x 11 (4) ditto

Now recall that J is 43 and observe
21414 = 83 x 258 = 3 x 86 = 2 x 43 (4) which makes B = 3 & P = 83

But K must be 47 or 53;
23214 = 73 x 318 = 3 x 106 = 2 x 53 (4) so K is 53 and N is 73

For A B P Q we have
44322 = 89 x 498 = 3 x 166 = 2 x 83 (4) Giving Q = 89

And we can ignore the remaining two possibilities
41322 = 97 x 426 = 3 x 142 = 2 x 71 (4)
14442 = 83 x 174 = 3 x 58 = 2 x 29 (4) because we have found all candidates

That the resulting products indeed fit validly on the tetris blocks is left as an exercise.


Be Well!
Dane Brooke