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