Helmut send a text file with all the solutions. Let W = number of wrapped pentominoes, S = number of solutions for this case. Then we have the following possibilities:
(W,S) = (4,5), (5,14), (6,33), (7,42), (8,10), (9,1).
You can find the solutions as above: Search for “4A” resp. “9A” to get the solutions with the fewest resp. most wrapping pentominoes.

Helmut made a similiar for 64. (October 10, 2014)
The following Sudoku has the additional property that the values in all coloured cells sum up to 64.



Correct solutions earn eternal fame.
Send your solution to: o.d.m@fulladsl.be

 Aad van de Wetering: "Voor de 64 (ook een idee) had ik geen software, met Xudoku alle oplossingen gemaakt en Excel laten uitrekenen welke som 64 had voor de gevraagde vakjes. Dat was er één, alle andere hadden een grotere som."
Sander Waalboer zal daar wel software voor hebben want op 10 min stuurde hij zowel een oplossing voor "64" als voor "70" terug
Nico Looije: "Dit is een leuk idee. Ik moest een minimale programmatische aanpassing doen om ze op te lossen"

Furthermore, there are 60 white cells (ignoring the isolated cell in the digit 6), so they can be covered by 12 pentominoes. Unfortunately, it is not possible to use each pentomino exactly once.


But there are some interesting possibilities:
1)Choose 6 pentominoes and use each one twice. There are many possibilities. Try a set without the L or the P pentomino.
Without  the L-pentomino: 2 combinations.
INPTWY: 3 solutions
INPUVY: 1 solution
Without the  P-pentomino : 2 combinations.
ILNTVY: 2 solutions
ILNTVW: 1 solution
2) Choose 4 pentominoes and use each one three times. The choice is unique, and there are three trivially connected solutions .
3)The biggest number of different pentominoes used is 10. There are many possibilities. Three among them are unique:
one without the L pentomino
one without the N
one with the X
4) The smallest number of pentominoes used is 4. There are many possibilities.:
 One among them is unique: it uses the Z pentomino.
5)The biggest number of copies of a single pentomino is 8. The remaining 4 pieces need not be different. There are three essentially different possibilities, but all with the same 8 copy pentomino.
There are 12 solutions.

Another idea is to glue the two horizontal and/or vertical edges together so to make a cylinder or a torus.
W = number of wrapped pentominoes, S = number of solutions for the corresponding case.
1) Connect the vertical edges (vertical cylinder): There are 22 solutions.
(W,S) = (2,3), (3,10), (4,7), (5,2).
2) Connect the horizontal edges (horizontal cylinder): There are 12 solutions.
(W,S) = (3,3), (4,9).
3) Connect the horizontal and vertical edges (torus): There are 6577 solutions (including the 34 cylinders).
(W,S) = (2,16), (3,179), (4,852), (5,2029), (6,1926), (7,1087), (8,434), (9,51), (10,3).

Helmut added the wrap information to the first line of each solution. Here is an example: "2768 HVH.H.V.BB.. 4V 5H 2B 7A"
2768 is the solution number.
The next string consists of 12 characters, one for each pentomino in alphabetical order (FILNPTUVWXYZ).
‘V’ = only vertical wrap, ‘H’ = only horizontal wrap, ‘B’ = both wraps, ‘.’ = no wrap.
The following four numbers are the numbers of pentominoes with the indicated wrap kind: ‘V’ = vertical, ‘H’ = horizontal, ‘B’ = both, ‘A’ = any. V includes also B (since a pentomino with both wraps does also wrap vertically), H includes also B (for the same reason), and A can be computed as A = V + H – B.
The given example means that I and U wrap only vertically, F, L and P wrap only horizontally, and W and X wrap in both ways. There are 4 pentominoes that wrap vertically (IUWX), 5 horizontally (FLPWX), 2 in both ways (WX) and 7 anyway (FILPUWX).

Large numbers in Sudoku
The following sudoku puzzles are not with pentominos but they are funny. The Sudoku can be used at one or other occasion.
Correct solutions earn eternal fame.
Send your solution to: o.d.m@fulladsl.be