Pentominoes Numbers
Can you better?
Mail a 'number' to o.d.m@fulladsl.be

Maybe we send you a chocolate 'number'

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 George Sicherman: "5 of your digits lack unique solutions. Permit me to remedy this."

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Large numbers in Sudoku and pentominos
We asked Helmut if he could think of something fun with "70" (for the anniversary of Aad van de Wetering November 16, 2014) and we got a wonderful sudoku connected with a lot pentominoproblems.
The following Sudoku has the additional property that the values in all coloured cells sum up to 70.


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

Name Country
Helmut Postl Austria
Aad Thoen The Netherlands
Aad van de Wetering The Netherlands
Odette De Meulemeester Belgium
Sander Waalboer The Netherlands
George Sicherman USA
Nico Looije The Netherlands
Bob Henderson USA


George Sicherman:"I solved them by computer. Doing it by hand would take too much trial and error, and I already have enough trial and error in my life!"

Bob Henderson: "I found the following results by modifying my Sudoku solver to run until the marked cells had the correct sums. I did not search for any more solutions: I have no doubt that Helmut designed them to have only one!"

Furthermore, there are 60 white cells, 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 4 pentominoes and use each one three times. The choice is unique, and there are two trivially connected solutions.

2) The biggest number of different pentominoes used is 9. There are several possibilities. One among them is unique: it is without the N pentomino. Here is the solution.
3)The smallest number of pentominoes used is 3. Each of them can be used in an arbitrary number of copies.  The solution is unique.
4) The biggest number of copies of a single pentomino is 8. The remaining 4 pieces need not be different. There are two trivially connected solutions.
Another idea is to glue the two horizontal edges together so to make a cylinder. Unfortunately, there are still no solutions. The same holds when the two vertical edges are glued together. But combining both (so to make a torus) yields 105 solutions. There are two unique ones among them: one with the fewest (4) and one with the most (9) pentominoes wrapping around.


one with the fewest (4) pentominoes wrapping around


one with the most (9) pentominoes wrapping around

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

Name Country
Helmut Postl Austria
Aad Thoen The Netherlands
Aad van de Wetering The Netherlands
Odette De Meulemeester Belgium
Sander Waalboer The Netherlands
George Sicherman USA
Nico Looije The Netherlands
Bob Henderson USA
Martin Friedeman The Netherlands

 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).


vertical cylinder


horizontal cylinder

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


The sum of the numbers in the seven  horizontal or vertical I-trominoes (three cells) is 15.
In the two vertical I-pentomino (
five cells) are the numbers form an increasing or decreasing sequence.

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Nico Looije - The Netherlands
Martin Friedeman - The Netherlands
Ilse De Boeck - Belgium
Sander Waalboer - The Netherlands
Aad Thoen - The Netherlands
George Sicherman - USA


The diagonals also contain 1 to 9.
In the colored cells are odd numbers

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Nico Looije - The Netherlands
Martin Friedeman - The Netherlands
Sander Waalboer - The Netherlands
Ilse De Boeck - Belgium
Aad Thoen - The Netherlands
Bob Henderson - USA
George Sicherman - USA


The diagonals also contain 1 to 9.
In the colored cells are odd numbers

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Nico Looije - The Netherlands
Martin Friedeman - The Netherlands
Sander Waalboer - The Netherlands
Aad Thoen - The Netherlands
Bob Henderson - USA
George Sicherman - USA


The diagonals also contain 1 to 9.
In the colored cells are odd numbers
Direct diagonal neighbours are different.

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Nico Looije - The Netherlands
Martin Friedeman - The Netherlands
Sander Waalboer - The Netherlands
George Sicherman - USA


In the colored cells are odd numbers
Direct diagonal neighbours are different.
.
Ten hints.

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Nico Looije - The Netherlands
Martin Friedeman - The Netherlands
Sander Waalboer - The Netherlands
George Sicherman - USA


The diagonals also contain 1 to 9.
Direct diagonal neighbours are different.
b9c9-g9h9 = 61
b5c5-g5h5=61
b1c1-g1h1=61

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Martin Friedeman - The Netherlands
Sander Waalboer - The Netherlands
Nico Looije - The Netherlands


The diagonals also contain 1 to 9
In the vertical I-pentomino (five cells) are the numbers form an increasing or decreasing sequence.
Also in the night  horizontal or vertical I-trominoes (three cells) are the numbers form an increasing or decreasing sequence.

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Nico Looije - The Netherlands
Martin Friedeman - The Netherlands
Sander Waalboer - The Netherlands
George Sicherman - USA


The diagonals also contain 1 to 9
In the colored cells are odd numbers

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Nico Looije - The Netherlands
Martin Friedeman - The Netherlands
Sander Waalboer - The Netherlands
Bob Henderson - USA


The diagonals also contain 1 to 9
In the colored cells are odd numbers

Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Nico Looije - The Netherlands
Martin Friedeman - The Netherlands
Sander Waalboer - The Netherlands
Bob Henderson - USA
George Sicherman - USA

In the splendid book "Exotische sudoku's" by Aad Thoen and Aad van de Wetering is more: 19, 24, 32, 37,44, 50, 52, 64, 66, 69, 83 and 88.

You can order this book e.g. at Standaard Boekhandel.

Here are two sudoku puzzles written by Aad van de Wetering to put someone in the spotlight .


The sum of the numbers in each of the eight trios is equal to 15.
Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Martin Friedeman - The Netherlands
Nico Looije - The Netherlands
George Sicherman - USA
Sander Waalboer - The Netherlands
Bob Henderson - USA


The sum of the numbers in each of the eight trios is equal to 15.
Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Martin Friedeman - The Netherlands
Nico Looije - The Netherlands
Sander Waalboer - The Netherlands
George Sicherman - USA
Bob Henderson - USA

For "65" Aad made this pento sudoku with 65 cells.

Three 5x5 concatenated sudoku's a1-e5, e1-i5, i1-m5
Each pentomino must have the numbers 1, 2,3,4 and 5
Solvers: Aad van de Wetering - The Netherlands
Odette De Meulemeester - Belgium
Martin Friedeman - The Netherlands
Nico Looije - The Netherlands
Bob Henderson - USA
George Sicherman - USA