Dear All,

This tribute article is dedicated specially to Hans Moors (23.6.1942-18.5.2022) and his wife, Heleen, who is still a very dear friend to my little family of Jenny, Michael and myself. We visited Heleen in The Netherlands last Saturday, and it was really good to all be together and to share lots of happy memories of Hans.

Our friendship began way back in 1996, when Hans & Heleen very kindly let me stay at their home for ten days, as I was playing in an international chess tournament taking place within the same city. The warm welcome and generous hospitality helped me greatly, and we enjoyed celebrating when I won the tournament! π

It wasn’t long before puzzles and jokes were spilling forth, and the probability of Hans solving/laughing regarding mine was always four times 1/4, given his terrific sense of humour and his superior mathematical abilities! Yes, Hans was a brilliant University Professor of Statistics, and I will always be grateful for everything that I learned from precious chats with him, from tricky puzzles that he shared with me, and from the wonderful friendship of his family. I am writing this article also to thank God for all those treasured gifts, and more β₯

Michael especially loved the many after-dinner games of monopoly at the home of Hans & Heleen, and he also remembers fondly an outing with their four grandchildren to *De Efteling* park.

Fast-forwarding to last Saturday with Heleen, she told us that she and Hans bought their house together in 1972, over half a century ago. The number on the door is 47, and now I’d like to share a few mathematical surprises which would certainly have made Hans smile π.

Hans lived to be 79, and that’s exactly the number of digits in 47 raised to the power of 47. For the record, the whole number is 3877924263464448622666648186154330754898344901344205917642325627886496385062863.

Here in Blog Post #151, it’s also fitting that the following super-long sentence is true…

*Hans and Heleen lived very happily together at house number forty-seven for fifty years, and this sentence will contain one hundred and fifty-one letters provided it stops right here!*

As a quick, fun puzzle π, can you find a slightly different number word other than * one hundred and fifty-one* which would still be completely correct in the

*italicised sentence*above?

The sum of my family’s house number 11 and Heleen’s house number 47 is: 11 + 47 = 58. In honour of lots of fabulous times together at the two homes, I present to you the following special 58-digit number: *1525423728813559322033898305084745762711864406779661016949*. If you multiply it by the number of letters in *FAMILY*, here’s what happens…

*1525423728813559322033898305084745762711864406779661016949* *x 6* = *9152542372881355932203389830508474576271186440677966101694* π

The next surprise is one of my favourites. It will feature the following numbers:-

2, in honour of β₯ Hans & Heleen β₯

9, the number of letters in HANS MOORS

11, the number of letters in HELEEN MOORS

79, the very good age that Hans reached

151, the blog post number of this tribute article

The product 2 x 9 x 11 x 79 x 151 = 2361942, and Hans was born on 23.6.1942 π

The cube of Hans’ birth year is interesting because 1942 x 1942 x 1942 = 7323988888, ending with five consecutive 8s.

(If we only square a whole number, the maximum group of repeated digits that can occur at the end of it is three 4s. The last previous year with that property was my birth year, 1962, and 1962 x 1962 = 3849444. Within the five hundred years from 1962 until the far-future time of 2462, only one other year can have a square which ends with 444, and that is 2038 for which 2038 x 2038 = 4153444.)

**Five Fun Puzzles in Honour of Hans** β€πβ€πβ€

- It’s of course very easy to make 24 using the numbers 1, 4 & 6 because 1 x 4 x 6 = 24.
You may freely use parentheses ( ) and operations involving +, -, x, Γ· as you wish.*Your fun challenge is to make 24 using the numbers 1, 4, 6 AND my favourite 3, once each in a calculation.*

2. Hans was a very experienced chess-player. You don’t need to be a player (or even to know the rules of ‘The Royal Game’ at all!) in order to solve the following puzzle… Imagine a magic box with an endless supply of pure white and pure black chess pieces, and a separate big bag containing an assortment of 1000 pure white chess pieces and 999 pure black ones. ** You’ll be taking two pieces at a time out of the bag**. If they’re both white, just one white piece will be put back into the bag; if instead the removed pieces were both black, just one white piece would be moved from the box to the bag; if instead the two removed pieces were of opposite colour (meaning: one white and one black), just the black piece would be put back into the bag.

**(taking two pieces at a time out of the bag, and reacting according to their colour(s)), until finally only one piece remains in the bag after completion of the final process.**

*You’ll keep repeating the whole process*

*Your fun puzzle is to figure out, with proof, what the colour of the last piece in the bag will be.*3. Imagine placing three coins randomly in three of the 64 little unit squares on a standard 8 x 8 square chess board. ** What is the precise probability that the three coins will all be on different horizontal ranks and vertical files from each other?** (Assume that the three coins are round, all of exactly equal size, and that they are placed centrally in the unit squares.)

4. Since Hans would win a prize for solving all the puzzles, I’m thinking of **A PRIZE MUST β TRAPEZIUMS**!

5. The following beautiful chess puzzle is dedicated not only to Hans, but to all fans of ‘The Royal Game’, including **Ralph Connell** (a former student of mine from the 1980s in Scotland!) who recently wrote me a really nice letter, and **Dr. Vipin Zamvar **who very kindly sent me a lovely gift book. Happy birthday also to Belgian chess friend **William Verleye** who has turned 62 today β€πβ€πβ€

It’s my intention to publish solutions to all the puzzles around the time that blog post #152 comes out, God-willing as always.

In the meantime, please do feel free to send me your best solutions to any or all of the puzzles, if you like π.

I would like to round off this article now by most sincerely wishing you a very blessed weekend, with lots of happiness in everything that you do β₯

With kindest wishes as always,

Paul Mπtwani β₯

Jesus said to Nathanael, βTruly, truly, I say to you, you will see Heaven opened, and the angels of God ascending and descending on the Son of Man.β–Bible verse, John 1:51 β₯

**P.S. = Puzzle Solutions** (being posted now on 12.03.2023)

The sentence

*Hans and Heleen lived very happily together at house number forty-seven for fifty years, and this sentence will contain one hundred and fifty-three letters provided it stops right here!*

is the alternative correct answer.

**6 Γ· (1 – 3 Γ· 4) = 24** π

The final chess piece in the bag will be a **black** one. Note that each complete process described in the puzzle reduces the total number of pieces in the bag by precisely one, and so the total does indeed keep diminishing, for sure. However, when black pieces are permanently removed from the bag, they are removed in pairs (or ‘twos’) and are replaced with one new white piece. Therefore, whenever the number of black pieces reduces permanently, it goes down by 2 each time. Since there were 99 (=an odd number of) black pieces in the bag at the start, **the number of black pieces can’t reach zero and stay at zero**. So, that proves that the very last chess piece in the bag must be a black one.

In the puzzle regarding 3 coins on an 8 x 8 chessboard, the probability that they will all be on different horizontal ranks and vertical files is 64 x 49 x 36 Γ· (64 x 63 x 62), which simplifies to the fraction **14/31**.

In the puzzle about the trapezium, first note that the two clear triangles (not coloured) have equal areas, each equal to C, say. To see that, observe that A + C on the left and A + C on the right give us big triangles with the same height and base length, so their areas must be equal. Next, note that the coloured triangles are mathematically similar because they have the same angles as each other (using standard facts regarding vertically opposite angles being equal, as are alternate interior angles). The ‘area scale factor’ relating triangles ‘A’ and ‘B’ is A/B, and so the ‘length scale factor’ is the square root of (A/B). Noticing that each of the ‘C’ triangles have a side in common with A and a shorter side in common with B, it can be deduced that each area C must equal A Γ· the square root of (A/B). That simplifies to C = sqrt (AB), where sqrt is short for ‘square root’. Therefore, the total trapezium area, A + B + 2C = **A + B + 2*sqrt (AB)**. A nice, neat, alternative expression for the total area is **(sqrt A + sqrt B) squared** π.

White wins beautifully with **1 Nxd5! exd5 2 Qxf7+!! Kxf7** (or 2…Kh8 3 Ne6) **3 Bxd5+ Kg6 **(or 2…Kf8 3 Ne6+)** 4 f5+ Kh5 5 Bf3+ Kh4 6 g3+ Kh3 7 Bg2+ Kg4 8 Rf4+ Kh5 9 Rh4+ Bxh4 10 Bf3#, checkmate!** π