Blog Post #151: Thank You for dear Hans & Heleen β™₯😊β™₯

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.

Lovely Yellow Roses represent Warm Friendship ❀

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! 😊

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™₯

Hans & Heleen with Michael in December 2016 β™₯😊β™₯
Hans once commented that “things are blossoming”, and he discovered the amusing fact
that the 11 letters in A TULIP WOMAN can be rearranged to make PAUL MOTWANI!!
Maybe there’s a message in there that my good wife, Jenny, really deserves a lot of the credit!

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 😊.

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 😎, 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?

A lovely picture at Heleen’s home β™₯

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 😊

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 😊

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 ❀😊❀😊❀

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

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. You’ll keep repeating the whole process (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. 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!

Your brainteaser is to find a mathematical expression for the Area of the entire Trapezium in terms of the areas A and B indicated in the picture.

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 ❀😊❀😊❀

It’s White to play and win in a stunning game from 1962, the year I was born.

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 😊.

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 M😊twani β™₯

Jesus said to Nathanael, β€œTruly, truly, I say to you, you will see Heaven opened, and the angels of God ascending and descending on the Son of Man.”–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 😊

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.

Your brainteaser is to find a mathematical expression for the Area of the entire Trapezium in terms of the areas A and B indicated in the picture.

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 😊.

It’s White to play and win in a stunning game from 1962, the year I was born.

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! πŸ’–

Blog Post #150: An Epicentre of Culture β™₯

Dear All,

I had the pleasure this past week of meeting Maximilian Gililov, the youngest son of world-famous classical pianist Pavel Lvovich Gililov. Max was visiting Musica Mundi School, which he described beautifully as “an epicentre of culture”.

Max and I enjoyed seeing fabulous concerts at Musica Mundi School, and we played & discussed some very interesting Chess, too! 😊😊

The following feast of fun puzzles is specially dedicated to Max 😊

  1. Of the three chess games that we played, the third game lasted for 38 moves, the second for 39 moves, and the first game was shorter. The average number of moves in the three games was a whole multiple of ten.

Part 1: Exactly how many moves were played in the first game?

Part 2: Imagine drawing a straight line across the stunning rectangular photo above, so that the line divides the photo into two parts of equal areas. How many such lines are possible? Is the correct answer 4, or slightly more, or infinitely more!? Can you prove that your answer is correct?

2. Rearrange the letters of US SECLUDED BAY or DEAD BUSY CLUES to get the name of Max’s favourite famous musical composer!

3. Max is an excellent pianist, but his other favourite instrument can be found by changing just one letter of HELLO to a different letter…

4. Choose well and replace one letter that is repeated in AIR STAR by the letter U, and then rearrange to make the seven-letter name of the beautiful country where Max lives and works.

5. Max and I thoroughly enjoyed the concerts that we saw. All the performances were superb! For every student who is practising to become better and better, Max’s honest words regarding himself can be inspirational. He said, “I’m an inferior version of my future self.”

I saw two concerts: one on Thursday in which every performance was on the piano; and one on Wednesday in which each performance was on either the harp, the oboe, or the cello.

For Thursday’s concert, the five-letter word PIANO was written 11 times (alongside 11 performers), making a total of 55 ‘piano’ letters there. For Wednesday’s concert, the total number of ‘HARP’, ‘OBOE’ & ‘CELLO’ letters was also 55, and there were more cellists than harpists or oboe-players.

Exactly how many cello performances were there in Wednesday’s concert? Also, what was the total number of performances (featuring harp, oboe, or cello) in that concert?

6. Max joked, “Life is not always black and white, but if you’re a pianist who likes to play chess, then it kind of is!!” 😊

OK, I’ve read some minds that have spotted black, white and red too, but have you read Max’s mind!? There’s a white knight and pawn beside his drinking glass, but exactly which white pieces &/or pawns are hiding behind the glass!?

7. Are you feeling nicely relaxed here in Blog Post #150?

Part 1: In our normal base ten, the number 150 ends with a zero. In which different number base does it end with two zeros, and exactly what other digits would be needed?

Part 2: The number 150 (base ten) consists of B digits when represented in base B. What is the value of B, and exactly what digits would be needed this time?

8. Imagine that Max and his parents have a water bottle each. All three bottles have the same capacity in litres: either 0.5, 0.75, 1, 1.5 or 2 litres.

Max’s bottle is full, but his parents’ bottles are empty. In what follows, the numbers of millilitres of water in the bottles are always whole numbers after each sharing action. (You can assume that everything is done very precisely with the aid of measuring cylinders and such items, if you like!) 😊

Max first pours some water from his own bottle to his father’s bottle until they both have equal amounts.

Max’s father then pours some water from his own bottle to his wife’s bottle until they both have equal amounts.

Max also pours some millilitres of water from his own bottle into his mother’s bottle, and then he gives his father the same nice bonus 😊.

All three bottles now contain exactly equal amounts of water.

Exactly how many millilitres of water did Max pour personally into his mother’s bottle? Also, what is the full capacity of each of the bottles?

9. Max was born in the year 2000. If you multiply the age he’ll be on his birthday this year by the number of days in his birth month, the three-digit result contains Max’s three favourite (positive one-digit) numbers 😊😊😊.

What are Max’s three favourite numbers?

10. The largest of Max’s three favourite numbers is his absolute favourite, and it’s also the month number for his birthday 😊. Now in 2023, day #1 was 1 January, day #42 is 11 February, and so forth. Max’s birthday in 2023 will be on day #N of this year, and N is a whole multiple of Max’s current age.

The fun challenge is to figure out Max’s exact date of birth.

This card is appearing here far in advance of Max’s actual birthday,
but I’m happy to be early and sure rather than risking missing the party! 😊

11. List Max’s three favourite positive numbers from smallest to largest. Think of them as being the first three terms of a never-ending sequence of numbers! In that sequence, each term (from the second term onwards) can be generated by Multiplying the term immediately before it by M and then Adding on A.

Exactly what numbers do M and A represent?

Also, what is the exact value of the term in the sequence that is closest to 1000000?

12. Max went to high school in the lovely Austrian village of St. Gilgen, where Mozart’s mother was born loooooooooooooooooooong ago! 😊

It’s time now for one of my favourite brainteasers of all time, which I solved really happily in my early teens at school in Scotland 😊β™₯😊.

As Max will be turning 23 later this year, I’m thinking of a very special 23-digit whole number which starts and finishes with 1 at both ends. If you divide the super-loooooooooooooooooooong number by 99, the result is a twenty-one digit whole number which looks absolutely identical to the starting number except that it’s missing the first and last ‘end 1s’.

Your super-fun brainteaser is to find the 23-digit number in honour of Max! 😊

Notes:- No computer program or calculator is needed at all. The brainteaser can be solved quickly on paper. I find it a delightful bonus detail that there are 21 and 23 letters respectively in the names of Wolfgang Amadeus Mozart and his mother, Anna Maria Walburga Mozart 😊β™₯😊.

13. Last week, I gave a ruler as a gift to a Maths student. Now, today, I’m thinking of one of the greatest mathematicians of all time (with initials LE) whose full name can be found by rearranging the letters of LE HAD ONE RULER. The world-famous mathematical genius was born in the year – – – – and the digits are the same as Max’s birthday when stated in the form ddmm (day number followed by month number) 😊

Can you identify the famous mathematician?

This lovely photo of Max & Chess sets galore in Istanbul reminds me in various ways of a time long ago when I played for Scotland at the 1986 World Chess Olympiad in another stunning location: Dubai! The Burj Khalifa wasn’t yet there, of course, but it’s going to star in Max’s next adventure, coming…now! 😊β™₯😊


Dubai’s Burj Khalifa skyscraper (2717 feet high) featured in one of Tom Cruise’s Mission Impossible movies, and now we’ve got Max and Superman scaling its heights in the brainteaser below!! 😊😊

Imagine tiny me sitting on flat ground at the same level as the base of the Burj Khalifa, but some distance from it. Superman has flown to a point at a certain height on the skyscraper, and the eye-to-eye distance between him and me is 1000 feet. Max has already climbed to a point on the skyscraper that is 1000 feet directly above where Superman is.

When I look up at Max, his angle of elevation is twice (or double) the angle of elevation involved when I look at Superman.

Your sky-high brainteaser is to figure out the number of feet that Max still has to climb to reach the very top. (Just assume that people’s heights are negligible in comparison to the other distances involved in the puzzle.)

Fun Note: In a way, the digits of my answer are a tribute to Max and to Jens Van Steerteghem, an excellent mathematical/scientific colleague of mine at Musica Mundi School, as will be explained when the solutions are published.

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

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 M😊twani β™₯


This chess puzzle is from a real Grandmaster battle in which it’s White to play and win by force in dazzling fashion! It’s dedicated to all fans of ‘The Royal Game’, and especially to Max Gililov, to Eric Van Steerteghem (whose birthday is tomorrow), and to International Masters Geert Vanderstricht and Roddy McKay who have their birthdays today!! 😊😊😊😊

“And the spirit of the Lord shall rest upon him, the spirit of wisdom and understanding, the spirit of counsel and might, the spirit of knowledge and of the fear of the Lord”–Bible Verse, Isaiah 11:2 β™₯

P.S. = Puzzle Solutions (being posted on 17.2.2023)

  1. Part 1: 13 moves; Part 2: The possibilities are infinite! For example, draw a straight line from a point on the baseline of the photo, at a distance d from the bottom left-hand corner, to a point along the top of the photo at a distance d from the top right-hand corner. The rectangular photo is thereby split into two trapeziums of equal areas; the area of either one is exactly half of the area of the entire rectangular photo.
  3. HELLO – H + C β†’CELLO.
  4. AIR STAR – R + U β†’AUSTRIA.
  5. There were 7 cello performances and 5 harp or oboe performances, making 12 performances in total.
  6. Hiding behind Max’s glass were the white queen and another white pawn. They were no longer on the board itself.
  7. Part 1: 150 (base ten) is equivalent to 1100 (base five) or, for Part 2, 2112 (base four); B = 4.
  8. Max personally poured 125 ml from his 1.5-litre bottle to his mother’s 1.5-litre bottle. Among the bottle-sizes mentioned in the puzzle, only the 1.5-litre option was wholly divisible by four and by three, so that Max’s parents could have a quarter-full bottle each (with 375 ml), before Max topped them up to one-third full (with 500 ml).
  9. 23 x 31 = 713; Max’s three different favourite numbers are 7, 3 and 1.
  10. Max’s exact date of birth was 17.7.2000. Note that, in 2023, July 17 will be day #198 of this non-leap year, and 198 = 9 x Max’s current age of 22.
  11. The numbers in the sequence 1, 3, 7, … can be generated by multiplying terms by 2 and adding on 1 to get the next term; M = 2 & A = 1. Note also that each term is always 1 less than some power of 2. For instance, 3 = 2 squared – 1 & 7 = 2 cubed – 1. The value of the term that is closest to 1000000 is 1048575, which is 1 less than 2 raised to the power of 20.
  12. The magical number from my early teens is 11123595505617977528091. Note that 11123595505617977528091 Γ· 99 = 112359550561797752809.
  13. LE HAD ONE RULER β†’LEONHARD EULER, born in 1707.
  14. In the puzzle, Superman is 500 feet up from the base of the Burj Khalifa at a 30° angle of elevation from where I am, while Max is 1500 feet up at a 60° angle of elevation with respect to me. Max still has a further 1217 feet to climb to reach the top, and the digits 12 & 17 are early celebrations of the birthdays of Jens Van Steerteghem & Max Gililov on July 12 and 17 respectively 😊😊
  15. In the Chess puzzle, White won beautifully with 1 Bxf5 exf5 2 Nh6+ Kh8 3 Rxg7! Kxg7 4 Rg1+ Kh8 5 Qe2!! Be6 (or 5…Qxe2 6 Nxf7#) 6 Nxf7+!, intending 6…Qxf7 7 Qe5+.

Blog Post #149: A Funny Tale of Pairs of Pairs of Furry Tails!! πŸ˜Šβ™₯😊β™₯

Dear All,

Mr. Jan Vanderwegen, an excellent colleague, IT expert and friend of mine at Musica Mundi School, enjoys being creative and thinking ‘out of the box’, just like his very clever trio of mathematical cats! 😊β™₯😊

Cute Coco likes square-based boxes…so we’ll certainly feature some sneaky puzzles about squares! 😊
There on the comfy sofa, Coco, Marron and Noisette are dreaming up some fun puzzles in early celebration for Jan turning N*(N+1)*(N+2) years old next month, on day number (N+2)*(N+2) + (N+4)*(N+4) of this year 2023 😊β™₯😊


  1. How old will Jan be on his birthday next month?
  2. What was Jan’s exact date of birth?
  3. Remove just one letter from the word FEELING and rearrange the remaining letters to make a proper, six-letter English word.
  4. Imagine me visiting the cats’ home as a – – – – -. Change the last letter of – – – – – to the letter immediately before it in the English alphabet. Can you guess what word you’ll have then?


5. What is the smallest positive whole number for which its square begins with the digits 222?

6. What is the smallest positive whole number (with more than one digit) which becomes a square number if its reverse is either added to or subtracted from it?

The next brainteaser is going to feature Jan’s three cats…and one of my own…
so that we can have a funny tale of pairs of pairs of furry tails!! 😊β™₯😊β™₯


Imagine that I write down the ages (which are all different whole numbers of years) of Jan’s three cats and of my cat. The range (or age difference) between the oldest cat and the youngest of the four is exactly double Jan’s favourite number.

For each possible pair of cats among the four, I write down the sum of the two cats’ ages, until the list of sums is complete. Next, for each possible pair of sums from that list, I calculate the sum of the two sums involved! I put the results of those particular calculations in a box. The smallest result in the box is 22, and the largest result in the box is 50.
Your fun brainteaser is to figure out Jan’s favourite number ! 😊😊

By the way, where does a cat go if it loses its tail…?!

…It goes to the retail store!!

Did you hear about the secret agent Maths teacher who couldn’t get on the plane?!…
…He let the cat out of the bag!!

Bright B😊nus: Suppose that the traditional colours of a rainbow, represented by ROY G BIV, correspond to the numbers 1, 2, 3, 4, 5, 6, 7 respectively. The product of the values of Jan’s two favourite colours is a square number.

What are Jan’s two favourite colours?

8. Jan’s second-favourite and third-favourite numbers are both odd whole numbers, greater than 1. One of them is larger than Jan’s favourite number which you (hopefully) found already.

With that given information, what is the smallest-possible product if we multiply Jan’s three favourite numbers together? (The true product may be higher, of course, but that can’t be confirmed without further information.)

9. The following neat chess puzzle is dedicated to Jan, and to James Gallagher–a former student of mine who is fascinated by ‘The Royal Game’.

Black is threatening …Bh4#, but…it’s White to play and force checkmate in just 3 moves! 😊

10. The Cats’ Sky-High B😊nus Birthday Brainteaser for Eric Van Steerteghem next Sunday!

As Eric’s birthday is coming in 7 days from now, on 12 February, Jan’s clever trio of cats have a sky-high bonus brainteaser involving a triangle and the number 84 = 7 x 12 for Eric 😊

Imagine a right-angled triangle in which the three side lengths (in centimetres) are each exact whole numbers. One of them is 84 cm.

Part 1 of the brainteaser is to figure out the maximum-possible perimeter of the triangle, and also its maximum-possible area.

Part 2 of the brainteaser is to figure out the minimum-possible perimeter of the triangle, and also its minimum-possible area.

It’s my intention to publish solutions to all the puzzles around the time that blog post #150 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 😊.

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

With kindest wishes as always,

Paul M😊twani β™₯

“You have already been pruned and purified by the message I have given you.”–Bible verse, John 15:3 β™₯

Here’s Wishing You a Happy Valentine’s Day in 3 x 3 days from now…😊β™₯😊

P.S. = Puzzle Solutions (being posted on 17.2.2023)

  1. Jan will be 3 x 4 x 5 = 60 years old on his birthday next month.
  2. Jan’s date of birth was 15.3.1963. Note that 15 March is day number 74 or (5 x 5) + (7 x 7), in non-leap years.
  4. GUEST – T + S β†’GUESS.
  5. 149 squared = 22201.
  6. 65 + 56 = 121 = 11 squared & 65 – 56 = 9 = 3 squared.
  7. Jan’s favourite number is 7. Jan’s favourite colours are Red and Green.
  8. 3 x 7 x 9 = 189.
  9. In the Chess puzzle, White forces checkmate with 1 Rh5+! Kxh5 2 Qh7+ Kg5 3 h4#.
  10. Part 1: The maximum possible perimeter is 84 + 1763 + 1765 = 3612 cm; the maximum possible area is 84 x 1763 Γ· 2 = 74046 square centimetres; Part 2: The minimum possible perimeter is 84 + 13 + 85 = 182 cm; the minimum possible area is 84 x 13 Γ· 2 = 546 square centimetres.