Blog Post #140: Every Precious Moment Counts β™₯

Dear Readers,

Today, 25 November, it’s exactly one month until Christmas, and I recently promised some friends that I would do my best to publish a nice, fresh article before then. I had intended to wait a few more days, but I try to always remember that every precious moment counts, and to never make any arrogant assumptions regarding the future. An important reminder of that is, “Do not boast about tomorrow, for you do not know what a day may bring”-Proverbs 27:1.

While it’s not wrong to have good, personal hopes and intentions, we need to know and respect that God’s perfect plans will always prevail, and not always in the way or at the time that any person can predict.

Many of my friends like puzzles very much as I do, and so I will try to offer now some lovely puzzles, starting with one for a lady who has her birthday on day number D x D in December, when her new age will be the two-digit number EF (in which E is smaller than F, and note also that D, E and F are all whole numbers bigger than 1). Here in Blog Post #140, it’s nice that D x D x E x F = 140. Your first fun puzzle here is to figure out the date of the lady’s birthday, and the new age that the lady will be.

Next, I’m remembering Max (a student from my previous school), who liked to remember facts such as: there are 1440 precious minutes in a day. Your second fun puzzle is to figure out (preferably without using a calculator), what is the smallest positive whole number of standard days for which the total number of minutes will be a whole multiple of 140?

As 3 is my absolute favourite number, I hope that you’ll particularly enjoy the third fun puzzle, coming in just a wee M😊ment… There was just one particular year in the 20th century such that its four-digit number could be expressed as the product A3 x B3, where A3 and B3 are both two-digit whole numbers. A few days ago, I discovered that A3 x B3 + 14 x 140 = C3 x D3. The fun brainteaser is to figure out the precise values of A, B, C & D. (Naturally, A & B are interchangeable in this puzzle, as are C & D.)

One of the most precious gifts that we can be granted is wisdom from God. King Solomon used that gift well for a while. ‘Solomon’ is also the name of one of my excellent, current colleagues at Musica Mundi School. I’m almost two months too early in writing this to Solomon, but I would like to wish him a wonderful, happy birthday for the date A3.01.20A3, when he’ll be A3 years old 😊😊😊.

Other colleagues with birthdays coming sooner include:-

Jan V.L. in 3 days’ time; Herman in A + A3 days’ time; Peter in 3 + A3 + A3 days’ time.

Have fun figuring out their birthday dates 😊😊😊.

Clearly, Jan V.L.’s birthday (3 days after 25 November) is 28 November, which brings to mind the number 2811. Here comes a very beautiful puzzle featuring that number… Suppose that J, A, N, V & L stand for certain consecutive whole numbers, in increasing order. The nice challenge is to discover their values such that (J x A x N x V x L) Γ· (J + A + N + V + L) + 3 = 2811 β™₯β™₯β™₯β™₯β™₯

One of my favourite chess puzzles was composed by W.Speckmann in the year 14 x 140 = 1960. I’m sure that FIDE Master Arben Dardha will love it, too. Happy birthday today, dear Ben! β™₯

It’s White to play and force checkmate in 4 moves β™₯

God-willing, I will publish solutions to all the puzzles around the time when I hope to share Blog Post #141 with you. In the meantime, you are more than welcome to send me your solutions to some or all of the puzzles, if you like. I wish you a very blessed, happy weekend now.

Special congratulations to Michael and Annie, two lovely young people who got engaged on Trent Bridge, Nottingham, tonight β™₯😊😊β™₯

Warm congratulations to Michael & Annie on getting engaged tonight β™₯β™₯

Congratulations also to Eric & Marianna (lovely former Maths students of mine) who recently married near Paris β™₯β™₯

Marianna and Eric, a really lovely couple β™₯β™₯

With love and kindest wishes as always,

Paul M😊twani β™₯

As I don’t want to waste this precious moment, I’ll leave you with even more than an extra couple of couples of nice bonus puzzles!!!! β™₯😊😊β™₯

First, suppose that the letters C, O, U, P, L, E and S each stand for different, positive, whole numbers. Then, what would be the minimum possible value for the sum C squared + O squared + U squared + P squared + L squared + E squared + S squared ?

Next from our huge chocolate box of puzzles, can you rearrange the letters of WAIST to make another proper English word?

Waiting now and looking forward to 25 December has inspired this penultimate wee puzzle today: (25 – W) x (25 – A) x (25 – I) x (25 – T) = 25, where W, A, I & T represent different integers. What is the value of the sum W + A + I + T ?

The final puzzle tonight is in honour of Alicia, the eldest daughter of Nathan Braude, a new colleague at Musica Mundi School. As Alicia’s birthday was on 5 October, I’m thinking of the three-digit number 510. The number 6 will also be starring in this wee puzzle, since the names ‘Alicia’ and ‘Nathan’ each have six letters. Now I’ll tell you that 6 x 5 x the age that Alicia will be on her birthday in (just under) five years from now equals 510. Can you be almost as fast as Alicia and figure out the year when she was born?!

By the way, though I’m now 60 years old, I still love to solve Junior Maths Challenge puzzles! I’m betting that Alicia will enjoy them, too! 😊 So, specially for Alicia and all Maths fans, here is a link to a further feast of wonderful puzzles:

A feast of fun puzzles is waiting for Alicia and all Maths fans 😊😊

For all Chess enthusiasts-including Daniel Mathiesen from HΓ€rnΓΆsand, Sweden-I’m including a link to two of my favourite sites: &

(I couldn’t resist giving everyone yet another b😊nus!!…and so the second link, just above, leads to a delightfully crisp case of ‘White to play and force checkmate in 3 moves’ even though the material is equal there β™₯)

Here’s wishing Daniel Mathiesen, and lots of other keen players, continued enjoyment from the Royal Game of Chess β™₯

P.S. = Puzzle Solutions (being posted on 2 December 2022)

2 x 2 x 5 x 7 = 140; the lady has her birthday on December 4 (= 2 x 2), and she’ll be turning 57.

In the second puzzle, 7 days works, because 1440 x 7 Γ· 140 = 144 x 7 Γ· 14 = 144 Γ· 2 = 72, a whole number.

23 x 83 = 1909, and 1909 + 14 x 140 = 3869 = 53 x 73.

Herman’s birthday is on December 20 (2 + 23 = 25 days after 25 November, when this article was first published).

Peter’s birthday is on January 13 (3 + 23 + 23 = 49 days after 25 November).

Solomon has his next birthday on 23.01.2023, when he’ll be 23 years old!

(9 x 10 x 11 x 12 x 13) Γ· (9 + 10 + 11 + 12 + 13) + 3 = 2811 β™₯

In the Chess study, White forces checkmate with 1 Qe6! Bg4 2 Qa2+! Ra4 3 Qg8 b4 4 Qc4#; beautiful play! 😊

The minimum possible value for the sum C squared + O squared + U squared + P squared + L squared + E squared + S squared = 1 squared + 2 squared + 3 squared + 4 squared + 5 squared + 6 squared + 7 squared = 1 + 4 + 9 + 16 + 25 + 36 + 49 = 140, nice for this blog post #140 β™₯😊β™₯

WAIST rearranges to make WAITS.

If (25 – W) x (25 – A) x (25 – I) x (25 – T) = 25 for different integers W, A, I & T, then, in some order, (25 – W), (25 – A), (25 – I) & (25 – T) must equal -1, 1, -5 & 5. So, in some order, W, A, I & T have to equal 26, 24, 30 & 20. Therefore, the sum W + A + I + T must be 20 + 24 + 26 + 30 = 100 exactly β™₯

In the puzzle about Alicia, 6 x 5 x 17 = 510; Alicia will be 17 in 2027 (five years from now); she was born in 2010.

Blog Post #139: Yo, Jens! πŸ˜ŠπŸ˜Š

Dear Readers,

This particular post is being written as a nice wee bonus surprise for Jens, a friend and brilliant colleague of mine at Musica Mundi School.

Jens solves lots of puzzles regularly, and he really – – – – – – them! The ‘blanks’ – – – – – – stand for a proper six-letter English word that can be made by rearranging the letters of the title ‘Yo, Jens! 😊😊’

The smiley faces 😊😊 also fit well with the answer, but as Jens especially enjoys mathematical puzzles, here comes a sneaky little one inspired by the name JENS. Regarding positions within the English alphabet, J=10, E=5, N=14 & S=19. As we’re now in blog post #139, your fun Maths challenge is to use the numbers 5, 10, 14 & 19 exactly once each in a calculation which leads to a result of 139. You may use the 5, 10, 14 & 19 in any order that you like, and you may also use +, -, x, Γ·, ( ) freely wherever you may wish to.

Via Blog Post #138, Jens discovered that my dad will turn 88 next month. I’m now thinking of two proper 2-digit numbers XY & YX such that the sum XY + YX = 88, while the product XY times YX is as small as possible (but neither X nor Y is zero). I’m also thinking of those particular numbers because XY times YX Γ· 100 gives the day & month numbers of Jens’ birthday in the form DD.MM. Now you can enjoy discovering when Jens’ next birthday will be! 😊

Jens also loves some good chess! In the position featured below (which I saw via Facebook earlier today), White is of course in a totally, easily winning position. However, what still makes for a delightful wee puzzle is: How can White (to move) force checkmate in only 2 moves…?

White is to move and force checkmate in only 2 moves β™₯

It’s my intention to wait at least a day or two before publishing solutions so that, in the meantime, readers can enjoy tussling with the puzzles. You are most welcome to send me any/all of your answers, if you like…and now I have a beautiful Bible reflection to conclude this article.

“Therefore, encourage one another and build one another up, just as you are doing”-1 Thessalonians 5:11

Here’s wishing everyone a very good and peaceful Sunday now,

Paul M😊twani β™₯

P.S. = Puzzle Solutions (being posted on 26.11.22)


5 x (10 + 14) + 19 = 139.

XY + YX = 17 + 71 = 88


XY times YX Γ· 100 = 17 x 71 Γ· 100 = 12.07 in honour of Jens’ birthday 0n 12 July.

In the chess puzzle, White can underpromote the c-pawn to a new bishop! So, instead of 1 c8=Q?? which would produce a stalemate position, White forces checkmate via 1 c8=B! Kxe8 2 Rg8# 😊

Blog Post #138: With One’s Whole Heart β™₯

Dear Readers,

A particularly beautiful Bible verse is “I give thanks to You, O Lord my God, with my whole heart, and I will glorify Your name forever”-Psalm 86:12. That verse is inspirational because it reminds us that we give glory to God by doing everything with gratitude to Him.

Earlier today, my son and I had a nice chat while I drove him to Brussels Zaventem Airport for a special weekend trip that he’s making. On my way back home afterwards, I spotted a vehicle with an interesting number plate which led me to making a fascinating investigation into properties of numbers. For me, it was like receiving a surprise gift. I’m going to share it with you now, and I wish you lots of enjoyment too β™₯😊😊β™₯

I like my house number, 11, but my absolute favourite number is 3. So, I invite you to write down a three-digit number 11N in which N is any digit that you like from 0 to 9. Now calculate N x 11N, and let’s call the result your Star Number.

Next, we’ll shake things around a wee bit 😊 because I’m going to ask you to calculate 1N x 10N (where 1N is a two-digit number and 10N represents the three-digit number with digits 1, 0 and N). That new result will be your Super Star Number.

Question 1.1: What is the relationship between your Super Star Number and your Star Number?

If N is now any whole number (which can even be as big as you like), then N x (110 + N) is clearly not prime because it’s a product of numbers N & (110 + N). Even 1 x (110 + 1) is not prime because 1 x 111 = 111 = 3 x 37, a composite product.

Question 1.2: Is 1000 + N x (110 + N) prime or not?

Part of the license plate number that I saw today was 819.

Question 1.3: Can you show why 819 is a possible Star Number?

Question 1.4: Can you show why 1819 is a possible Super Star Number?

Here in Blog Post #138, let’s note that 138 = 2 x 3 x 23.

Question 1.5: 1819 raised to the power of P results in a 23-digit whole number. What is the value of P? Also, can you calculate the exact 23-digit result!?

Question 1.6: (138 x P Γ· 3) raised to the power of (3 x 3) also results in a 23-digit whole number. Can you calculate that 23-digit result precisely, too!?

Question 1.P: Can you spot a special connection between the 23-digit results of questions 1.5 and 1.6 ?

It’s my intention to wait a day or two before publishing solutions so that, in the meantime, readers can enjoy tussling with the puzzles. You are most welcome to send me any/all of your answers, if you like…and now I have more goodies for you!

My dad is not a chess player, but I think that he’ll like the fact that I’m now going to feature a truly fantastic old chess study from Germany (in which it’s White to move and force checkmate in 5 moves) in early celebration of his birthday coming up soon, on 5 December β™₯😊β™₯

Question 1.8: By the way, how old will my dad be then (on 5 December) if I tell you that his new age + his age 38 years before then will equal 138 ?

Brilliant Old Chess Study from Germany (also shared recently by others on Facebook)

It’s White to move and force checkmate in 5 moves !

Early Christmas Word Puzzle β™₯😊β™₯

Remove a particular letter from GERMANY and then rearrange the remaining six letters to make a proper 6-letter English word which relates to Christmas. There is just one, unique solution.

I will finish now by wishing you and everyone a wonderful, happy weekend.

With kindest wishes as always,

Paul M😊twani β™₯

P.S. = Puzzle Solutions (being posted now on 13.11.22)

1.1 The Super Star Number is always exactly 1000 more than the Star Number.

1.2 Since 1000 + N x (110 + N) is actually the same as the product (10 + N) x (100 + N), it’s always composite; never prime.

1.3 819 = 7 x 117, and is therefore one possible Star Number (with N = 7).

1.4 1819 = 17 x 107, and is therefore one possible Super Star Number (also with N = 7).

1.5 P = 7, and 1819 raised to the power of 7 gives the 23-digit result 65891424018613967932339.

1.6 (138 x 7 Γ· 3) raised to the power of (3 x 3) gives the 23-digit result 37213699403613156884992.

1.7 It’s not so easy to spot at a glance that the two 23-digit results actually contain exactly the same 23 digits (in two different orders, of course) ! β™₯

1.8 88 + (88 – 38) = 138. My dad will be turning 88 on his birthday next month β™₯

Brilliant Old Chess Study Solution

The incredible, most powerful, centralising move for White from the chess study position given earlier is 1 Qe4!!, intending Qb1 followed by Qb5# or Qb6#.

Note that 1…Bxe4 would allow 2 Rc5#,
while after 1…Rxe4 (or 1…Qxe4) 2 Rh8, Black cannot prevent 3 Rc8# !

Black sees that 2 Qb1 is threatened, and 1…Nd2 loses to 2 Qe3, for example.

My favourite variation in this stunning study is probably 1…Qf3 (to answer 2 Qb1? with 2…Qb3!) 2 Qd4!! Rxd4 3 Rh8 Be4 4 Rc8+ Kd5 5 Rc5# β™₯😊β™₯

Another line is 1…Qg1 2 Qb1 Qc5 3 Qb4!! (even stronger than 3 Nxc5) 3…Qxb4 4 axb4 and 5 b5# cannot be stopped!

Early Christmas Word Puzzle Solution

GERMANY – Y gives GERMAN which rearranges nicely to make MANGER for Christmas coming soon β™₯

Special congratulations to my friend and brilliant colleague, Jens Van Steerteghem (one of the Science teachers at Musica Mundi School), who sent in terrific answers to many of the mathematical puzzles. Feel free to now enjoy Blog Post #139: Yo, Jens! 😊😊

Blog Post #137: VIPs Young and Old β™₯

Dear Readers,

Whether we have already met yet, or not, I think of you as a VIP because we are all children of God, and that automatically makes you a Very Important Person indeed.

Everyone who took part in a fun chess talk/’simul’ at Edinburgh Chess Club three days ago (in early celebration of the club’s 200th anniversary coming three days from now), received lovely prizes such as books, magazines and chocolates to let them all feel like the VIPs that they truly are β™₯

Thanks to Sue Loumgair, Dr. Vipin Zamvar, Ian Whittaker and FM Neil Berry (the President of Edinburgh Chess Club) for having sent nice photos from the event.

Dr. Vipin Zamvar is a really good-hearted gentleman who was particularly kind to my family during our final evening in Edinburgh β™₯

In addition to enjoying seeing very long-time friends Lindsay McGregor, David Montgomery, FM Craig Thomson, FM Neil Berry, Jonathan Grant, GM Keti Arakhamia-Grant & their daughter Elena Grant, it was lovely to get to make many new friends during the club’s joyful celebrations 😊😊

I would like to thank all who attended the chess talk for being a wonderfully receptive audience, and I congratulate everyone from the ‘simul’ for playing so many good moves. Special congratulations to Warrick Campbell and Neil Irving for taking full advantage of their opportunities and playing better than I did!

The title of ‘Funniest Chess Friend in Edinburgh’ probably has to go to Craig Thomson! He knows that 3 is my favourite number, but as I hadn’t seen him for 13 years (since 2009 !) and as his email address contains the number 17, I offer these amusing thoughts…I was obliged to score 17/19 in your honour, Craig…and also 1303 x 3 x 3 x 17 = 199359 in recognition of the fact that the club was precisely 199 years & 359 days old when we met three days ago! 😊😊

I promised Lindsay McGregor that I would publish the moves of a personal game that I also shared during the chess talk, as he was very interested in that game. Here it is:-

P.A.Motwani vs. My Computer (Training Game played at home on 27.8.2022)

1 e4 g6 2 d4 Bg7 3 Nc3 d6 4 Be3 Nf6 5 Qd2 0-0 6 Nf3 a6 7 Bh6 Bg4 8 h4 c5 9 h5 Bxh5 10 Bd3 Bxf3 11 gxf3 cxd4 12 Bxg7 Kxg7 13 Qh6+ Kg8 14 e5 dxe5 15 Ne4 Re8 16 Ng5 Qa5+ 17 Ke2 e4 18 fxe4 e5 19 Nxh7 Nxh7

20 Rag1 Qc7 21 Rxg6+ fxg6 22 Qxg6+ Kf8 23 Rxh7 Qxh7 24 Qxh7 Rd8 25 Bc4 Rd7 26 Qh8+ Ke7 27 Qxe5+ Kd8 28 Be6 Nc6 29 Qh8+ Kc7 30 Qxa8 Rd8 31 Qxd8+ Kxd8 32 Bd5 and White (then 2 pawns ahead) soon won… 1:0.

Here now is a selection of positions that occurred in the ‘simul’. In each case, it’s White to play and win.

At the moment, I can still recall the moves from the 19 games in the ‘simul’. After a while, I will probably forget some of the details, but I will always treasure having met all the lovely people at the club β™₯

Chess Position Solutions

  1. White obtained a decisive advantage via 1 Bxf6 gxf6 2 Qh6 f5 3 Qf6.
  2. White obtained a decisive material advantage via 1 Qd2 Ned7 (1…Ng6 2 f5 Ne5 3 h3 Bh5 4 g4 also traps Black’s queen’s bishop) 2 h3 Bh5 3 g4.
  3. White won with 1 Rxd7 Re4 (or 1…Qxd7 2 Qh6) 2 Rc7 intending 2…Rxh4 3 Rxc6 or 2…Qxc7 3 Qh6 or 2…Qd6 3 Qxe4.
  4. White won with 1 Nxd5, intending 1…exd5 2 Rxe8+ Qxe8 3 Bxf6 gxf6 4 Qxd5+.

After all that chess, let’s round off with a nice wee dose of Maths here in Blog Post #137…

Not only is 137 a prime number, but so is 13, 17 and 37. 137 is the smallest whole number with digits possessing such properties!

In honour of Edinburgh Chess Club’s 200th anniversary, 200 is the smallest whole number which cannot be made prime by changing one of its digits.

I would like to wish everyone a very happy All Saints Day now on 1 November β™₯

As there must surely be lots of singing in Heaven, the saying “Music is the Medicine of the Mind” comes happily to mind β™₯

With kindest wishes as always,

Paul M😊twani β™₯