## Blog Post #149: A Funny Tale of Pairs of Pairs of Furry Tails!! πβ₯π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! πβ₯π

COOL CATS CHALLENGES!

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?

MR. Mπ’s FUN BRAINTEASERS

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?

7.

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

…It goes to the retail store!!

Bright Bπ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’.

10. The Cats’ Sky-High Bπ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 π

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 π.

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 Mπtwani β₯

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

## Blog Post #148: Giant Rescue Operation Codename R-E-V-S! πβ₯π

Dear All,

This action-packed article is specially dedicated to Mr. Eric Van Steerteghem, the father of Jens Van Steerteghem who is an excellent colleague of mine at Musica Mundi School π

For several weeks, I have been preparing nice, early surprises for Eric’s birthday coming soon, but due to a dramatic mathematical malfunction of my time machine, Eric got transported back to a year in the 16th century!!

Very fortunately for me, and for Eric, I received invaluable help from the Mathematical Murphy Family in Operation ‘REVS’: Rescue Eric Van Steerteghem!

Continue reading “Blog Post #148: Giant Rescue Operation Codename R-E-V-S! πβ₯π”

## Blog Post #147: Cascade de Surprises! πβ₯π

The title of this particular post was suggested 3 days ago by my ‘big brother’, Jan Van Landeghem, just after he and I and RaphaΓ«l (a super student at Musica Mundi School) had enjoyed discussing a stunningly beautiful chess study containing many surprises to delight us!

It’s 13 January, the birthday of Peter, and 13 is his favourite number! π

Alexander, another great colleague, has a smaller, positive, favourite whole Number. Let’s call it N.

Brainteaser starring Peter and Alexander the Great! ππ

Imagine that Peter has many music CDs, and on each CD there are 13 completely different songs.

Also suppose that Alexander has the same number of music CDs as Peter has, and on each CD there are N completely different songs (and different from any of Peter’s songs too).

The total number of songs on all their CDs together is 2023.

Part 1: What is Alexander’s favourite number, N?

Part 2: Exactly how many CDs does Alexander have?

Part 3: Imagine now that Peter has an unlimited supply of CDs, some with 13 songs and some with N songs. What would be the minimum total number of such CDs needed so that the total number of songs would be exactly 2023?

13 is the total number of letters in…

Cheers to Peter, Brainteaser! πβ₯π

The bottle on the left used to be full up to the top, but I enjoyed a nice drink and said, “Cheers to Peter!” The bottle is now only 65.57% full regarding the volume remaining, as a percentage of the original full volume.

Your brainteaser is to figure out, to the nearest whole millimetre, what should be the height indicated over on the right of the picture just above? (It’s the same bottle, flipped over.)

Brainteaser in Honour of Peter’s life prior to today!! π (The reason will be clear when solutions are posted later on!)

I’m thinking of a particular whole number which ends with the digits 23 on the right. Let’s call the entire number P. If P is multiplied by my favourite number, 3, the result will be Q, say. A remarkable detail is that P & Q together feature all the digits from 1 to 9 inclusive, once each, with no zeros. Furthermore, P is the smallest whole number such that it and its triple together feature 1, 2,…,9 once each in some order.

Your fun brainteaser is to discover the exact values of P & Q.

Another PETER Brainteaser! π

Imagine that A=1, B=2,… and so on. The multiplicative value of PETER would be found by substituting the appropriate letter values in P x E x T x E x R.

Without even needing to do any calculations at all, can you read my mind and say which proper English word I’m thinking of which is longer than PETER and yet has exactly the same multiplicative value?

Time for a Brainteaser about Time!! ππ

I’m thinking right now about two positive whole numbers… Let’s call them A and B.

(A raised to the power B) Γ· (B raised to the power A) results in a decimal number that looks just like a time that I noticed on a clock this morning. That particular time is also the very first time after 7.00 for which the sum of all the digits (for the hour and minutes) equals 7, and all the digits are different from each other.

Your fun brainteaser is to discover my numbers A and B.

Brainteaser in Honour of Haiyue, Defne G. and Uriel πππ

Three of my younger students voluntarily did an extra 100 minutes each of Maths study and practice late into Wednesday evening, a couple of days ago!! They’re all preparing diligently for a test that’s coming soon.

One of the questions that Uriel asked about inspired me to do a further Maths investigation myself yesterday, and I discovered a formula which may possibly be brand new.

Imagine a ship at a position S. It sails a distance of x kilometres on an acute angle bearing of yΒ° from S to T. It then sails a further x kilometres due East from T to U, in honour of Uriel!

Your brainteaser is to find a neat, simple expression in terms of y for the bearing of U from S.

It’s something of beauty to the mathematical mind that the final, simplified expression will be independent of x. (In other words, after simplifying the algebraic terms involved in the problem, x will not appear in the final result.) πβ₯π

Brainteaser with Happy Memories of my previous school in Belgium π

I’m now thinking of a particular whole number. Let’s call it S, in honour of my previous School. S is actually the sum of 22 consecutive whole numbers (but note that I haven’t said which 22 numbers are involved!). It’s also the sum of the next 21 whole numbers (meaning the ones following right after the earlier 22 numbers).

S is also the product of several whole numbers which are all bigger than 1:-

My favourite number x The number of letters in my family name x My house number x The age I was when I started working at my previous school.

Even if you didn’t know any of those numbers in advance, it’s still possible for you in this brainteaser to…

Figure out the age I was when I started working at my previous school πβ₯π

It’s my intention to publish solutions to all the puzzles around the time that blog post #148 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 Mπtwani β₯

As 51 x 7 x 17 gives the same as 2023 tripled, let’s conclude with the following powerful Bible verse:

P.S. = Puzzle Solutions (being posted on 17 January)

My thanks and congratulations to CΓ©cile Gregoire and Jens Van Steerteghem who sent me kind messages regarding the articles and very good solutions to the puzzles. ππ

In the Chess study, White triumphs with 1 Nb6+ Ka7 2 Ra2! Qg8 (2…Qxg3 3 Kb4+ Kb8 4 Ra8+ Kc7 5 Rc8#) 3 Kb2+ Kb8 4 Ra8+ Kc7 5 Kc1!! (not 5 Rxg8 which produces stalemate, as does 5 Kc3 Qb3+! 6 Kxb3) 5…Qe6 (or 5…Qh8 6 Nd5+! cxd5 7 Rxh8, with an easy win) 6 Rf8! Qg8 7 Na8+! (‘breaking’ the stalemate situation) 7…Kd7 8 Rxg8.

In the puzzle with Peter & Alexander The Great, the key is that 2023 = 7 x 17 squared, and so 17 is the factor of 2023 that is more than 13 but less than double 13. Since 2023 = 17 x (7 x 17) = 17 x 119, Peter & Alexander will have 119 CDs each, and the number of songs on each of Alexander’s CDs will be 17 – 13 = 4.

In Part 3, when Peter has some CDs with 13 songs and some CDs with 4 songs, then (155 x 13) + (2 x 4) reaches a total of 2023 songs with just 155 + 2 = 157 CDs.

Imagine that the bottle could be swapped for a shorter wholly cylindrical bottle with the same capacity (and with the base radius unchanged). Its new full height would be 12 Γ· 65.57 x 100 = 18.30cm, correct to four significant figures. So, the part without liquid would be equivalent to a column of height 18.30-12 = 6.30cm. Therefore, the height indicated in the right-hand picture would be 21 – 6.30 = 14.70cm, or 147mm correct to the nearest millimetre. That’s nice here in Blog Post #147 π

(Note: Since we were given that the bottle is 65.57% full, the part with air represents the other 34.43% of the bottle’s capacity. So, an alternative way to calculate the equivalent cylindrical column height of the part without liquid is, using proportion, 12 x 34.43 Γ· 65.57 = 6.30cm, approximately.)

In the next puzzle, 5823 x 3 = 17469. Q = 17469 & P = 5823 containing the digits 58 in honour of Peter turning from 58 to 59 on his birthday last Friday π

Words with the same multiplicative value as PETER are REPEAT or RETAPE.

The morning clock time was 10.24, and that’s like 1024 Γ· 100, or (2 to the power of 10) Γ· (10 to the power of 2). A = 2 & B = 10.

In the puzzle about angle bearings, sketching a diagram helps with finding that the bearing of U from S equals (45 + y/2)Β°.

(Bonus Note: The Average or Mean value of (45 + y/2)–taken over all values in the interval from 0 to 90–is 67.5. A funny detail in anticipation of Blog Post #148 next is that 67.5 multiplied by the infinitely recurring decimal 1.48148148148148… equals 100 exactly! πβ₯π)

In the brainteaser about S = the sum of 22 consecutive whole numbers from n to n+21, say, then S = 22 x (n+21 + n) Γ· 2, which simplifies to S = 22n + 231.

S is also the sum of the 21 consecutive whole numbers from n+22 to n+42, and so S = 21 x (n+42+n+22) Γ· 2, which simplifies to S = 21n + 672.

Equating the bold-type expressions for S gives 22n+231 = 21n+672, and so n=441.

Therefore, S = 22 x 441 + 231 = 9933.

The prime factorisation of S is 9933 = 3 x 7 x 11 x 43.

The only factor there that could reasonably be the age of a qualified school teacher (adult) is 43. I did indeed begin teaching at my previous school in Belgium when I was 43 years old, after having done other work before, including teaching in schools in Scotland. πβ₯π

## Blog Post #146: Three Books πβ₯π

Out of the many millions of books that have ever been written, if I had to pick just 3–my absolute favourite number!–of them to keep enjoying forever, then my top selection would probably be the following:-

1. The Holy Bible, because it’s a perfect book revealing to us the Word of God, which can be trusted totally and is of supreme importance.

2. After my clear first choice above, it’s not at all easy for me to pick a second book in preference to all other books, but I’m sure that a very strong candidate would be: ’15 Minutes Alone With God’ by Bob Barnes.

If I fast-forward to pages 185-188 of the book, there’s a four-page article entitled I Didn’t Believe It Anyway, which includes the following powerful poem:

‘Twas the night before Jesus came and all through the house

Not a creature was praying, not one in the house.

Their Bibles were lain on the shelf without care

In hopes that Jesus would not come there.

The children were dressing to crawl into bed,

Not once ever kneeling or bowing a head.

And Mom in her rocker and baby on her lap

Was watching the Late Show while I took a nap.

When out of the East there arose such a clatter,

I sprang to my feet to see what was the matter.

Away to the window I flew like a flash

Tore open the shutters and threw up the sash!

When what to my wondering eyes should appear

But angels proclaiming that Jesus was here.

With a light like the sun sending forth a bright ray

I knew in a moment this must be THE DAY!

It was Jesus! Returning just like He had said.

And though I possessed worldly wisdom and wealth,

I cried when I saw Him in spite of myself.

In the Book of Life which He held in His hand

Was written the name of every saved man.

He spoke not a word as He searched for my name;

When He said, “It’s not here,” my head hung in shame.

The people whose names had been written with love

He gathered to take to His Father above.

With those who were ready He rose without a sound

While all the rest were left standing around.

I fell to my knees, but it was too late;

I had waited too long and thus sealed my fate.

I stood and I cried as they rose out of sight;

In the words of this poem the meaning is clear;

The coming of Jesus is drawing near.

There’s only one life and when comes the last call

We’ll find that the Bible was true after all!

3. No further book is really needed, but still I thank God every day for having let me enjoy many thousands of fascinating puzzles in my life so far. For me, a compilation of all those puzzles, about Chess, Mathematics, Words and more, would certainly be a treat! I believe that the puzzles in store in Heaven will be better and more magical than I can possibly imagine. For the moment, I can only offer what I know right now. So, I would like to share some surprises with you, since God gives us good thoughts to be shared. Here comes fresh puzzle ideas that came yesterday evening and in the morning today…with some extra bonuses this evening!

I would like to specially dedicate the puzzles to my excellent colleague Jens Van Steerteghem, his brother Nick, and their father Eric, as all three gentlemen are passionate about puzzles and have great talent for solving them!

A SNEAKY SPEED BRAINTEASER πβ₯π

The name ‘Eric’ always makes me think of the famous missionary Eric Liddell–affectionately known as ‘The Flying Scotsman’–who won the Gold Medal in the 400m race at the 1924 Paris Olympics. Fast-forwarding 99 years to the present 2023…imagine that Eric Van Steerteghem runs a long distance from A to B at an average speed of 3 metres per second. On the way back from B to A (following exactly the same route as before, only in the opposite direction, and naturally more tired than before), Eric’s average speed is 2 metres per second.

The brainteaser is to figure out Eric’s average speed for his entire run from A to B to A. (Being an expert in Physics, Mathematics and more, Jens could tell you immediately that the average speed will not be 2.5 metres per second! Can you do like Jens and figure out the correct value?)

CHINESE FOOD WORD PUZZLE πβ₯π

CUBOID BRAINTEASER πβ₯π

Part 1: As a very quick warm-up before the main Part 2, imagine a cube with its dimensions (equal length, width and height) in centimetres (cm).

If a particular cube’s volume (in cubic centimetres) is numerically equal to its total surface area (in square centimetres), then what must be the cube’s exact dimensions?

Part 2: If the length, width and height of a certain cuboid are all exact whole numbers of centimetres, and if the cuboid’s volume (in cubic centimetres) is numerically equal to its total surface area (in square centimetres), then what is the maximum possible height of the cuboid?

Multiplication Magic Square, Beautiful Billion Brainteaser! πβ₯π

RECIPROCALS BRAINTEASER πβ₯π

For this puzzle, we need to know that, in Mathematics, the reciprocal of any non-zero number n is 1 Γ· n.

Imagine that a lady on her birthday today said, “The difference between the reciprocal of my new age now and the reciprocal of the age I’ll be in a year from now is equal to the reciprocal of the year when my younger sister was born.”

Your brainteaser is to figure out the lady’s new age now, and figure out the exact year when the lady’s younger sister was born.

You know that I like the number 141, as it features in one of my email addresses, pmotwani141@gmail.com. Here in blog post #146, I should give a special mention to the number 14641, which equals the fourth power of my house number 11 π

146 BRAINTEASER πβ₯π

Part 1: In our normal base ten, the number 146 = 6 x 1 + 4 x 10 + 1 x 10 squared.

However, in another base B (not base ten), 222 (base B) = 146 (base 10).

Figure out the value of B.

Part 2: This involves a new base, N. We are told that

222 (base N) = xyz (base 10),

where xyz represents a proper three-digit whole number.

The brainteaser is to figure out the maximum possible value of xyz, and the corresponding value of N.

A Wee Dose of Chess To Finish! πβ₯π

It’s my intention to publish solutions to all the puzzles around the time that blog post #147 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 Mπtwani β₯

P.S. = Puzzle Solutions!

TIME RAN = MARTINE

Eric’s average speed for the entire run was 2.4 metres per second. That can be verified using the formula Average Speed = 2vw Γ· (v+w), in which v=3 and w=2, the respective speeds for the outward and return runs covering equal distances.

PRC CALORIES can turn one’s diet upside down because they make RECIPROCALS !! π

Regarding length L, width W and height H, when a cube has L = W = H = 6cm, then its volume = 6 x 6 x 6 = 216 cubic centimetres, and the total surface area of its six faces is 216 square centimetres because each one of the faces has an area of 6 x 6 = 36 square centimetres.

In the cuboid part of the brainteaser, a maximum whole number height of 42cm is achievable when the length and width are 3cm and 7cm in either order.

Given that the volume was numerically equal to the total surface area, I used LWH = 2LW + 2LH + 2WH and then 1 = 2/H + 2/W + 2/L.

Letting L=3 helps to βuse upβ two thirds of the 1, leaving only one third or 2/6. So W canβt then be 6, but it can be 7, letting us solve directly for the optimal H.

(If the length and width are 4cm and 5cm in either order, we would find that H = 20cm; smaller than our optimal 42cm.)

In the Multiplication Magic Square, we must have the number 10 in the central box, and all the other factors of one hundred can be filled in the rows in (for example) this order (starting from the top-left box):- 20, 1, 50; 25, centre 10, 4; 2, 100, 5.

In the reciprocals brainteaser, the lady is 44 and her younger sister was born in the year 1980.

It makes use of the fact that 1/n – 1/(n+1) = 1/(n(n+1)). In the puzzle, n(n+1) has to be the year when the younger sister was born. The only value for n that gives a suitable value for n(n+1) in the reasonably recent past is n=44, and then n(n+1) = 44 x 45 = 1980.

In the number bases brainteaser, 222 (base N) has the value of 2 + 2 x N + 2 x N squared. If you were to generate an accurate table of different values for N and the corresponding values of 2 + 2 x N + 2 x N squared, it would show, for example, that 2 + 2 x N + 2 x N squared = 146 when N = 8 and 2 + 2 x N + 2 x N squared = 926 when N = 21 and 2 + 2 x N + 2 x N squared = 1014 when N = 22.

So, the answers asked for in the puzzle are:- B = 8; xyz = 926; N = 21.

In the Chess puzzle, 1 Bg5+ Kg8 2 Qh7+ Kf8 3 Qh8# is the fastest win for White.

If it were actually Black’s turn to move, then (though it’s true that 1…Qxg3+ would win easily) the quickest forced win is 1…Rh1+! 2 Kxh1 Qg1#, a key checkmating pattern π

## Blog Post #145: Happy New Year Brainteasers! πβ₯π

My family and I hope that this message finds you keeping well, and we would like to wish you a very happy new year, 2023 β₯

Puzzle #1: A Magical New Year Brainteaser!

Michael and Jenny write down two positive whole numbers. They calculate the sum by adding their two numbers together, and they also calculate the product by multiplying their two numbers together.

Guess how many whole numbers there are from their sum up to and including their product as well…there’s exactly 2023 whole numbers !

Your fun challenge is to discover Michael & Jenny’s numbers!

There are three different possible solutions! πβ₯π

1. The purposeful meaning behind the words ‘as well’ above is that the sum and the product are both included in the 2023 whole numbers, with the sum at the very start and the product at the very end. Be extra-careful when considering the difference between the product and the sum…

2. Very sincere thanks to Teun Spaans (on 6 January 2023) for having kindly mentioned this brainteaser on his site justpuzzles.wordpress.com, but please note well point 1, just above.

Some Maths students know that 5! means 1 x 2 x 3 x 4 x 5 = 120.

Here in blog post #145, it’s nice to note that the special number 145 = 1!+4!+5!

Puzzle #2: A Good Book Puzzle β₯

A boy is enjoying currently reading two of the 66 distinct books of The Bible that he has been given as a gift β₯

Your puzzle is this: If you wanted to choose two out of 66 distinct books to read, how many different selections would be possible?

Puzzle #3: A Champion’s Challenge π

I recently received a happy message from Cansu, an excellent former Maths student of mine whom I always think of as ‘Champion Cansu’! π

In honour of C CANSU, use the numbers 3, 3, 1, 14, 19, 21 (in any order that you want) to make the target number 2023. You may also use parentheses ( ) and any of the operations +, -, x, Γ· as you wish. π

Puzzle #4: A Neat Word Puzzle

Rearrange the letters of the word LISTEN to make another proper six-letter English word.

There are four different possible solutions! β₯ππβ₯

Puzzle #5: Dedicated to Elton (a former student of mine from 2012-2013 who likes the Royal Game of Chess) π

Puzzle #6: OUR ANGLE Brainteaser π

Your brainteaser is to figure out the maximum possible size (in degrees) for angle RAG that fits correctly with the given information.

It’s my intention to publish solutions to all the puzzles around the time that blog post #146 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 2023, with lots of happiness in everything that you do β₯

With kindest wishes as always,

Paul Mπtwani β₯

“You have decided the length of our lives. You know how many months we will live, and we are not given a minute longer. You set the boundary, and no one can cross it.”–Job 14:5

Joke: Why should you keep your left foot still at the beginning of January?

You’ll start off the New Year on the right foot!

P.S. = Puzzle Solutions (being posted now on 6 January 2023)

Magical New Year Brainteaser

If we let Michael & Jenny’s positive whole numbers be x and y, then their sum is x+y and their product is xy, which are respectively the very first and very last of 2023 consecutive whole numbers. Therefore, the difference xy – (x+y) = 2022 (not 2023). However, the puzzle can be solved really neatly by noting that xy – x – y + 1 = 2023, because we can then use factorisation to get that (x-1)(y-1) = 2023. In other words, 2023 is the product of whole numbers (x-1) and (y-1) which must both be factors of 2023 ππ

2023 = 1 x 2023 or 7 x 289 or 17 x 119 (because 2023 = 7 x 17 x 17), and so we get that x-1 = 1 & y-1 = 2023 or x-1 = 7 & y-1 = 289 or x-1 = 17 & y-1 = 119, leading to

(x, y) = (2, 2024) or (8, 290) or (18, 120). Those are our 3 distinct solutions β₯

Naturally, x & y are interchangeable, but in the context of this number puzzle we wouldn’t count 2024 & 2 (for example) as being a different pair from 2 & 2024.

A Good Book Puzzle

The number of different possible selections of two books from 66 distinct books is 66 x 65 Γ· 2 = 2145, which is nice here in blog post #145 π

A Champion’s Challenge

We saw in the first puzzle that 2023 = 119 x 17, and having that target in mind can help us to use 3, 3, 1, 14, 19, 21 to easily make 2023 as follows:

((19 + 21) x 3 – 1) x (3 + 14) πβ₯π

A Neat Word Puzzle

Listen = Silent = Tinsel = Enlist = Inlets β₯

In the chess puzzle, if Black’s invisible knight is on f4, then 1…Ne2# delivers checkmate instantly!

Alternatively, if the invisible knight is on f2, then Black wins beautifully with 1…Rh1+! 2 Bxh1 Nh3+ 3 Kh2 Qe2+! and then either 4 Bg2 Qxg2# or 4 Kxh3 Qh5#, a very pretty checkmate! πβ₯π

OUR ANGLE Brainteaser

Note: In all of the following steps of working, the angle sizes are in degrees.

Step 1: Angle NRG = 180 – (2x + y) & angle NGR = 180 – (2y + x).

Step 2: Angle RNG = 180 – angle NRG – angle NGR; after now using results from Step 1 above, and simplifying the algebraic terms, we get that angle RNG = 3x + 3y – 180 or, in factorised form, angle RNG = 3(x + y – 60).

Step 3: From the result of Step 2 above, we can conclude that x + y must be greater than 60.

Step 4: Angle ARG = 2x & angle AGR = 2y, so angle RAG = 180 – 2x – 2y, which can also be written as 180 – 2(x + y).

Step 5: Since x + y is greater than 60 (from Step 3 above), angle RAG must be less than 180 – 2 (60); so angle RAG must be less than 60 degrees. That would normally be our final conclusion… However…

Step 6: Since we were given that x and y are positive whole numbers, then the minimum possible value for x + y is 61, to be greater than 60.

Therefore, when x & y are whole numbers,

the maximum possible measure for angle RAG is 180 – 2 (61) = 180 – 122 = 58Β°.

(It’s also worth noting that, when angle RAG = 58Β°, angle RNG = 3Β°.)

Very warm congratulations to everyone who enjoyed trying and solving some or all of the puzzles πβ₯π

## Blog Post #144: A Taste of Heaven β₯

When this day (19 December 2022) began, there were exactly six days = 144 hours until the start of Christmas Day β₯

By midday at the centre or heart of each new day, God has granted us another 144 super-precious 5-minute gifts of time (as 144 x 5β720 minutes = 12 hours), and it’s good to aim to use all of them to thank and honour Him in everything that we do β₯

I found today that ABCDF contains more than one interesting idea. Thinking of the product of the numbers that correspond to the normal positions of the letters, 1 x 2 x 3 x 4 x 6 = 144, right here in Blog Post #144. I count myself and everybody as being the missing ‘E’ for ‘EVERYONE’… I know that we will all go on to G, H, I, J, K, L…with God in Heaven provided that we believe gratefully In Jesus, King of Love β₯

We could note very briefly in passing that Lβ12, the (square) root of 144, but it’s really much more important to recognise properly that God’s Love for all of us is at the root of our salvation.

As many of my students–and my colleagues–like puzzles, I’ll offer several now for everyone’s enjoyment π

Word Puzzles

Rearrange the letters of NEAR GIANT to make the name of a 9-letter country.

Rearrange the letters of MR TALENTS IN U to make a proper 11-letter word that all the students of Musica Mundi School use πβ₯π

Numbers Puzzle

Special Puzzle in honour of my colleague Jens Van Steerteghem

JENS is a brilliant physicist, chemist & mathematician, too. Suppose that J=10, E=5, N=14 & S=19. Now here comes a super-fun puzzle…Add up the values of any two or any three of the four letters, then multiply by the left-over total, and divide by 4 in honour of JENS.

What is the maximum possible final result?

It’s a delightful wee Christmas present to Jens and everyone who loves mathematical puzzles, and in fact it’s possible to figure out the answer mentally using nice logic πβ₯π

Puzzle Regarding A Lovely Lady (use A=1, B=2,…,Z=26 in this puzzle)

A former colleague of mine from my previous school in Belgium sent me a lovely Christmas card by email. In the lady’s six-letter first name, there’s no A. The product of the values of a certain four of the letters is 100. The product of the values of a certain five of the letters equals 5 x 144.

What is the exact product of all six letters? π

BπNUS: Can you make a smart conjecture to guess the lady’s well-known first name, that many ladies have had?

A Really Beautiful Brainteaser β₯πβ₯ (use A=1, B=2,…,Z=26 in this brainteaser)

RaphaΓ«l writes a proper six-letter English word that uses 5 distinct letters, including an R. The product of the values of all the letters in the word uses 4 distinct digits and begins with 5703…

Your mega-fun brainteaser is to figure out RaphaΓ«l’s six-letter word π

Chess Game in honour of my colleague Emile Daems

Emile Daems–a great colleague of mine at Musica Mundi School in Waterloo, Belgium–has been enjoying discussing some famous Chess openings with me. So, I’m including a photo of a recent game for Emile and other fans of The Royal Game β₯

A Whisky Puzzle in honour of my friends ‘Happy’ & Mandi!! ππ

Many people from Scotland and elsewhere enjoy some whisky! The word ‘whisky’ also brings ‘malt’ to mind, or the adjective ‘malty’. Here’s one description that I came across: “Generally speaking, a malt taste can be described as having a combination of flavours. It tastes sweet and nutty, but is also described as tasting similar to toast, caramel, coffee or fruits like raisins. The reason for its sweet, almost dessert-like taste has to do with how malt is made from barley.”

Here’s the puzzle now: Use all the letters of MALTY + H to make the name of a six-letter town where my friends ‘Happy’ & Mandi live, in the English county of Lancashire π

A Maths Mega-Brainteaser is coming next, in honour of my students and my brilliant colleague Jens & his ingenious brother Nick Van Steerteghem (who wrote a computer program specifically to solve another recent brainteaser!–special congratulations also to RaphaΓ«l Murphy who solved the brainteaser directly himself!!)

Part 1

Imagine that I put the numbers 1, 2, 3, 4, 5, 6 & 7 (one of each) in a bag.

RaphaΓ«l, Tarik and Wout each take a number out of the bag.

Damla, Sophie and Jens also each take a number out of the bag.

I then announce, “The total sum of RaphaΓ«l’s, Tarik’s and Wout’s numbers is exactly equal to the total sum of Damla’s, Sophie’s and Jens’ numbers! Furthermore, the total sum of the squares of RaphaΓ«l’s, Tarik’s and Wout’s numbers is exactly equal to the total sum of the squares of Damla’s, Sophie’s and Jens’ numbers!!”

Your fun brainteaser is to figure out exactly which number is still in the bag.

Part 2 (Super-Tricky!!)

A large group of Maths fans goes with me to visit an old Spiritual Maths monk at a monastery on a high hill. The number on the monastery building is a proper three-digit number. When we get to the door and see the number, I tell everyone that the sum of the squares of its digits equals the monk’s age!

The monk’s age is also equal to the sum of the fourth powers of three distinct positive whole numbers (just meaning different from each other).

Your mega-challenge brainteaser is in three parts:-

2.1: How old is the monk?

2.2: What is the smallest possible three-digit number that could be on the monastery building, given the clues above?

2.3: What are the fourteen different possible three-digit numbers that could be on the monastery building, consistent with the clues? (Note: That’s fourteen answers including the answer to part 2.2.)

Note also that the Spiritual Maths monk is not hundreds of years old like Yoda!! The monk is at an age that many other people have reached, too, in human history. Of course, I didn’t strictly need to mention that. Even if the monk’s current age could have been, say, 144–or 22 years more than the oldest person on record to-date π!!–the first clue relating his age to the sum of the squares of the digits on the three-digit door number meant that he couldn’t be more than 9 squared + 9 squared + 9 squared, or 243 years old now!! Still, my sincere answer to the question, “What could be better than living to be 144+99 years old?” is: “Living forever in Heaven.”

It is my intention to publish full solutions (God-willing, as always) to Blog Posts #142 & 144 before 9 January 2023, when the next semester’s lessons at Musica Mundi School begin.

In the meantime, dear students, colleagues and other readers, please do feel free to send in your best solutions to any or all of the puzzles, if you like β₯

The school’s cats visit the Maths classroom frequently, and they’re already pondering the fresh puzzles now!! πππ

My family and I would like to wish you and everyone a very blessed, merry Christmas soon, followed by a wonderful, happy New Year β₯β₯β₯

With kindest wishes as always,

Paul Mπtwani β₯

Meanwhile, here’s wishing Headmaster Herman a very happy birthday tomorrow!

Certainly let any nice school cat add the missing ‘E’ (from much earlier in the article), because SCHOOL CAT + E = CHOCOLATES

for everyone…

…and add comes from 144 π

P.S. = Puzzle Solutions (being posted on 31.12.2022)

NEAR GIANT = ARGENTINA

MR TALENTS IN U = INSTRUMENTAL

Using the numbers on Cleo’s sweater, (6×6)(1+3) = 36 x 4 = 144

In the puzzle about JENS, since J=10, E=5, N=14 & S=19 have a total sum of 48, we can achieve the optimal result with (10+14)(5+19)Γ·4 = 24 x 24Γ·4 = 144 again! ππ

In the puzzle about the lady’s six-letter name, the product of all her letter values = 3600, the LCM of 720 (= 5 x 144) & 100

Her name is DEBBIE (with 4 x 5 x 2 x 2 x 9 x 5 = 3600) π

RaphaΓ«l’s six-letter word is MEMORY, for which the product of the letter values is 13 x 5 x 13 x 15 x 18 x 25 = 5703750

(There were some cases to check, but since we were given that the word included an R–with letter value 18–the overall product had to be an even number and also a multiple of 9, and so in this particular puzzle the last digit had to be 0 & the sum of all the digits had to be a multiple of 9, which helped enormously to narrow down the cases for checking π)

MALTY + H = LYTHAM, where ‘Happy’ & Mandi live ππ

In part 1 of the Maths Mega-Brainteaser, 2+3+7=1+5+6 and, crucially, 2 squared + 3 squared + 7 squared = 1 squared + 5 squared + 6 squared; the unused number 4 is left in the bag π

In part 2, we don’t have to consider 4 to the power of 4 because that’s 256, which is a bit too old!! Instead, 1 to the power of 4 + 2 to the power of 4 + 3 to the power of 4 = 1 + 16 + 81 = 98, the monk’s true age β₯

Now we know that 98 = the sum of the squares of the digits of the monastery’s three-digit door number, which could be any of:-

149, 194, 358, 385, 419, 491, 538, 583, 707, 770, 835, 853, 914, 941

since 1 squared + 4 squared + 9 squared = 3 squared + 5 squared + 8 squared = 7 squared + 0 squared + 7 squared = 98 in those cases or permutations of them; clearly 149 would be the smallest possible proper three-digit door number meeting the monk’s requirements! π

## Blog Post #143: Infinitely Precious β₯

Dear Colleagues, Students, Friends and All Readers,

Whether we already know each other, or not, my wish this Christmas for every human being is that each person will know that he/she is infinitely precious to God, our Creator.

Even if some of you might be younger than me, God thought of you (and me) long before I was born, way back on 13 June 1962. For simplicity here, if I were to count that official date on my birth certificate as my day #1, then today, 16 December 2022, is my day #22102. It’s a nice day on which to reflect back on 2022 and make sure that God is #1 at the centre–at the heart of my life–today and always.

Going along the cobbled Rue de la Croix that leads to the stunningly beautiful Musica Mundi School (MMS, where I am privileged to work as the Mathematics teacher with lovely students and colleagues) in Waterloo, Belgium, I paused yesterday morning to take some photos, because the breathtaking views of the land, the school, and the sky had God’s perfect stamp everywhere. Of course some things are man-made using gifts and talents that we have been given, but it’s good to recognise the true Creator.

Now, what about all the lovely people of the school community…? We currently have 40 students and around 43 staff members (including instrumental, theoretical, conducting and choir Music teachers, Academic subject teachers, and staff for Administration, Operations & Student Welfare). We surely have at least 100 other very good people to also thank for their wonderful support to the school β₯

I also had many other ideas in mind when, just a couple of days ago, I sent out the fun, open-ended, thought-provoking question, why do I write that “MATHS IS 183“?

Here in Blog Post #143, it’s fitting that 1 funny ‘Big Brother’ + 4 other teachers + 3 students will be starring with the creative answers that they sent in before I started writing this article πβ₯π. Further down, I’ll offer my personal first idea, and some other still quite nice ones.

This kind of Maths can be very refreshing, because practically any idea from thoughtful children or adults is interesting to hear and to reflect on. There is certainly not just one ‘right answer’; the truth is that the good possibilities are infinite!

Guillaume is the school’s youngest current student, yet he was thinking until late yesterday evening, when he sent me his ‘answer’… Guillaume wanted MATHS to become a product M x A x T x H x S, with 183 as the target result. He realised very maturely that he could, for example, freely pick any non-zero number values for four of the five letters in MATHS, then calculate their product, after which the value of the one remaining letter would have to be 183 Γ· the calculated product. Guillaume liked the choice of starting with the whole numbers 1, 2, 3 and 5, after which the one other number value has to be 6.1, and it’s not far off being a whole number. His idea would work with an infinite number of different target numbers. Good thinking, Guillaume! Well done! π

Sophie noticed that the total number of characters in MATHS IS 183 is 5 + 2 + 3, or 10, and 3+8-1 also makes 10. Sophie’s idea there would work with lots of other numbers, too. Good job! π

Defne G. focused on 183 and wondered, had I turned 18 when I chose 3 as my favourite number?! I absolutely loved that thought πβ₯π, and Defne was the only person to write in with the idea of splitting 183 into a two-digit number and a one-digit number. Especially at Christmas, I think of Joseph & Mary (two) doing their best to care for baby Jesus (the One to actually save everyone). You may ask, “Why particularly 18 and 3?”… An answer is coming soon…but apart from 183 in Maths, the Bible verse Matthew 18:3 is super-important: Jesus called a little child to His side and set him on his feet in the middle of them all. βBelieve me,β He said, βunless you change your whole outlook and become like little children, you will never enter the Kingdom of Heaven. It is the man who can be as humble as this little child who is greatest in the Kingdom of Heaven.”

My ‘Big Brother’ Jan V.L. knows that 3 is my absolute favourite number, and Jens (another excellent colleague) remembered that I’m due to turn 61 on my next birthday. 183=3 x 61. Also, 3+6+1=10 & 1+0=1β”MATHS IS NUMBER ONE”! π (Jens is actually a brilliant physicist, chemist and mathematician. We’ll hear of him again when I publish my solutions to Blog Post #142 in the coming days. Also, I’m intending to feature a beautiful brainteaser within Blog Post #144, soon π)

The Headmistress and Kate and David (also lovely colleagues at MMS) had essentially the same idea as each other, and the following words are from Kate:

“If you add the number of the position of each letter of the alphabet of the word MATHS together it equals 61. Mr. Mo is our MATHS teacher and his favourite number is 3. If you multiply 61 by 3, you get 183.”

(13 + 1 + 20 + 8 + 19) x 3 = 61 x 3 = 183.

Those are certainly fun ideas, but several of the adults quite rightly felt that there ought to be more than that to MATHS IS 183. After all, Galileo famously said, “Mathematics is the language with which God created the Universe.”

183 is a bit special because it’s actually the smallest positive whole number such that it, when concatenated with its next big brother 184, makes a perfect square number: 183184 = 428 squared. That’s a cute detail too, but still, is it really of significance at Christmas?

David (already mentioned above) was the very first to write in, and with two distinctly different ideas. His second one is particularly nice β₯β₯β₯

1 represents the Oneness of God

8 represents the Infiniteness of God

3–in Music, the ‘musical third’ is the interval symbolising Love = God

Personally, my principal idea was the following… Remembering that Maths is actually a universal language, instead of replacing letters by numbers…we can alternatively replace numbers by letters. I told you that I really liked Defne’s thought of seeing 183 as 18 3. Then, MATHS IS 18 3 becomes MATHS IS R C…to wish you and everyone a very blessed, merry CHRISTMAS β₯β₯β₯

With lots of love and kindest wishes as always,

Paul Mπtwani β₯β₯β₯

P.S.

1. Everyone at the school who sent in a nice answer in time will receive a ‘thankyou’ prize on Monday π
2. All MMS colleagues and students are very warmly appreciated, and so there will actually be something nice for everyone there on Monday β₯
3. Thanks to Wout for one of his lovely new thoughts this evening, that MATHS can stand for Maths At The High School, and in reply I congratulate Wout on being a really great MMS of MMS = Marvelous Mathematical Student of Musica Mundi School! π

## Blog Post #142: Good Knowledge β₯

Theodore Roosevelt famously said, “A thorough knowledge of The Bible is worth more than a college education”, and I’m certain that he was right. Personally, I’ll always be truly grateful for all my excellent teachers of Mathematics, Physics, Chess, and many other fascinating subjects, but it’s still much more important to come to know God’s Word well through The Bible, which contains all the examples, truths, and commands that God wants us to know, believe, respect, and live by. In prayer, we can ask God to grant us increased understanding, wisdom and good powers of discernment. Mark 12:30 gives a clear, unmistakable message: “Love the Lord your God with all your heart and with all your soul and with all your mind and with all your strength.”

There are many people who want to have faith in God, but may be struggling with doubts. Three lines below here, there follows a free link directly to a really helpful recent article by Rev. John Piper (the founder of http://www.desiringgod.org)β

https://www.desiringgod.org/messages/the-purposes-of-god-in-the-pain-of-the-world

It’s well worth thinking about that article carefully. The good author has done a superb job of helping to answer questions which have troubled billions of people in the past, right up to the present time.

I am also thankful for every moment I have as the Mathematics teacher at Musica Mundi School (MMS), where I have the privilege and pleasure of working with wonderful students and colleagues. Since many of them enjoy puzzles, I will do my best to offer a bumper-sized gala feast now near Christmas! β₯πβ₯

First, rearrange the letters of now King to make a proper seven-letter English word.

Also, rearrange the letters of xy gala to make a proper six-letter English word.

Next, here comes a wee joke… What might someone say if he/she didn’t know the meaning of the word ‘AXON’ in Biology?…”It’s getting on my nerves!!” π

Thinking of miracles performed by Jesus, use the letters of GLORY HEAL AXON to make OR + a proper 11-letter Maths-related English adverb. There’s a clue in the lovely photo just below…

Just before we meet more puzzles, I’d like to share a happy, musical memory with you… One evening in the school a couple of years ago, I was sitting doing Maths work in the dining hall, while simultaneously enjoying listening to a beautiful music album. A Turkish student named Idil (a fabulous cellist) came into the hall and immediately asked about the lovely album. She liked it so much that I sent her a link to the full album, right away. I think that it’s perfect for Christmas, and so I’m sharing it now with you, too! πβ₯π

Duration: Just under 44 minutes. Happy listening!

I would also like to take this opportunity to wish all MMS students lots of enjoyment and success in all the concerts in which they are performing during this wonderful, festive period β₯

Meanwhile, more puzzles are coming in fast now!

Imagine that a lady once declared on a weekday, “I know that I was working 142 days ago because it was a weekday back then. My birthday is 142 days from now, and that will also be a weekday; not during a weekend.”

Your fun challenge puzzle is to figure out what day of the week the lady’s birthday was going to be on. Also, on what day of the week did the lady make her statement?

Next, can you think of seventy-one consecutive integers which have a total sum of 142 ?

Here comes another interesting puzzle… The total sum of all the prime numbers less than N is a whole multiple of 71. If the prime number N is included too, then the new total sum is also a proper multiple of 71. Exactly what number is N ?

And now I’ll share a very short, true story with you… Just a few days ago, while driving on my way to work, I found myself behind a vehicle with a number plate that included 2 β₯ 7 1, where β₯ stands for a particular digit. Given the value of β₯ that I saw, I realised that the four-digit number 2 β₯ 7 1 must be a proper multiple of 99. Your fun challenge is to figure out the value of β₯. Also, in general, if ABCD is a proper multiple of 99, then what is the special connection between AB and CD ? (One exception to the general rule is 9999, in which AB = 99 = CD.)

BRAINTEASER

Imagine that XYZ stands for a proper three-digit whole number.

1. Prove that XYZ times 3 can never equal YZX.
2. Find two solutions to the equation XYZ times 3 equals YZW, where the only restriction is that W does not equal X.

By the way, let’s call RaphaΓ«l’s favourite whole number R. It’s a single-digit number. Here in blog post #142, it’s nice that 1 Γ· R gives the recurring decimal 0.142857142857… Also, 2 Γ· R or 3 Γ· R or 4 Γ· R or 5 Γ· R or 6 Γ· R all produce decimals with exactly the same recurring digits, in different orders. If you’ve already figured out the value of R, then it will only take an extra moment to figure out the date this month of Headmaster Herman’s birthday, coming in R+1 days from now π

It’s my hope and intention (God-willing as always) to publish solutions to the puzzles around the time of the next blog post. In the meantime, please do feel free to send in your best solutions to some/all of the puzzles, if you like π

My family and I would like to wish you and everyone a very blessed, merry Christmas and a happy New Year, coming soon β₯πβ₯

Jesus said, “In My Father’s house are many rooms. If that were not so, I would have told you, because I am going there to prepare a place for you.” John 14:2

With love and kindest wishes as always,

Paul Mπtwani β₯

P.S. = Puzzle Solutions (being posted on 30.12.2022)

now King = Knowing

xy gala = galaxy

GLORY HEAL AXON = OR + HEXAGONALLY

2 x 71 = 142, the number of this blog post

142 days = 20 weeks + 2 days, so the lady was speaking on a Wednesday, looking forward to her birthday to come on a Friday (20 weeks + 2 days later), while also thinking back to a Monday when she was working, 20 weeks + 2 days earlier

The seventy-one consecutive integers from -33 to 37 have a total sum of 71 x the number at the very middle of the long list of numbers = 71 x 2 = 142, as required π

The prime number N = 71

β₯ = 8; 2871 = 99 x 29; note that 28 + 71 = 99, and in general AB + CD = 99 when ABCD is a multiple of 99 (except in the case of 9999)

Consideration of the place values of the digits in XYZ shows that XYZ = 100X + 10Y + Z, while YZX = 100Y + 10Z + X;

it’s not possible that 3(100X + 10Y + Z) = 100Y + 10Z + X because that would imply that 300X + 30Y + 3Z = 100Y + 10Z + X, leading to 299X = 70Y + 7Z, which is a clear contradiction since 70Y + 7Z = 7(10Y + Z), a multiple of 7, but 299 is not a multiple of 7 and if X=7 then 299X = 299 x 7 = 2093 which is far too big to equal 70Y + 7Z, given that digits Y & Z can’t be bigger than 9; so the proof is complete π

In part 2 of the brainteaser, the requirement boils down to

300X – W = 7(10Y + Z);

if X = 1 & W = 6, then 300X – W = 300 – 6 = 294 = 7 x 42, a multiple of 7, so we’ve found that it works with 3 x 142 = 426;

similarly, if X = 2 & W = 5, then 300X – W = 600 – 5 = 595 = 7 x 85, a multiple of 7, so we’ve also found that it works with 3 x 285 = 855;

there are no higher solutions, because if X = 3 or more, then 300X – W would be far too big to equal 7(10Y + Z), given that digits Y & Z can’t be bigger than 9 π

In the chess puzzle, White forces checkmate with 1 b6! cxb6 2 a7 followed by 3 a8=Q#, and ‘underpromoting’ to a rook at the end is also perfectly sufficient π

RaphaΓ«l’s favourite number is R=7

Headmaster Herman’s birthday came on December 20, which was 7+1 days after this blog post was first published (on December 12) β₯

## Blog Post #141: Christmas Celebration of Love β₯

Christmas (no less than any other time) offers you, me, and the whole world a perfect opportunity to recognise clearly, and celebrate with joy and thanks, God’s perfect love for us in having sent β₯ Jesus as the one true Saviour β₯ that we all need if we are to go to Heaven after our short lives now. In comparison to that all-important truth, I realise that anything that I might know about Mathematics, Chess, or other subjects, is practically insignificant.

Still, God purposely gave us all personal gifts to be used well, and so I will now offer some beautiful puzzles which I hope that many readers will enjoy solving.

The first puzzle (below) is one that I thought of just yesterday, and of all the puzzles that I have ever offered, it’s definitely one of my absolute favourites.

β₯ Christmas Love Celebration Brainteaser β₯

Your Christmas Love brainteaser challenge is this: If the total sum of all the fifteen number values (for C, H, R, I, S, T, M, A, L, O, V, E, and also 52, 31 and the red heart) equals 141, then what must be the exact number value of the red heart?

NOTE WELL: Each item anywhere above the bottom row of the pyramid always equals the sum of the two items just below it. For example, I = M + A, because M & A are just under the letter I there. Similarly, R = S + T, and so on.

The number 141 has specially appeared in my life on literally hundreds of occasions within my sixty years, so far, and I couldn’t help noticing that the total sum of the eleven numbers in the display column of the next photo (here in blog post #141) is again 141 π

β₯ A Second Lovely Puzzle β₯

From the display board, I’m thinking of two particular numbers. If I increase the numbers by 10 each, their product increases by 300. What are my two starting numbers?

The word ‘CARDS’ may be closely connected to ‘CARES’, and ‘REASON’ is certainly connected to ‘SEASON’ through the perfect truth given in the next photo β₯

β₯ A Wonderful Chess Study β₯

It’s now 22 days until Christmas 2022. Sometime before then, I hope and intend (God-willing, as always) to publish solutions to the puzzles within this post, but please do feel free to send me your best solutions to any/all of the puzzles, if you like π

Here’s a beautiful detail regarding the birthday tomorrow (4 December, 21 days before Christmas) of a lady I know, born in 1965… I shared with her that 21 x 1965 = 41265. We noted that, even I were just to write 21 x 19XY = D12XY, it can be proven that the day number D has to be 4 and the year has to be 1965; that’s the unique solution β₯

To everyone whose birthday is yet to come this month, I wish you a wonderful, happy time. Most of all, I wish everyone a very blessed Christmas soon.

With love and kindest wishes as always,

Paul Mπtwani β₯

P.S. = Puzzle Solutions (being published now on 12.12.2022)

The Addition Pyramid Brainteaser can actually be solved rather quickly by considering the sums of the entries, one row at a time

Top Row: 52

2nd Row: 52

(because C + H must exactly match the number 52 positioned right above the letters C & H)

3rd Row: 52 – β₯

(because R + β₯ + I = (R + β₯) + (β₯ + I) – β₯ = C + H – β₯ = 52 – β₯)

4th Row: 52 – β₯ – β₯

(because S + T + M + A = (S + T) + (M + A) = R + I = (R + β₯ + I) – β₯ = (52 – β₯) – β₯)

Bottom Row: 52 – β₯ – β₯ – β₯ + 31

(because L + O + 31 + V + E = (L + O) + (V + E) + 31 = S + A + 31 = (S + T + M + A) – (T + M) + 31 = (52 – β₯ – β₯) – β₯ + 31)

So, the grand total sum of all the rows is

52 + 52 + 52 – β₯ + 52 – β₯ – β₯ + 52 – β₯ – β₯ – β₯ + 31 = 141 (given information)

β(52 x 5) -6 x β₯ + 31 = 141

β260 – 6 x β₯ + 31 = 141

β291 – 6 x β₯ = 141

β6 x β₯ = 150

ββ₯ = 150 Γ· 6 = 25.

β₯ = 25, perfect for Christmas π

In the second puzzle, suppose that X & Y represent the two numbers that I was thinking of. Then, from the information given in the puzzle,

(X+10)*(Y+10)=X*Y+300

βX*Y+10*X+10*Y+100=X*Y+300

β10*X+10*Y+100=300 (after cancelling X*Y terms)

β10*X+10*Y=200

βX+Y=20 (after Γ· by 10)

The only numbers on the display board which could satisfy the requirement X+Y=20 were 9 & 11.

Warm congratulations to Jens Van Steerteghem, who solved that puzzle and the Addition Pyramid β₯ Brainteaser! πβ₯π

In the Chess study, White wins with 1 c4! (threatening 2 Qd5#) 1…Qe4 2 d4+!, after which either 2…cxd4 or 2…Qxd4 is met by 3 Qe7#.

Alternatively, 1 c4! Qh1 2 d4+! cxd4 3 Qe7#, or 2…Ke4 and Black’s queen will be lost after the skewering check 3 Qd5+.

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

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 Mπ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 πππ.

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 πππ.

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₯

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₯ππβ₯

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

With love and kindest wishes as always,

Paul Mπ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₯ππ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! π So, specially for Alicia and all Maths fans, here is a link to a further feast of wonderful puzzles: https://www.ukmt.org.uk/competitions/solo/junior-mathematical-challenge/archive

For all Chess enthusiasts-including Daniel Mathiesen from HΓ€rnΓΆsand, Sweden-I’m including a link to two of my favourite sites: http://www.chess.com/ccc & http://www.ideachess.com/chess_tactics_puzzles/checkmate_n/34288

(I couldn’t resist giving everyone yet another bπ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₯)

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! π

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₯π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.