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!

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! ππ

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! πβ₯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!! ππ

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 πππ

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.) πβ₯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 πβ₯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 Mπtwani 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)

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 CDseach, 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! πβ₯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. πβ₯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:-

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 thenight 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!

The light of His face made me cover my head.

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;

Oh, if only I had been ready tonight.

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 πβ₯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 πβ₯π

CUBOID BRAINTEASER πβ₯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?

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 πβ₯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! πβ₯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 Mπtwani 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 βuse upβ two thirds of the 1, leaving only one third or 2/6. So W canβt 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

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! πβ₯π

Additional Notes:

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₯ππβ₯

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 Mπtwani 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 ππ

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) πβ₯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! πβ₯π

OUR ANGLE Brainteaser

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

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 πβ₯π