Dear Readers,

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)

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. ππ

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! 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