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 Raphaël (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! 😊
Alexander, another great colleague, has a smaller, positive, favourite whole Number. Let’s call it N.
Brainteaser starring Peter and Alexander the Great! 😊😊
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! 😊♥😊
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!! 😊 (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! 😊
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!! 😊😊
I’m thinking right now about two positive whole numbers… Let’s call them A and B.
(A raised to the power B) ÷ (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 😊😊😊
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 y° 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.) 😊♥😊
Brainteaser with Happy Memories of my previous school in Belgium 😊
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 😊♥😊
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 😊.
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 ♥
With kindest wishes as always,
Paul M😊twani ♥
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 Cécile Gregoire and Jens Van Steerteghem who sent me kind messages regarding the articles and very good solutions to the puzzles. 😊😊
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 ÷ 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 😊
(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 ÷ 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 😊
Words with the same multiplicative value as PETER are REPEAT or RETAPE.
The morning clock time was 10.24, and that’s like 1024 ÷ 100, or (2 to the power of 10) ÷ (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)°.
(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! 😊♥😊)
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) ÷ 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) ÷ 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. 😊♥😊
Paul Motwani 