## Blog Post #157: Nice Prize Puzzles in Honour of AndrΓ©e!! ππππ

Dear All,

Last year, Jenny and I were delighted that our very dear friend, AndrΓ©e, was able to come from Luxembourg to celebrate my turning 60. Now we are really looking forward to going in two weeks’ time to join in lots of great fun for AndrΓ©e’s upcoming birthday! πππ

I decided to offer a fresh collection of prize puzzles in honour of AndrΓ©e! π

The people who send me the best answers to all of the puzzles will receive a nice gift βππ I would still warmly encourage you to not be shy to send in your solutions even if you might not be sure about some of them.

I look forward hopefully to giving lovely gifts to some people at AndrΓ©e’s birthday party, to some students and colleagues at Musica Mundi School, and to other readers who may send in very good solutions π

Solutions should be sent to pmotwani141@gmail.com by midnight (CET) on Friday 19 May, or as early as you like! I will personally let the winner(s) know about his/her/their prize(s), and I intend to also place a nice announcement along with complete solutions in the next blog post around the start of June, God-willing as always π

Prize Puzzle #1: Anagram starring Harriet & AndrΓ©e π

This puzzle also celebrates the fact that today is the birthday of Harriet, a lovely colleague at Musica Mundi School. Happy birthday, dear Harriet! Have a really wonderful time!! π

Can you read my mind and tell me exactly what is the missing six-letter word in the puzzle below?

HARRIET SENDS

ANDREE – – – – – –.

Harriet might say, “Why would I send those to AndrΓ©e?!” and Jenny could reply jokingly, “We’ve got Paul’s back! Let’s keep the word a secret…Sh….” π

Prize Puzzle #2: Happy House Number π

Not counting my own house number, 11, what is the smallest whole two-digit house number which can be multiplied by another whole number to produce a result that consists entirely of nines? Exactly what number would you multiply the house number by in order to get the string of nines result?

Prize Puzzle #3: Starring AndrΓ©e, Anthony & Zoe πππ

AndrΓ©e, Anthony & Zoe each write down a different whole number and then put the three numbers in order from smallest to largest. The average of the squares of the first two numbers exactly equals the third number. Here in blog post #157, it’s nice that the average of the squares of the second and third numbers equals 157.

What exactly are the three numbers that AndrΓ©e, Anthony & Zoe must have written down? (As an extra, optional challenge for Maths fans, enjoy proving algebraically that the solution is unique. π)

Prize Puzzle #4: Thinking of AndrΓ©e’s Party πΆππ

Thinking ahead to AndrΓ©e’s party, I’ve now got a particular English six-letter plural word in mind. If I remove the penultimate letter of the word, the remaining five-letter word is still a proper English word, and its last two letters are the same as each other.

I would love to hear your answers regarding what six-letter word you think I’m thinking of!! π (You can even send me more than one answer, if you want to, though I just have one particular word in mind now.)

Prize Puzzle #5: A Puzzle about most of Your Ages!! π

I’m now thinking of a very particular, mystery whole number, M. If I were to ask lots of people about their ages, and then divide each of them by M, most of the results would be decimal numbers containing the digits 1, 5, 7 in some order within the first six digits after the decimal point.

What exactly is my mystery number, M?

Well, dear friends, I wish you lots of fun with the puzzles. My first thought had been to give three…then a bonus to make it four…but it’s very fitting to have a high-five π in honour of AndrΓ©e, especially because 3 x 4 x 5 = 60 π

With lots of love always

From Jenny, Michael and me xxx

Paul π

“A sweet friendship refreshes the soul”–Bible verse, Proverbs 27:9 π

P.S. = Puzzle Solutions (being posted now on Saturday 20 May 2023)

What’s infinitely more important than the actual solutions are all the people who enjoyed trying the puzzles and who then took the time to write really nice, personal messages in which they communicated their answers, their enjoyment and appreciation πππ I loved reading your delightful messages, and so it’s with great pleasure and happiness now that I announce that all of the following people (mainly in alphabetical order) will be receiving a prize at AndrΓ©e’s party next Saturday, 27 May:-

π AndrΓ©e, Anna, Anthony, Chantal & John, Deborah, Hilary & Richard, Susan, Zoe π

During the coming week, I will also give a prize in person at Musica Mundi School to my terrific puzzle-solving colleague, Jens Van Steerteghem, and I’ll certainly include a gift for Jens’ father, Eric, who loves having fun with all the blog posts π

It’s time now to share the puzzle solutions:-

1. HARRIET SENDS is an anagram of ANDREE SHIRTS π
2. Various people noted that 33 x 3 = 99 and 27 x 37 = 999, but Anna and Jens discovered that 13 x 76923 = 999999. Therefore, not counting my own house number 11, AndrΓ©e’s house number 13 is the smallest whole two-digit number which can be multiplied by another whole number to produce a result consisting entirely of nines. No even house numbers could work, because all whole products involving them would also be even numbers (so they couldn’t ever equal 999…). Note that 999999 = 999 x 1001 which equals 999 x 7 x 11 x 13, and since 999 x 7 x 11 = 76923, that demonstrates one way of finding out that 13 x 76923 works βπ Here’s a fun extra-bonus challenge π: Not counting my own house number, 11, or AndrΓ©e’s house number, 13, what is the next smallest whole two-digit house number which can be multiplied by another whole number to produce a result that consists entirely of nines? Exactly what number would you multiply the house number by in order to get the string of nines result? π
3. Suppose that A, B & C are the positive whole numbers that AndrΓ©e, Anthony & Zoe wrote down, put in increasing order from smallest to largest. We were basically given that (A2+B2)/2=C & (B2+C2)/2=157. So, A2+B2=2C & B2+C2=314. Noting that B2=2C-A2βC2+2C-A2=314βC2 < 314βC β€ 17. Also, C2+2C+1-A2=315β C2+2C+1=315+A2β(C+1)2 Λ 315βC+1β₯ 18βCβ₯17. Having proved that C β€ 17 and Cβ₯17, the unique solution has to be C=17 exactly. Since A2=(C+1)2-315, we get A2=182-315=324-315=9βA=3. Recalling that B2=2C-A2 β B2=34-9=25βB=5. Summary: A=3, B=5, C=17 is the unique solution. Many congratulations indeed to everyone who found the numbers 3, 5 & 17. That puzzle was really well done πππ
4. “Thinking ahead to AndrΓ©e’s party, I have a particular English six-letter plural word in mind. If I remove the penultimate letter of the word, the remaining five-letter word is still a proper English word, and its last two letters are the same as each other.” I intended the last part to refer to the remaining five-letter word, but if it would be referring instead to the original six-letter word, then there’s a lovely possibility which Zoe thought of: the word CARESSβCARES. The word that I actually had in mind was GUESTSβGUESS, and Anna and Deborah both got it! ππ
5. My mystery number, M = 7. When whole numbers are divided by 7, the remainder is either 0, 1, 2, 3, 4, 5 or 6. The remainder is zero only when the dividend (the number being divided) is a whole multiple of 7. Otherwise, when the remainder is 1, 2, 3, 4, 5 or 6, the quotient involves the fractions 1/7, 2/7, 3/7, 4/7, 5/7 or 6/7. Their decimal forms are 0.142857…, 0.285714…, 0.428571…, 0.571428…, 0.714285…, 0.857142… which all involve 1, 5, 7 in some order and fit nicely here in Blog Post #157 π AndrΓ©e’s party is starting in 007 days from now…πππ

## Blog Post #156: You Are A Lovely, Truly Delightful Miracle!! ππ

Dear All,

“He will be a joy and a delight to you, and many will rejoice because of his birth.”–Bible verse, Luke 1:14 π

Given that we are children of God, created in His image, it’s lovely to get to know one another, and gently help each other to become ever more what our Creator made us to be. When we use well and share our God-given gifts for the benefit of others, there is no limit to the good that can be accomplished.

Staying ‘tuned in’ to the Holy Spirit and coming to know God’s word and His will through Bible study and prayer also helps us to understand more about ourselves and each other. When we then interact with gentle, loving patience and kindness, problems are often solved very happily with miraculous ease! ππ

Nice Word Puzzle π

Use all the letters of ARC SMILE to make a proper 8-letter English word.

There are (at least) three possible solutions!

I recently promised a friend that this article being offered freely here would include some good, fun number/number-theory puzzles. I wish you lots of enjoyment with them now. πβ

Power of Friends Brainteaser π

The ages of five special friends are consecutive whole numbers, represented here by A, B, C, D and E. Andy is the youngest, followed by Beth, Chris, Dee and Eric, the oldest. The five friends are staying together at a large holiday house until 31 May. The house number (on the front door) is represented here by H Λ 1.

A remarkable detail is the fact that 31 x 5 x (HA + HB + HC + HD) = HE β HA.

Your super-fun brainteaser is to figure out, with proof, the exact value of the house number H. π

New 2023 Brainteaser π

The 2023 brainteaser is to figure out, with proof, all possible positive whole number values for N such that N – 1 is a factor of N2 + 2023.

Double All-Play-All Chess Brainteaser (with no Chess moves!! πβπ€£)

Suppose that P players are having a double all-play-all Chess tournament, and P Λ 2. Each player will play against each other player twice: once with White, and once with Black. In each game, if there’s a winner, he/she gets 1 point and the loser 0 points; if the game ends in a draw, then the two players involved receive 0.5 points each. At the end of the tournament, each player’s final score will be the total sum of the points that he/she received in his/her games.

Suppose that the gold medal winner in 1st place actually achieves the very highest possible score π, and the silver medal winner in 2nd place achieves what is then the highest-possible runner-up score (lower than the winner’s score, though).

It turns out that the number obtained by dividing the winner’s score by the runner-up’s score does not involve more than one distinct, different digit when you write it out.

The brainteaser is to figure out, with proof, three possible values for P, the number of players. ππ

A Double High-Five Brainteaser!! ππ

Consider all non-negative whole numbers n with not more than D digits.

The brainteaser is to figure out (as an expression, with proof) how many of them are such that n10 + 1 is exactly divisible by 10.

Advanced Algebra Triple Brainteaser!!! πππ

Suppose that ‘a’ and ‘b’ are distinct non-zero numbers such that

a2 + 2ab – 3b2 = p & 2a2 – 3ab + b2 = q.

The triple challenge is to find a neat, simplified expression for ‘a’ in terms of p, q & b AND to discover three values for p Γ· q which are NOT possible in this brainteaser AND to find a proper, mathematical, English word which contains a, b, p and q (not necessarily in that order) at least once each among its letters!!!

Chess Puzzle (with moves this time!! ππ)

It’s my intention to publish solutions to all the puzzles around the time that blog post #157 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 by most sincerely wishing you a very blessed month of May and a wonderful weekend now, with lots of happiness in everything that you do β€.

Beautiful Bonus! πππ

Extra-special wishes to lots of people I know who are celebrating their birthdays today! You can write down any proper three-digit whole number that you like (e.g. 523). Repeat it to get a six-digit number (e.g. 523523). Divide by 2 in honour of our friendship. Now divide by 7, the number of letters in FRIENDS. Also divide by 13, the total number of letters in HAPPY BIRTHDAY. To finish, divide by the three-digit number that you started with…

The final result of…π 5.5…is to wish you a beautiful birthday now on 5 May πππ

With kindest wishes as always,

Paul Mπtwani β€

“For shoes, put on the peace that comes from the Good News, so that you will be fully prepared.”–Bible verse, Ephesians 6:15 π

P.S. = Puzzle Solutions (being posted on 12.5.2023)

ARC SMILE rearranges to make MIRACLES (and you could claim that it makes CLAIMERS or RECLAIMS too!! π).

In the big H brainteaser, we were given that

31 x 5 x (HA + HB + HC + HD) = HE β HA

where A, B, C, D & E are consecutive whole numbers in increasing order, and so

155 x HA x (1+H+H2+H3) = HA x (H4 – 1)

and ‘cancelling’ the common factor HA on both sides of the equation gives

155 (1+H+H2+H3) = (H4 – 1)

and then factorising the expression on the right-hand side leads to

155 (1+H+H2+H3) = (H-1) (1+H+H2+H3)

which simplifies to

155 = H-1

so H = 156, which is nice here in blog post #156 π

In the ‘2023 Brainteaser’, we were given that

N – 1 is a factor of N2 + 2023, and since N2 + 2023 = (N-1)(N+1)+2024,

N-1 must also be a factor of 2024β1, 2, 4, 8, 11, 22, 23, 44, 46, 88, 92, 184, 253, 506, 1012 or 2024 for N-1,

and so

N can be any of the sixteen numbers 2, 3, 5, 9, 12, 23, 24, 45, 47, 89, 93, 185, 254, 507, 1013 or 2025.

In the Double All-Play-All Chess Tournament with P players,

the maximum-possible score (for someone who wins every single game) is 2(P-1), and in that case the maximum-possible runner-up score (for someone who wins every single game apart from the two games lost against the super champion!) is 2(P-2).

The winner’s score divided by the runner-up’s score simplifies to (P-1)/(P-2).

There are three possible values of P for which (P-1)/(P-2) produces positive results that in each case don’t require more than one distinct, different digit when written out. They are P = 3 or 11 or 12, and the corresponding (P-1)/(P-2) values are 2, 1.111… repeated and 1.1 exactly.

In the Double High-Five Brainteaser, n10 + 1 is exactly divisible by 10

βn10 + 1 ends with a digit 0βn10 ends with a 9βn5 ends with a 3 or a 7βn itself ends with a 3 or a 7; that is precisely two out of each ten numbers ending with 0 to 9, because only the ones ending in 3 or 7 actually work here. Now, the total number of non-zero whole numbers with not more than D digits is 10D. So, the total number of suitable numbers for n is 10D x 2 Γ· 10,

which simplifies to 2 x 10D-1 πβ

In the advanced Advanced Algebra Triple Brainteaser, where

‘a’ and ‘b’ are distinct non-zero numbers such that

a2 + 2ab – 3b2 = p & 2a2 – 3ab + b2 = q,

factorisation leads to

(a-b)(a+3b) = p & (a-b)(2a-b) = q

so p/q = (a+3b)/(2a-b). Since a is not equal to zero, p/q is not equal to -3; since b is not equal to zero, p/q is not equal to 1/2; since a is not equal to b, p/q is not equal to 4. In summary, p/q cannot equal -3, 1/2 or 4.

Also, the equation p/q = (a+3b)/(2a-b) can be rearranged to make ‘a’ the subject. We get a = (p+3q)b/(2p-q). That equation helps to confirm again that, since a is not zero, p+3q is not zero, so p/q is not -3. Similarly, 2p-q can’t be zero, so p/q can’t equal 1/2. Also, a is not allowed to equal b in this puzzle, so (p+3q)/(2p-q) can’t equal 1βp+3q β  2p-qβ4q β  pβp/q β  4.

Mathematical words containing the letters a, b, p and q include equiprobable and equiprobability.

In the Chess puzzle,

White wins beautifully with 1 Qxh6+!! Kxh6 2 Rh3+ Kg5 (2…Kg7 3 Rh7#) 3 f4+! Kxf4 (or 3…Kf5 4 g4+ Kxf4 5 Rf1+ Kg5 6 Rh7 and 7 h4#) 4 Rf1+ Kg5 5 Rh7! followed by 6 h4#! π

## Blog Post #155: Fun Birthday Surprises for Jens, just 3 x 3 x 3 x 3 + 4 days early! ππππ

Dear All,

I’m not just 3 or 4 days early with surprises here…it’s actually 34 + 4 days in advance of the 12 July birthday of Jens Van Steerteghem, a brilliant colleague who loves solving fun brainteasers, as do many of our students at Musica Mundi School πΆπ

Just yesterday evening, a special idea which involves “Jens’ 12/7″π came to me, and so I decided happily to share this surprise π super-early now!! π It was tempting to wait until tomorrow and then be precisely 12 x 7 days before 12 July…but instead we’re (12 – 7 + 12) x (12 – 7) days early! π

OK, here we go… There’s literally an infinite choice for pairs of numbers a and b such that a2 β b2 = 12/7. Imagine that I pick such a pair, and I whisper the numbers to Eric and Martine (Jens’ lovely parents) ππ.

To turn up the sneaky-level button π, Eric then calculates a2 + 2ab – 3b2 and Martine calculates 2a2 – 3ab + b2.

Next, Nick (Jens’ brother) multiplies Eric’s and Martine’s results together.

The Advanced Algebra Birthday Brainteaser for Jens (and other great solvers!) to figure out is this: what is the most extreme final result that Nick could get?

In other words, the ‘most extreme’ possibility is either a minimum value or a maximum value. The fun challenge is to determine which of those two is correct, and also to figure out the actual extreme value.

To make the challenge all the more interesting and surprising, it can be completely solved without using Calculus at all! So, it really is an Advanced Algebra brainteaser coming far in advance of Jens’ 12/7 birthday!

Compared to that one, a very wee snack bonus puzzle, here in Blog Post #155, is to use all six of the numbers 3, 4, 3, 4, 7 & 12 to make 155. Parentheses ( ) and operations involving +, -, x, Γ· can be used freely, as you wish.

Another quick one is: how can the digits within 12 and 7 combine in a somewhat different type of calculation to get the result 127?

And here’s quite a nice puzzle: how many whole numbers from 1 to 1000 inclusive do not contain Jens’ favourite digit 7?

The following neat brainteaser gives me happy memories… π How old was I when I could then truthfully say (all in relation to that time), “The product of my age n years ago and my age n years from now is 1207”? π

Joke Time! π

With apologies to lovely lady Martine, why should men be good at solving the earlier algebra brainteaser?!

Answer: Martine Eric & Nick (MEN) were starring in it! π€£

To make up for that one…I’ve got another algebra brainteaser for Jens’ family and everyone to enjoy!

It’s my intention to publish solutions to all the puzzles around the time that blog post #156 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 week, with lots of happiness in everything that you do β€.

With kindest wishes as always,

Paul Mπtwani β€

βI am the vine; you are the branches. If you remain in me and I in you, you will bear much fruit; apart from me you can do nothing.”–Bible verse, John 15:5

P.S. = Puzzle Solutions (being posted on 12.5.2023)

Let a2 + 2ab – 3b2 = p and 2a2 – 3ab + b2 = q. Then 3p + 2q = 3(a2 + 2ab – 3b2) + 2(2a2 – 3ab + b2), which simplifies to 7a2 – 7b2 or 7(a2 – b2). We were given that a2 – b2 = 12/7, and so 7(a2 – b2) = 7*12/7, which is 12. We’ve proved that 3p + 2q must have the constant value 12. That is, 3p + 2q = 12 is absolutely forced, given that a2 – b2 = 12/7.

So, q = (12 – 3p)/2, and the product pq = p(12-3p)/2 or -1.5p2 + 6p. Using standard facts regarding quadratic expressions, the graph of the function f(p) = -1.5p2 + 6p would be a parabola with a maximum value when p = -6/(2*-1.5)βp = 2. The actual maximum value of the function is then -1.5*22 + 6*2β-6 + 12β6 is the maximum-possible value π

One may ask, how can we be sure that it’s really possible to have p = 2 and q = 3 in order to achieve pq = 6, the claimed maximum-possible value of pq when a2 – b2 = 12/7 ? Do suitable values exist for ‘a’ and for ‘b’ which would get us to p = 2 and q = 3 ? Well, if you check out the solutions to the next blog post (#156), you’ll see a proof showing that a = (p+3q)b/(2p-q). Requiring p = 2 & q = 3 also then requires that a = 11b. So, (11b)2 – b2 = 12/7β121b2 – b2 = 12/7β120b2 = 12/7βb2 = 1/70βb = 1/β(70) or -1/β(70) and, since a = 11b, we find that a = 11/β(70) or -11/β(70) πβ

Here in Blog Post #155, we were invited to use all six of the numbers 3, 4, 3, 4, 7 & 12 to make 155. Parentheses ( ) and operations involving +, -, x, Γ· could be used freely, wherever we wish. One solution is (3+3+7) x 12 – (4Γ·4) = 155. Note that it’s also perfectly sufficient to write (3+3+7) x 12 – 4Γ·4 = 155 if we’re familiar with the standard, internationally-accepted rules regarding ‘Order of Operations’ πβ

In the next puzzle, 27 – 1 = 127 π

Just as there are one thousand whole numbers from 1 up to and including 1000, there’s also one thousand such possibilities from 000 to 999. However, if we exclude numbers containing any 7, then for each of the three – – – place-value positions, we only have nine (instead of ten) options (namely 0, 1, 2, 3, 4, 5, 6, 8, 9 but not 7), and so the total number of such possibilities is 9 x 9 x 9β729 π

How old was I when I could then truthfully say (all in relation to that time), “The product of my age n years ago and my age n years from now is 1207”? Note that 1207 = 17 x 71, and the number mid-way between 17 & 71 is (17+71)/2 which is 44. Note also that 44 – 27 = 17 & 44 + 27 =71. So, I was 44 when I could truthfully say, “The product of my age 27 years ago and my age 27 years from now is 1207.” π

We use the facts that each interior angle of a regular hexagon, regular pentagon and a square is 120Β°, 108Β°, 90Β° respectively. We also make repeated use of the fact that the sum of the angles in a triangle is 180Β°. So, the angles immediately to the left and to the right of the top vertex of the pentagon are (60-a)Β° and (60-b)Β° respectively, with a 108Β° angle between them. Therefore, (60-a)Β° + (60-b)Β° + 108Β° = 180Β°βa + b = 48.

Similarly, the angles just above the top edge of the square are (18+c)Β° & (18+d)Β°. So, (18+c)Β° + (18+d)Β° + 108Β° = 180Β°βc + d = 36.

Therefore, a + b + c + d = 48 + 36 β84.

Also, the maximum-possible value for the product a*b*c*d is 24*24*18*18β186624 π

White wins beautifully with 1 Qh6+!! Kxh6 2 Nf5++ Kg6 3 Rh6# or via the similar 1…Kh8 2 Ng6+ & 3 Qxh7# or 2 Qxh7+ Kxh7 3 Nf5+ Kg6 4 Rh6# π

## Blog Post #154: Happy Easter, Celebrating Jesus’ Triumph Over Death β₯

Dear All,

Here in Blog Post #154, with the coming (in just a few days’ time) of Easter–which celebrates Jesus Christ’s triumph over death, and the gift of eternal life to faithful followers–a perfect Bible verse to recognise properly with gratitude and joy is: “Then, when our dying bodies have been transformed into bodies that will never die, this Scripture will be fulfilled: Death is swallowed up by triumphant Life.”–1 Corinthians 15:54 β₯

The above truly is infinitely more important than anything at all relating to Chess or Mathematics, for example, but still I’m now going to happily share a feast of nice, fresh puzzles ππ

FUN BRAINTEASER π

A girl makes a lovely cuboid-shaped gift box for her father for Easter. The total outer surface area of the closed box is 386 cm2. Its dimensions (length, width and height) in centimetres are each different positive whole numbers, which are also the ages (in years) of the girl and her two younger siblings.

Your fun brainteaser is to figure out all of their ages. The puzzle can be solved using neat algebra πβ

A 2023 PUZZLE π

If S is the smallest positive whole number for which the sum of the digits equals 2023, then what is the sum of the digits of S+1 ?

A PIECE OF CAKE BRAINTEASER! π

On his birthday next year, in 2024, Nathan will be N years old. He looks forward to then having a cake with N candles on it. Imagine making triangles by connecting every possible distinct trio of three candles on the cake. Then there would be precisely 364 different triangles, to match the days in 2024 ! π

The brainteaser is to figure out the value of N.

JOKE π

What did the Chinese philosopher Li Ping say about the other two days…?

“Did you miss today’s joke?”

but I thought she said, “Did you miss two days…joke?!!” π

AS EASY AS ABC…! π

Which three-digit positive whole number ABC is exactly equal to AA + BB + CC? Can you prove that the solution is unique?

A THREE-PEOPLE PUZZLE FROM 11 YEARS AGO πππ

I am six years older than Jenny, and she is thirty years older than Michael. How old will Michael be when (in future, God-willing) the total of our three ages will be as close as possible to 154?

NUMBER WORD PUZZLE π

The number-word ONE HUNDRED AND FIFTY-FOUR contains all the five official English alphabet vowels, A, E, I, O, U. What is the smallest positive whole number for which its number-word contains all the vowels A, E, I, O, U exactly once each? (Hint: The correct number is greater than 154.)

FAST EASTER WORD PUZZLE π

Rearrange the letters of EASTER to make a different, proper six-letter English word that does not begin with E.

AN ELEGANT CHESS PUZZLE β

It’s my intention to publish solutions to all the puzzles around the time that blog post #155 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 month of April, with lots of happiness in everything that you do β€.

With kindest wishes as always,

Paul Mπtwani β€

P.S. = Puzzle Solutions (being posted on 19.4.2023)

In the cuboid brainteaser, 2(LW+LH+WH)=386βLW+LH+WH=193. So, 193+H2 = LW+LH+WH+H2 = L(W+H)+H(W+H) = (L+H)(W+H), a neat and noteworthy algebraic trick πβ

Now trying H=1: 194 = (L+1)(W+1)β2 x 97 = (L+1)(W+1), but then L = 1 and W = 96 (or vice-versa), and that’s not OK because H & L should be different (like the siblings’ ages), and also none of them should be 96 in this puzzle!!

Trying H=2: 197 is a prime number; it can’t equal (L+2)(W+2), since L & W have to be positive. (We can’t allow L+2=1 & W+2=197, for example, because we’d get L=-1.)

Trying H=3: 202 = 2 x 101, which can’t equal (L+3)(W+3) in any suitable way.

Trying H=4: 209 = 11 x 19 = (L+4)(W+4) with L=7 & W=15, for example.

So, a solution for the siblings’ ages is 4, 7 and 15, with the ‘main’ girl in the puzzle being 15 (the oldest one).

If we were to investigate further, we wouldn’t find any truly different solutions, though of course H=7 or H=15 are possible, and would lead to L & W being 4 & 15 or 4 & 7. We still end up with the same set of dimensions (and siblings’ ages): 4, 7 & definitely 15 for the girl π

The ‘2023’ puzzle can be solved nicely like this: since 2016 is a multiple of 9 (224 x 9=2016) and 2023 = 7 + 2016, then the number S = 7999…999 (starting with a 7 followed by 224 nines) will be the smallest number–using the least-possible number of digits–whose digits sum equals 2023. Therefore, S+1 = 8000…000 (starting with an 8 followed by 224 zeroes), and so its digits sum is simply 8. π

When there are N candles on Nathan’s cake, the number of different trios of candles that we could connect to get triangles is N x (N-1) x (N-2) Γ· 6. (We divide by 6, because we don’t want to miscount any same trio 6 times just because that same trio could be listed in 6 different orders.) We were basically given that N x (N-1) x (N-2) Γ· 6 = 364, and so N x (N-1) x (N-2) = 2184. We observe that N-1 should be pretty close to the cube root of 2184, which is 12.974…, very close to 13. That suggests N = 14, and indeed 14 x 13 x 12 = 2184. So, it’s confirmed: N = 14. π

In the ‘ABC’ puzzle, ABC = 198, the unique solution π

Proof: Given ABC = AA + BB + CC in this particular puzzle, then consideration of the true ‘place values’ of the digits leads toβ100A + 10B + C = 10A + A + 10B + B + 10C + C, which rearranges toβ89A = B + 10C. Since 10C + B can’t be big enough to be a three-digit number (because C & B are numbers not bigger than 9), it follows that 89A has to just be a two-digit number, and therefore A = 1. Then 10C + B = 89, which forces C = 8 & B = 9. It’s confirmed: ABC = 198. π That lovely, special number always reminds me of 1998, the year when my son, Michael, was born π

In my family puzzle, when Michael is 29 (God-willing), Jenny 59 & Paul 65, the total of the three ages will be 153, which is the closest it can get to 154 before shooting past there! π

TWO HUNDRED AND SIX contains vowels A, E, I, O & U exactly once each. π

EASTERβTEASER π

In the Chess puzzle…

White forces checkmate quickly, with the ‘main’ line being 1 Qg8+ Kd6 2 Qd8+ Ke6 3 Qc8+ Kd6 4 Bc5#, provided that Black’s invisible b-pawn is not on b6 !! ππ€£

## Blog Post #153: A Super High Five For All MMS Musicians ππΆπ

Dear All,

Just a few hours ago, hundreds of happy people were enthralled by all the marvellous student chamber music performances that they witnessed in the Bach Concert Hall within Musica Mundi School (MMS), Waterloo. Everyone who was involved deserves really warm congratulations for yet another stunningly beautiful production at the school! ππ

After the concert, lots of people enjoyed mingling together and chatting at a very nice reception π

Word Puzzle

Hidden in the word ‘LATENT’ is another proper six-letter English word which all MMS students have… Have you found it already!? π

I would like to take this opportunity to wish ThaΓ―s and Liav lots of enjoyment and success this coming Sunday when they’ll be performing together in Gent as part of a huge music festival πΆπ.

The Chess set in the background of the above photo was well used this past Tuesday evening when TimothΓ©e and I played a ‘friendly’ game as a practice warm-up for a fun Chess event that we’re going to be having at the school later today, open to everyone in the MMS Family π

A ‘FEEL’ Number Puzzle

ThaΓ―s told me that her favourite number is a particular two-digit whole number which she likes for the ‘feel’ of it. Let’s represent that two-digit number by EL. It’s special because when it’s squared the result is the four-digit number FEEL, in which FE is exactly three-quarters of EL.

Your fun puzzle is to figure out the exact value of ThaΓ―s’s favourite number π

If I tell you that what’s coming next is an advanced algebra brainteaser at the request of a certain brilliant colleague, you’ll probably either react with something like, “Oh yes, that’s exactly what I wanted too ππ!!”

OR

you might say, “Can I have a drink instead?!” ππ

If you chose the latter, then…

If you opted bravely for the brainteaser, then brace yourself…it’s coming now!

Here in Blog Post #153, we ought to note that 1 cubed + 5 cubed + 3 cubed = 153, and so it’s fitting that the brainteaser will feature cubes in it… Liav will be starring too, to make it nicer! π

BRAINTEASER π

Liav knows that there’s an infinite choice of pairs of real numbers such that the sum of each pair equals his favourite number (which is positive). For each such pair, imagine calculating the sum of the cubes of the two numbers in the pair. It turns out that the absolute minimum possible sum of the two cubes equals Liav’s favourite number precisely!

Your brainteaser is to figure out, with proof, what Liav’s favourite number is π

It’s my intention to publish solutions to all the puzzles around the time that blog post #154 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 month of April coming soon, with lots of happiness in everything that you do β€.

With kindest wishes as always,

Paul Mπtwani β€

The angel said to Mary, βThe Holy Spirit will come to you, and the power of the Most High God will cover you. The baby will be holy and will be called the Son of God.”

–Bible verse, Luke 1:35 β₯

P.S. = Puzzle Solutions (being posted on 4.4.2023)

LATENT β TALENT

Considering the place value of each digit in FEEL, the total value of FEEL = FE00 + EL = FE x 100 + EL, and (given information that FE = 3/4 of EL) that’s EL x 3/4 x 100 + EL, which simplifies to EL x 75 + EL β EL x 76. So, EL squared = EL x 76, and therefore EL equals 76, ThaΓ―s’ favourite number.

Suppose that a and b are real numbers whose sum equals Liavβs favourite positive number, L. That is, a + b = L Λ 0. Weβll now use the very handy mathematical βidentityβ a3 + b3 = (a+b)3 β 3ab(a+b), which becomes a3 + b3 = L3 β 3PL, where P = the product ab. Itβs a well-known result that (when numbers have a fixed sum) the maximum-possible value of their product P occurs when a = b, and here that would mean a = b = L/2, and Pmax = L/2 x L/2 = L2/4. So, the minimum-possible value of a3 + b3 is L3 β 3(L2/4)L β L3/4. We were also given information implying that (a3 + b3)min = L, and therefore L3/4 = L β L2/4=1βL2=4βL=2 (not -2, since we require L Λ 0). Liavβs favourite number is indeed 2 ππ

## Blog Post #152: A Good Heart π

Dear All,

This article is specially dedicated to Dr. Vipin Zamvar, a consultant cardiothoracic surgeon who certainly has a really good heart, in the kindest, very best sense β€. Vipin is originally from Mumbai, India, but now lives in Edinburgh, and my family and I had the pleasure of meeting the gentleman doctor in Scotland’s beautiful capital city through an event at Edinburgh Chess Club last October. We thoroughly enjoyed chatting with Vipin during dinner that evening, and when he kindly gave us a lift to our hotel afterwards. In addition to sharing the ‘Royal Game’ of Chess as a fine hobby, we also like Mathematics, and I am currently reading ‘The Music of the Primes’ which Vipin sent as a lovely gift book.

I would like to now offer several fresh brainteasers for the enjoyment of Vipin and all puzzle fans! πβ€π

1. Start with the number 152 here in Blog Post #152.

Multiply it by my favourite number, 3, and then add 3.

If you divide the result by Vipin’s favourite whole number, you’ll then have a prime number.

What exactly is Vipin’s favourite whole number (given that it is not more than 152) ?

2. Rearrange the letters of CREMONA (a beautiful city in Italy) to make a proper 7-letter English word. The nice seven-letter word – – – – – – – has a connection to Vipin because he chose his favourite number for the reason that it was part of the date on which he first met his wife β€

3. We already encountered the number 459 in the first puzzle (when doing 152 x 3 + 3), and now imagine that Vipin selects either 4 or 5 or 9. Let’s call his selected number V. Vipin will raise V to the power of his wife’s age now, and he’ll note the result. Vipin will also raise V to the power of his wife’s future age on her next birthday, and again he’ll note the result. Vipin will add his two results together to get a new, larger result, Z.

What exactly will be the units (or ‘ones’) digit of the number Z?

Can you prove what the digit will be?

4. Imagine a long bus travelling at a constant speed through a tunnel in India that is nearly 1km long. (The tunnel length is in fact a whole number of metres between 900 and 1000. The length of the bus is also a whole number of metres.) From the moment that the front of the bus enters the tunnel, the time taken for the entire bus to be inside the tunnel is t seconds. However, the time taken for the entire bus to pass through the tunnel is t minutes.

What is the exact length of the tunnel?

5. Now it’s time for an ABCD puzzle to wish you A Beautiful Creative Day!

πβ€β€π

If BC is not less than AB + CD,

then

what is the maximum-possible value for AD Γ· r ?

6. People don’t normally like going round in circles, but still…

…this next puzzle is actually lots of fun, too!! π

What is the value of X?

Also, what is the maximum-possible value for P?

7. In Chess, Vipin and I both like playing the Caro-Kann Defence as Black. So, let’s now enjoy seeing it in action in a super-fast victory π from Kiev in 1965, the year when Vipin was born. π

Mnatsakanian vs. Simagin, Kiev 1965.

1 e4 c6 2 Nc3 d5 3 d4 dxe4 4 Nxe4 Nf6 5 Nxf6+ exf6 6 Bc4 (6 c3 followed by Bd3 is more popular nowadays) 6…Be7 (Several decades ago, super-GM Julian Hodgson told me that he likes 6…Qe7+, especially if White responds with 7 Be3?? or 7 Ne2?? which lose to 7…Qb4+! π) 7 Qh5 0-0 8 Ne2 g6 9 Qh6 Bf5 10 Bb3 c5 11 Be3 Nc6 12 0-0-0? (White’s king castles into an unsafe region where it will be attacked very quickly indeed…) 12…c4!! 13 Bxc4 Nb4 14 Bb3 Rc8 (the point of Black’s energetic pawn-sacrifice at move 12 has become clear along the opened c-file) 15 Nc3 Qa5 (also good is 15…b5, intending 16 a3 Rxc3!! 17 bxc3 Nd5 with an enduring attack for Black in addition to having enormous positional compensation for the sacrificed material) 16 Kb1? (It’s often difficult to defend well against a sudden attack, but this move simply loses by force; 16 Bd2 is more tenacious)

16…Rxc3! (16…Bxc2+! 17 Bxc2 Rxc3 also works) 17 bxc3 Bxc2+! 0:1. White resigned, in view of 18 Bxc2 Qxa2+ 19 Kc1 Qxc2# or 18 Kc1 Nxa2+ with decisive threats against the fatally exposed White monarch.

I’m pretty sure that Scott Fleming (who recently sent me a really nice letter from Arbroath, Scotland) will also enjoy that very neat, crisp win for Black, as will FIDE Master Craig SM Thomson, who has played lots of wonderful games with the Caro-Kann Defence for nearly 50 years already!! ππ

It’s my intention to publish solutions to all the puzzles around the time that blog post #153 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€.

Special congratulations to my friend James Pitts who has turned 53 today.

πππ

With kindest wishes as always,

Paul Mπtwani β€

“He has dethroned rulers and has exalted humble people.”

–Bible verse, Luke 1:52 β₯

P.S. = Puzzle Solutions (being posted on 4.4.2023)

1. 459 Γ· 27 = 17, a prime number. Vipin’s favourite number is 27.
2. CREMONA β ROMANCE ! β₯
3. Note that VA + V(A+1) = VA(1+V). If V=4, then (1+v) = 5, and VA(1+V) will end with a digit zero. If V=5, then (1+v) = 6, and VA(1+V) will be even and will end with a digit zero (rather than a 5). If V=9, then (1+v) = 10, and again VA(1+V) will end with a digit zero.
4. If B and T represent the respective lengths (in metres) of the long bus and the tunnel, then (since t minutes is 60 times longer than t seconds), we can deduce that B+T = 60 x B and so T = 59 x B, a whole multiple of 59. The only such value for T between 900 and 1000 is 59 x 16 = 944. The tunnel is 944 metres long (and the length of the long bus is 16 metres).
5. If AD = L, then AB = AD – BD = L – 2r, and similarly CD = L -2r. Also, BC = L – 2(L-2r), which simplifies to 4r – L. So, for BC to be not less than AB + CD, we require that 4r – L β₯ 2 (L – 2r). That inequality can be simplified to 8r β₯ 3L, and so L/r β€ 8/3. Therefore, if BC is not less than AB + CD, the maximum-possible value of AD Γ· r is 8/3.
6. The only possible value for X is 10, and a maximum-possible product of 120 along each edge can be achieved by putting the numbers 4, 6 and 10 in the corner positions. Then, place 3 between 4 & 10; place 2 between 6 & 10; place 5 between 4 & 6. The products P each become 120.

## Blog Post #151: Thank You for dear Hans & Heleen β₯πβ₯

Dear All,

This tribute article is dedicated specially to Hans Moors (23.6.1942-18.5.2022) and his wife, Heleen, who is still a very dear friend to my little family of Jenny, Michael and myself. We visited Heleen in The Netherlands last Saturday, and it was really good to all be together and to share lots of happy memories of Hans.

Our friendship began way back in 1996, when Hans & Heleen very kindly let me stay at their home for ten days, as I was playing in an international chess tournament taking place within the same city. The warm welcome and generous hospitality helped me greatly, and we enjoyed celebrating when I won the tournament! π

It wasn’t long before puzzles and jokes were spilling forth, and the probability of Hans solving/laughing regarding mine was always four times 1/4, given his terrific sense of humour and his superior mathematical abilities! Yes, Hans was a brilliant University Professor of Statistics, and I will always be grateful for everything that I learned from precious chats with him, from tricky puzzles that he shared with me, and from the wonderful friendship of his family. I am writing this article also to thank God for all those treasured gifts, and more β₯

Michael especially loved the many after-dinner games of monopoly at the home of Hans & Heleen, and he also remembers fondly an outing with their four grandchildren to De Efteling park.

Fast-forwarding to last Saturday with Heleen, she told us that she and Hans bought their house together in 1972, over half a century ago. The number on the door is 47, and now I’d like to share a few mathematical surprises which would certainly have made Hans smile π.

Hans lived to be 79, and that’s exactly the number of digits in 47 raised to the power of 47. For the record, the whole number is 3877924263464448622666648186154330754898344901344205917642325627886496385062863.

Here in Blog Post #151, it’s also fitting that the following super-long sentence is true…

Hans and Heleen lived very happily together at house number forty-seven for fifty years, and this sentence will contain one hundred and fifty-one letters provided it stops right here!

As a quick, fun puzzle π, can you find a slightly different number word other than one hundred and fifty-one which would still be completely correct in the italicised sentence above?

The sum of my family’s house number 11 and Heleen’s house number 47 is: 11 + 47 = 58. In honour of lots of fabulous times together at the two homes, I present to you the following special 58-digit number: 1525423728813559322033898305084745762711864406779661016949. If you multiply it by the number of letters in FAMILY, here’s what happens…

1525423728813559322033898305084745762711864406779661016949 x 6 = 9152542372881355932203389830508474576271186440677966101694 π

The next surprise is one of my favourites. It will feature the following numbers:-

2, in honour of β₯ Hans & Heleen β₯

9, the number of letters in HANS MOORS

11, the number of letters in HELEEN MOORS

79, the very good age that Hans reached

151, the blog post number of this tribute article

The product 2 x 9 x 11 x 79 x 151 = 2361942, and Hans was born on 23.6.1942 π

The cube of Hans’ birth year is interesting because 1942 x 1942 x 1942 = 7323988888, ending with five consecutive 8s.

(If we only square a whole number, the maximum group of repeated digits that can occur at the end of it is three 4s. The last previous year with that property was my birth year, 1962, and 1962 x 1962 = 3849444. Within the five hundred years from 1962 until the far-future time of 2462, only one other year can have a square which ends with 444, and that is 2038 for which 2038 x 2038 = 4153444.)

Five Fun Puzzles in Honour of Hans β€πβ€πβ€

1. It’s of course very easy to make 24 using the numbers 1, 4 & 6 because 1 x 4 x 6 = 24. Your fun challenge is to make 24 using the numbers 1, 4, 6 AND my favourite 3, once each in a calculation. You may freely use parentheses ( ) and operations involving +, -, x, Γ· as you wish.

2. Hans was a very experienced chess-player. You don’t need to be a player (or even to know the rules of ‘The Royal Game’ at all!) in order to solve the following puzzle… Imagine a magic box with an endless supply of pure white and pure black chess pieces, and a separate big bag containing an assortment of 1000 pure white chess pieces and 999 pure black ones. You’ll be taking two pieces at a time out of the bag. If they’re both white, just one white piece will be put back into the bag; if instead the removed pieces were both black, just one white piece would be moved from the box to the bag; if instead the two removed pieces were of opposite colour (meaning: one white and one black), just the black piece would be put back into the bag. You’ll keep repeating the whole process (taking two pieces at a time out of the bag, and reacting according to their colour(s)), until finally only one piece remains in the bag after completion of the final process. Your fun puzzle is to figure out, with proof, what the colour of the last piece in the bag will be.

3. Imagine placing three coins randomly in three of the 64 little unit squares on a standard 8 x 8 square chess board. What is the precise probability that the three coins will all be on different horizontal ranks and vertical files from each other? (Assume that the three coins are round, all of exactly equal size, and that they are placed centrally in the unit squares.)

4. Since Hans would win a prize for solving all the puzzles, I’m thinking of A PRIZE MUST β TRAPEZIUMS!

5. The following beautiful chess puzzle is dedicated not only to Hans, but to all fans of ‘The Royal Game’, including Ralph Connell (a former student of mine from the 1980s in Scotland!) who recently wrote me a really nice letter, and Dr. Vipin Zamvar who very kindly sent me a lovely gift book. Happy birthday also to Belgian chess friend William Verleye who has turned 62 today β€πβ€πβ€

It’s my intention to publish solutions to all the puzzles around the time that blog post #152 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₯

Jesus said to Nathanael, βTruly, truly, I say to you, you will see Heaven opened, and the angels of God ascending and descending on the Son of Man.β–Bible verse, John 1:51 β₯

P.S. = Puzzle Solutions (being posted now on 12.03.2023)

The sentence

Hans and Heleen lived very happily together at house number forty-seven for fifty years, and this sentence will contain one hundred and fifty-three letters provided it stops right here!

is the alternative correct answer.

6 Γ· (1 – 3 Γ· 4) = 24 π

The final chess piece in the bag will be a black one. Note that each complete process described in the puzzle reduces the total number of pieces in the bag by precisely one, and so the total does indeed keep diminishing, for sure. However, when black pieces are permanently removed from the bag, they are removed in pairs (or ‘twos’) and are replaced with one new white piece. Therefore, whenever the number of black pieces reduces permanently, it goes down by 2 each time. Since there were 99 (=an odd number of) black pieces in the bag at the start, the number of black pieces can’t reach zero and stay at zero. So, that proves that the very last chess piece in the bag must be a black one.

In the puzzle regarding 3 coins on an 8 x 8 chessboard, the probability that they will all be on different horizontal ranks and vertical files is 64 x 49 x 36 Γ· (64 x 63 x 62), which simplifies to the fraction 14/31.

In the puzzle about the trapezium, first note that the two clear triangles (not coloured) have equal areas, each equal to C, say. To see that, observe that A + C on the left and A + C on the right give us big triangles with the same height and base length, so their areas must be equal. Next, note that the coloured triangles are mathematically similar because they have the same angles as each other (using standard facts regarding vertically opposite angles being equal, as are alternate interior angles). The ‘area scale factor’ relating triangles ‘A’ and ‘B’ is A/B, and so the ‘length scale factor’ is the square root of (A/B). Noticing that each of the ‘C’ triangles have a side in common with A and a shorter side in common with B, it can be deduced that each area C must equal A Γ· the square root of (A/B). That simplifies to C = sqrt (AB), where sqrt is short for ‘square root’. Therefore, the total trapezium area, A + B + 2C = A + B + 2*sqrt (AB). A nice, neat, alternative expression for the total area is (sqrt A + sqrt B) squared π.

White wins beautifully with 1 Nxd5! exd5 2 Qxf7+!! Kxf7 (or 2…Kh8 3 Ne6) 3 Bxd5+ Kg6 (or 2…Kf8 3 Ne6+) 4 f5+ Kh5 5 Bf3+ Kh4 6 g3+ Kh3 7 Bg2+ Kg4 8 Rf4+ Kh5 9 Rh4+ Bxh4 10 Bf3#, checkmate! π

## Blog Post #150: An Epicentre of Culture β₯

Dear All,

I had the pleasure this past week of meeting Maximilian Gililov, the youngest son of world-famous classical pianist Pavel Lvovich Gililov. Max was visiting Musica Mundi School, which he described beautifully as “an epicentre of culture”.

The following feast of fun puzzles is specially dedicated to Max π

1. Of the three chess games that we played, the third game lasted for 38 moves, the second for 39 moves, and the first game was shorter. The average number of moves in the three games was a whole multiple of ten.

Part 1: Exactly how many moves were played in the first game?

2. Rearrange the letters of US SECLUDED BAY or DEAD BUSY CLUES to get the name of Max’s favourite famous musical composer!

3. Max is an excellent pianist, but his other favourite instrument can be found by changing just one letter of HELLO to a different letter…

4. Choose well and replace one letter that is repeated in AIR STAR by the letter U, and then rearrange to make the seven-letter name of the beautiful country where Max lives and works.

5. Max and I thoroughly enjoyed the concerts that we saw. All the performances were superb! For every student who is practising to become better and better, Max’s honest words regarding himself can be inspirational. He said, “I’m an inferior version of my future self.”

I saw two concerts: one on Thursday in which every performance was on the piano; and one on Wednesday in which each performance was on either the harp, the oboe, or the cello.

For Thursday’s concert, the five-letter word PIANO was written 11 times (alongside 11 performers), making a total of 55 ‘piano’ letters there. For Wednesday’s concert, the total number of ‘HARP’, ‘OBOE’ & ‘CELLO’ letters was also 55, and there were more cellists than harpists or oboe-players.

Exactly how many cello performances were there in Wednesday’s concert? Also, what was the total number of performances (featuring harp, oboe, or cello) in that concert?

6. Max joked, “Life is not always black and white, but if you’re a pianist who likes to play chess, then it kind of is!!” π

7. Are you feeling nicely relaxed here in Blog Post #150?

Part 1: In our normal base ten, the number 150 ends with a zero. In which different number base does it end with two zeros, and exactly what other digits would be needed?

Part 2: The number 150 (base ten) consists of B digits when represented in base B. What is the value of B, and exactly what digits would be needed this time?

8. Imagine that Max and his parents have a water bottle each. All three bottles have the same capacity in litres: either 0.5, 0.75, 1, 1.5 or 2 litres.

Max’s bottle is full, but his parents’ bottles are empty. In what follows, the numbers of millilitres of water in the bottles are always whole numbers after each sharing action. (You can assume that everything is done very precisely with the aid of measuring cylinders and such items, if you like!) π

Max first pours some water from his own bottle to his father’s bottle until they both have equal amounts.

Max’s father then pours some water from his own bottle to his wife’s bottle until they both have equal amounts.

Max also pours some millilitres of water from his own bottle into his mother’s bottle, and then he gives his father the same nice bonus π.

All three bottles now contain exactly equal amounts of water.

Exactly how many millilitres of water did Max pour personally into his mother’s bottle? Also, what is the full capacity of each of the bottles?

9. Max was born in the year 2000. If you multiply the age he’ll be on his birthday this year by the number of days in his birth month, the three-digit result contains Max’s three favourite (positive one-digit) numbers πππ.

What are Max’s three favourite numbers?

10. The largest of Max’s three favourite numbers is his absolute favourite, and it’s also the month number for his birthday π. Now in 2023, day #1 was 1 January, day #42 is 11 February, and so forth. Max’s birthday in 2023 will be on day #N of this year, and N is a whole multiple of Max’s current age.

The fun challenge is to figure out Max’s exact date of birth.

11. List Max’s three favourite positive numbers from smallest to largest. Think of them as being the first three terms of a never-ending sequence of numbers! In that sequence, each term (from the second term onwards) can be generated by Multiplying the term immediately before it by M and then Adding on A.

Exactly what numbers do M and A represent?

Also, what is the exact value of the term in the sequence that is closest to 1000000?

12. Max went to high school in the lovely Austrian village of St. Gilgen, where Mozart’s mother was born loooooooooooooooooooong ago! π

It’s time now for one of my favourite brainteasers of all time, which I solved really happily in my early teens at school in Scotland πβ₯π.

As Max will be turning 23 later this year, I’m thinking of a very special 23-digit whole number which starts and finishes with 1 at both ends. If you divide the super-loooooooooooooooooooong number by 99, the result is a twenty-one digit whole number which looks absolutely identical to the starting number except that it’s missing the first and last ‘end 1s’.

Your super-fun brainteaser is to find the 23-digit number in honour of Max! π

Notes:- No computer program or calculator is needed at all. The brainteaser can be solved quickly on paper. I find it a delightful bonus detail that there are 21 and 23 letters respectively in the names of Wolfgang Amadeus Mozart and his mother, Anna Maria Walburga Mozart πβ₯π.

13. Last week, I gave a ruler as a gift to a Maths student. Now, today, I’m thinking of one of the greatest mathematicians of all time (with initials LE) whose full name can be found by rearranging the letters of LE HAD ONE RULER. The world-famous mathematical genius was born in the year – – – – and the digits are the same as Max’s birthday when stated in the form ddmm (day number followed by month number) π

Can you identify the famous mathematician?

14.

Imagine tiny me sitting on flat ground at the same level as the base of the Burj Khalifa, but some distance from it. Superman has flown to a point at a certain height on the skyscraper, and the eye-to-eye distance between him and me is 1000 feet. Max has already climbed to a point on the skyscraper that is 1000 feet directly above where Superman is.

When I look up at Max, his angle of elevation is twice (or double) the angle of elevation involved when I look at Superman.

Your sky-high brainteaser is to figure out the number of feet that Max still has to climb to reach the very top. (Just assume that people’s heights are negligible in comparison to the other distances involved in the puzzle.)

Fun Note: In a way, the digits of my answer are a tribute to Max and to Jens Van Steerteghem, an excellent mathematical/scientific colleague of mine at Musica Mundi School, as will be explained when the solutions are published.

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₯

BπNUS CHESS PUZZLE π

“And the spirit of the Lord shall rest upon him, the spirit of wisdom and understanding, the spirit of counsel and might, the spirit of knowledge and of the fear of the Lord”–Bible Verse, Isaiah 11:2 β₯

P.S. = Puzzle Solutions (being posted on 17.2.2023)

1. Part 1: 13 moves; Part 2: The possibilities are infinite! For example, draw a straight line from a point on the baseline of the photo, at a distance d from the bottom left-hand corner, to a point along the top of the photo at a distance d from the top right-hand corner. The rectangular photo is thereby split into two trapeziums of equal areas; the area of either one is exactly half of the area of the entire rectangular photo.
2. US SECLUDED BAY or DEAD BUSY CLUES βCLAUDE DEBUSSY !
3. HELLO – H + C βCELLO.
4. AIR STAR – R + U βAUSTRIA.
5. There were 7 cello performances and 5 harp or oboe performances, making 12 performances in total.
6. Hiding behind Max’s glass were the white queen and another white pawn. They were no longer on the board itself.
7. Part 1: 150 (base ten) is equivalent to 1100 (base five) or, for Part 2, 2112 (base four); B = 4.
8. Max personally poured 125 ml from his 1.5-litre bottle to his mother’s 1.5-litre bottle. Among the bottle-sizes mentioned in the puzzle, only the 1.5-litre option was wholly divisible by four and by three, so that Max’s parents could have a quarter-full bottle each (with 375 ml), before Max topped them up to one-third full (with 500 ml).
9. 23 x 31 = 713; Max’s three different favourite numbers are 7, 3 and 1.
10. Max’s exact date of birth was 17.7.2000. Note that, in 2023, July 17 will be day #198 of this non-leap year, and 198 = 9 x Max’s current age of 22.
11. The numbers in the sequence 1, 3, 7, … can be generated by multiplying terms by 2 and adding on 1 to get the next term; M = 2 & A = 1. Note also that each term is always 1 less than some power of 2. For instance, 3 = 2 squared – 1 & 7 = 2 cubed – 1. The value of the term that is closest to 1000000 is 1048575, which is 1 less than 2 raised to the power of 20.
12. The magical number from my early teens is 11123595505617977528091. Note that 11123595505617977528091 Γ· 99 = 112359550561797752809.
13. LE HAD ONE RULER βLEONHARD EULER, born in 1707.
14. In the puzzle, Superman is 500 feet up from the base of the Burj Khalifa at a 30Β° angle of elevation from where I am, while Max is 1500 feet up at a 60Β° angle of elevation with respect to me. Max still has a further 1217 feet to climb to reach the top, and the digits 12 & 17 are early celebrations of the birthdays of Jens Van Steerteghem & Max Gililov on July 12 and 17 respectively ππ
15. In the Chess puzzle, White won beautifully with 1 Bxf5 exf5 2 Nh6+ Kh8 3 Rxg7! Kxg7 4 Rg1+ Kh8 5 Qe2!! Be6 (or 5…Qxe2 6 Nxf7#) 6 Nxf7+!, intending 6…Qxf7 7 Qe5+.

## 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₯

P.S. = Puzzle Solutions (being posted on 17.2.2023)

1. Jan will be 3 x 4 x 5 = 60 years old on his birthday next month.
2. Jan’s date of birth was 15.3.1963. Note that 15 March is day number 74 or (5 x 5) + (7 x 7), in non-leap years.
3. FEELING – G βFELINE.
4. GUEST – T + S βGUESS.
5. 149 squared = 22201.
6. 65 + 56 = 121 = 11 squared & 65 – 56 = 9 = 3 squared.
7. Jan’s favourite number is 7. Jan’s favourite colours are Red and Green.
8. 3 x 7 x 9 = 189.
9. In the Chess puzzle, White forces checkmate with 1 Rh5+! Kxh5 2 Qh7+ Kg5 3 h4#.
10. Part 1: The maximum possible perimeter is 84 + 1763 + 1765 = 3612 cm; the maximum possible area is 84 x 1763 Γ· 2 = 74046 square centimetres; Part 2: The minimum possible perimeter is 84 + 13 + 85 = 182 cm; the minimum possible area is 84 x 13 Γ· 2 = 546 square centimetres.

## 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”