Around 35 years ago, I had the great honour and pleasure of doing some chess with Ally, a daughter of Mrs. Mo Brodie who was then a very dear Mathematics teacher colleague of mine in Scotland. Mo was surprised when I said to her one day that I learned from the chess games with her young daughter; but I really meant it! I am a teacher, yet a life-long student too.
Grandmaster Glenn Flear expressed it well, for he said that he had the feeling that, because I don’t actually play many games of chess, each game that I do play becomes like a ‘Cup Final’, a special event to remember and to learn from, as much as possible.
At the wonderful Musica Mundi School where I work as the Maths Teacher in Waterloo, Belgium, I sometimes get the chance to enjoy a ‘friendly game’ with colleagues and students during our lunch break. Yesterday, it was a lovely treat for me to play 12-year-old Wout Callens, who is extremely gifted in Music and Mathematics as well as having a passionate interest in the royal game of Chess.
Big thanks to Christophe Gillain for the following photos.
A superb team of dedicated, hardworking organizers led by Dirk Flamée really did extremely well to run the 36th International Open Chess Tournament of Geraardsbergen across five locations from August 8-12 to cater for 214 players who were happy and grateful for getting to take part in a wonderful over-the-board, in-person competition, overcoming the many extra challenges of managing this in these difficult Corona times.
For very many people, this highly memorable event was their first return to chess after a necessarily long time-gap due to the global situation.
Lots of keen and talented young players performed magnificently, and special congratulations must go to Dutch FIDE Master Onno Elgersma who, though still only a teenager, won the tournament impressively with 6 wins, 3 draws and no losses, thereby amassing a score of 7.5/9.
Among the older players (like myself!), Dutch International Master Herman Grooten did particularly well to equal Onno’s score of 7.5 points and finish 2nd on tie-break. (Though I don’t personally know Herman’s son, Tommy Grooten, I would still like to add sincere congratulations to young Tommy on scoring five points and winning a prize for being a high-achieving youth-player.)
Within the past few days, lots of people from numerous countries were captivated yet again by wonderful concerts seen live inside the Bach Concert Hall at Musica Mundi School, Waterloo, Belgium, or enjoyed online via top-quality broadcasts. My personal appreciation of the beauty of music, and its profound power to touch and heal, continues to grow.
I am especially grateful to Leonid Kerbel and Hagit Hassid-Kerbel, Musica Mundi School’s founders, who hired me three years ago to be the Mathematics teacher in their magical environment.
I am very grateful for still being able to hear pretty well, thanks to having wonderful hearing aids from Amplifon.
Living in Belgium, I have already received lots of excellent, expert care from former/current Amplifon employees including Anais, Elena L., Romy H. and Tessa VG.
Let’s wish Romy a super happy birthday today,
and we can all enjoy an original puzzle involving Romy’s favourite one-digit number together with Tessa VG’s different favourite single-digit number. Let’s denote the favourite numbers by R and T respectively.
In the spirit of good fun, the ladies have allowed me to tell you a bit more…! The two-digit number RT (meaning not R x T, but rather the two-digit number with digits R and T in normal base ten) is equivalent to 111 in base T.
One possibility might have been 13 (in normal base ten) equals 111 in base 3, because 1113=1+3+32=13ten. However, the true solution for the ladies’ RT is not 13, but something else…
Your fun challenge is to now figure out the precise values of R and T. Then calculate R2 to discover Romy’s new age today!
The puzzle answers will be given in just a few M😊ments…
First, something I left until right here, here are a couple of classic ear jokes to hear!!
What three ears did Captain Kirk have?
His left ear, his right ear, and the final frontier!
How much does it cost a pirate to pierce his ears? – A buccaneer!!
All the posts are free for everyone to enjoy, and today’s one is specially dedicated to Super Sigurd, the youngest of three children in a lovely Norwegian family with whom I often had the pleasure of doing fun Maths and Chess.
Sigurd is turning 11 today, and so we have lots of nice, sneaky surprises…!
It’s Wednesday, and W is the 23rd letter of the English alphabet. So, let’s start with the number 23.
Multiply by 19, because S (for Super Sigurd!) is the 19th letter of the alphabet.
Multiply by my favourite number, 3.
To crown this surprise, multiply by 11, Sigurd’s new age.
The result of 14421 is to wish Sigurd an unforgettable, happy birthday on 14.4.21=today !
Also, 14421 ÷ 11 = 1311,
like saying to Sigurd, “HAPPY BIRTHDAY (13 letters), 11 !”
Furthermore, it’s fun to note that, by changing just one digit, we have two ways of turning 1311 into a pretty palindrome…
Either change it to 1331, which equals 11 x 11 x 11
OR change it to 1111, which equals this blog post number times Sigurd’s new age: 101 x 11 = 1111.
It’s time now for Sigurd’s birthday chess puzzle…
Before revealing the chess solution, here’s a simple-looking yet rather enjoyable wee ‘word sum’ to increase the fun today…!
In ONE + ONE = TWO, each different letter stands for a different digit and each same letter stands for the same digit. The general digit options are 0-9, but neither O nor T can be 0 because ONE and TWO stand for proper three-digit numbers.
Your double fun challenge is to figure out the minimum possible value of ONE + ONE and also the maximum possible value of ONE + ONE if it equals TWO !
That’s a beautiful one + one for Sigurd’s 11th birthday, and yet it’s just a tiny illustration showing that, in Maths and everything created by God, there’s a limitless universe of surprises infinitely beyond our imagination still waiting to delight us for eternity in Heaven.
Sigurd’s Birthday Chess Puzzle Solution
Sigurd’s ONE + ONE = TWO Sneaky Birthday Maths Brainteaser Solution
In this TWO-story,
the minimum possible value of ONE + ONE is 412
the maximum possible value of ONE + ONE is 964.
Very warm congratulations for finding either or both of those values.
To conclude this article, I would like to wish Sigurd and everyone a really wonderful, happy day now.
With kindest wishes as always,
Paul M☺twani xxx
P.S. Even when he was a grade 1 primary school student several years ago, young Sigurd already loved to do great mental calculations, and he understood mathematical concepts such as squaring, with 122 = 144 being one of his favourite examples, not only on 14/4 ! So, as an extra birthday treat, we have the following b☺nus brainteaser f☺r Sigurd…
Can you discover a future year in which
the total of the year + Sigurd’s age on his birthday in that year
would together make a Square Number ?
God-willing as always, it is my hope and intention to publish a solution at the time of the next blog post, or possibly even sooner.
SOLUTION to Super Sigurd’s Bonus Brainteaser (being posted on 14 May 2021)
In the year 2063, Sigurd will be 53. Then, 2063+53=2116, which is equal to 46 squared, since 46 x 46 = 2116.
No securer intro. is possible than to say honestly and happily that I am celebrating with gratitude an event which can be discovered by rearranging the letters of OUR CENTRE SIR (or of securer intro).
The true message of Easter celebrates the RESURRECTION of Jesus, triumphing over death and marking the beginning of a new covenant between God and people, through which everyone who believes will enjoy eternal life.
The Good News Translation of John 3:16 in The Bible is: “For God loved the world so much that He gave His only Son, so that everyone who believes in Him may not die but have eternal life.”
That is really clear and wonderful news for everyone who will simply recognise and accept with sincere thanks the priceless gift.
The most important news has rightly been shared first.
I do have a lovely, fresh puzzle to offer now, too.
It involves a true story about 12 people who are all either students or colleagues of mine at the magical Musica Mundi School in Waterloo, Belgium. (Some of them are pictured in this article.)
I asked each person to pick his/her absolute favourite number. The 12 people picked 7, 8, 9, 11, 13, 16, 19.
As only seven different numbers were picked by the 12 people, we can realise that some people must have picked the same number as each other.
I will now add a couple of very significant extra facts. The numbers 7, 8 and 9 were all equally popular. Also, the grand total sum of the 12 people’s numbers was 133.
Your fun brainteaser is to figure out precisely how many people picked each of the listed numbers 7, 8, 9, 11, 13, 16, 19.
I hope to be back in person at the school on 19 April, and on that day I would love to award a special prize to the student(s) and/or colleague(s) who might send me good solutions by email before this coming Easter Monday, 5 April. Aim high, dear friends!
In the meantime, please enjoy a delightful selection of photos which shows some of the Musica Mundi School Family almost literally reaching for the sky!
God-willing as always, a full solution to the brainteaser prize puzzle will be published at the time of the next blog post, or possibly even sooner.
Update being posted now on Easter Monday, 5 April 2021
Special congratulations to Elaine G., a dear Scottish friend who solved the brainteaser. A scrumptious Easter gift is already on its way to Elaine’s home in Edinburgh.
Here is a complete, clear, step-by-step method for finding the unique solution to the brainteaser.
Step 1: The sum of the seven numbers 7, 8, 9, 11, 13, 16, 19 is 7 + 8 + 9 + 11 + 13 + 16 + 19 = 83. That would be the total if each of the seven numbers occurred just once each, picked by seven people.
Step 2: The grand total sum of 133 (for 12 people’s numbers) was included in the info. given in the puzzle, and is 50 more than 83. So, five people’s numbers must have a total sum of 50.
Step 3: At least one of those five people’s numbers must be 9 or less, since if they were all 11 or more, then their total would be greater than 50.
Step 4: However, we were also given (in the info. with the puzzle) that the “numbers 7, 8 and 9 were all equally popular.” So, since we know (from Step 3) that at least one of the numbers 7, 8, 9 must occur twice, it follows that all three of them occur twice, in order to be “equally popular.” (Note that 7, 8, 9 cannot occur three times each, together with 11, 13, 16, 19 at least once each, because that would take more than 12 people to have picked the numbers!)
Step 5: Having established that ten of the 12 people must have picked 7, 7, 8, 8, 9, 9, 11, 13, 16, 19, the total sum so far is 107. To reach the final target of 133, the remaining two people’s numbers must total 26, and must be from 11, 13, 16, 19, since we already know that none of 7, 8, 9 can feature a third time. The only way to finish the job correctly is with 13, 13 for the ‘missing’ two numbers.
In summary, the solution is: 7, 7, 8, 8, 9, 9, 11, 13, 13, 13, 16, 19.
In her email to me several hours ago, Elaine presented her solution succinctly and prettily like this:
1 x 11 16 19 = 46
2 x 7 8 9 = 48
3 x 13 = 39
– – – –
Elaine has sometimes said that Maths is not really her thing, but it’s great that she was willing and not afraid to have a good go at the brainteaser…and she succeeded in style! Congratulations again!
Blog Post #101 is very likely to include an original chess brainteaser that I composed recently…
Right now, though, I would like to wish everyone a wonderful, happy weekend.
Currently, my youngest Musica Mundi School Maths students are a super ‘High Five’ of Barbara, Jan, Julija, Sophie and—last, but certainly not least!—Hoi Yuet who turns from 11 to 12 tomorrow! They’ve all been happily and successfully going round in circles recently!!
Yes, the young stars have been solving numerous problems involving Circle Theorems, and have therefore encountered the number 360 more than a few times! They should all be awarded Degrees in Circles Maths!
To celebrate Hoi Yuet marching on and up from 11, feel free to grab a calculator and work out the high-five product 11 x 12 x 13 x 14 x 15.
The result of 360360 will hopefully bring a nice, round smiley to some faces!
The meaning of the name Helen is ‘bright, shining light’, and that well describes Mrs. Helen Coyne-Wincott, a lovely English teacher colleague of mine at Musica Mundi School.
In addition to being fluent in several languages, Helen is also very good at Mathematics, and she’s a highly talented musician, too, with special expertise on a particular instrument… You’ll discover which one, and lots more, in the following feast of fun puzzles that is an early celebration for Helen’s birthday coming in 3 days’ time! I have Helen’s permission to post numerous surprises. So, let’s magnify Helen’s enjoyment, OK!