Blog Post #101: Super Sigurd advances from 10 to 11 !

Dear Friends,

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.

Happy memories with dear friends four years ago

Sigurd is turning 11 today, and so we have lots of nice, sneaky surprises…!

Happy Birthday, Sigurd!

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…

Where exactly is White’s invisible queen if it’s White to play and force checkmate in two moves!?

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.

Write any whole number you want. The ‘sky’s the limit’ to how big it can be! Just when you think you’re finished writing it, on the right-hand end, start writing it all again in reverse order! For Super Sigurd’s 11th birthday, your final number will be an exact multiple of 11 ! For example, Sigurd’s big sister and brother have their birthdays in the 11th month (November) on day 15 and 22 respectively. 1551 ÷ 11 = 141 exactly, a nice number from pmotwani141@gmail.com ! 2222 ÷ 11 = 202, which is exactly double the number of this Blog Post 101 !

Sigurd’s Birthday Chess Puzzle Solution

With the ‘invisible queen’ now visible on f7, White can use the g5-pawn to capture Black’s knight with 1 gxh6, and there’s simply no way to prevent checkmate coming on White’s very next move!

Sigurd’s ONE + ONE = TWO Sneaky Birthday Maths Brainteaser Solution

In this TWO-story,

the minimum possible value of ONE + ONE is 412

and

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 Mtwani 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…

Handy chart of the first hundred positive Square Numbers

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.

Blog Post #100: A Priceless Gift of Love for Easter

Dear Friends,

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!

Aiming High towards Heaven

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.

Easter Eggs for Elaine!

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

                          – – – – 

                           133 👌

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.

With kindest wishes as always,

Paul Motwani xxx