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 ❤. 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! 😊❤😊
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 ❤
What a lovely couple! We see Vipin and his wife, Usha, pictured in Coimbatore, India 💖💖
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!
😊❤❤😊
The diagram shows two overlapping circles of equal radii, r, say. The points A, B, C and D are collinear, and all lie on the line which passes through the centres of the circles.
If BC is not less than AB + CD,
then
what is the maximum-possible value for AD ÷ r ?
6. People don’t normally like going round in circles, but still…
…this next puzzle is actually lots of fun, too!! 😂
Imagine that the distinct positive whole numbers 2, 3, 4, 5, 6 and X are going to be placed in the six rings; one number per ring. The products of the numbers on each of the three edges of the triangular array are to be equal to each other, and will each be P, say.
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 😎 from Kiev in 1965, the year when Vipin was born. 💖
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+! 😁) 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)
Get ready for a beautiful Chess combination! 😍
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!! 👍😊
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 😊.
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 ❤.
Special congratulations to my friend James Pitts who has turned 53 today.
🎂💖😊
With kindest wishes as always,
Paul M😊twani ❤
“He has dethroned rulers and has exalted humble people.”
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.
Lovely Yellow Roses represent Warm Friendship ❤
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! 😊
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 ♥
Hans & Heleen with Michael in December 2016 ♥😊♥ Hans once commented that “things are blossoming”, and he discovered the amusing fact that the 11 letters in A TULIP WOMAN can be rearranged to make PAUL MOTWANI!! Maybe there’s a message in there that my good wife, Jenny, really deserves a lot of the credit!
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 😊.
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 😎, 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?
A lovely picture at Heleen’s home ♥
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…
The next surprise is one of my favourites. It will feature the following numbers:-
2, in honour of ♥ Hans & Heleen ♥
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 😊
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 ❤😊❤😊❤
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, ÷ 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 → TRAPEZIUMS!
Your brainteaser is to find a mathematical expression for the Area of the entire Trapezium in terms of the areas A and B indicated in the picture.
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 ❤😊❤😊❤
It’s White to play and win in a stunning game from 1962, the year I was born.
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 😊.
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 ♥
Jesus said to Nathanael, “Truly, truly, I say to you, you will see Heaven opened, and the angels of God ascending and descending on the Son of Man.”–Bible verse, John 1:51 ♥
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 ÷ (1 – 3 ÷ 4) = 24 😊
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 ÷ (64 x 63 x 62), which simplifies to the fraction 14/31.
Your brainteaser is to find a mathematical expression for the Area of the entire Trapezium in terms of the areas A and B indicated in the picture.
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 ÷ 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 😊.
It’s White to play and win in a stunning game from 1962, the year I was born.
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”.
Max and I enjoyed seeing fabulous concerts at Musica Mundi School, and we played & discussed some very interesting Chess, too! 😊😊
The following feast of fun puzzles is specially dedicated to Max 😊
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?
Part 2: Imagine drawing a straight line across the stunning rectangular photo above, so that the line divides the photo into two parts of equal areas. How many such lines are possible? Is the correct answer 4, or slightly more, or infinitely more!? Can you prove that your answer is correct?
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!!” 😊
OK, I’ve read some minds that have spotted black, white and red too, but have you read Max’s mind!? There’s a white knight and pawn beside his drinking glass, but exactly which white pieces &/or pawns are hiding behind the glass!?
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!) 😊
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 😊.
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 😊😊😊.
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 😊. 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.
This card is appearing here far in advance of Max’s actual birthday, but I’m happy to be early and sure rather than risking missing the party! 😊
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! 😊
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 😊♥😊.
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! 😊
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 😊♥😊.
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) 😊
Can you identify the famous mathematician?
This lovely photo of Max & Chess sets galore in Istanbul reminds me in various ways of a time long ago when I played for Scotland at the 1986 World Chess Olympiad in another stunning location: Dubai! The Burj Khalifa wasn’t yet there, of course, but it’s going to star in Max’s next adventure, coming…now! 😊♥😊
14.
Dubai’s Burj Khalifa skyscraper (2717 feet high) featured in one of Tom Cruise’s Mission Impossible movies, and now we’ve got Max and Superman scaling its heights in the brainteaser below!! 😊😊
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 😊.
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 ♥
B😊NUS CHESS PUZZLE 😊
This chess puzzle is from a real Grandmaster battle in which it’s White to play and win by force in dazzling fashion! It’s dedicated to all fans of ‘The Royal Game’, and especially to Max Gililov, to Eric Van Steerteghem (whose birthday is tomorrow), and to International Masters Geert Vanderstricht and Roddy McKay who have their birthdays today!! 😊😊😊😊
“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 ♥
P.S. = Puzzle Solutions (being posted on 17.2.2023)
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.
US SECLUDED BAY or DEAD BUSY CLUES →CLAUDE DEBUSSY !
HELLO – H + C →CELLO.
AIR STAR – R + U →AUSTRIA.
There were 7 cello performances and 5 harp or oboe performances, making 12 performances in total.
Hiding behind Max’s glass were the white queen and another white pawn. They were no longer on the board itself.
Part 1: 150 (base ten) is equivalent to 1100 (base five) or, for Part 2, 2112 (base four); B = 4.
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).
23 x 31 = 713; Max’s three different favourite numbers are 7, 3 and 1.
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.
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.
The magical number from my early teens is 11123595505617977528091. Note that 11123595505617977528091 ÷ 99 = 112359550561797752809.
LE HAD ONE RULER →LEONHARD EULER, born in 1707.
In the puzzle, Superman is 500 feet up from the base of the Burj Khalifa at a 30° angle of elevation from where I am, while Max is 1500 feet up at a 60° 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 😊😊
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+.
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! 😊♥😊
Cute Coco likes square-based boxes…so we’ll certainly feature some sneaky puzzles about squares! 😊There on the comfy sofa, Coco, Marron and Noisette are dreaming up some fun puzzles in early celebration for Jan turning N*(N+1)*(N+2) years old next month, on day number (N+2)*(N+2) + (N+4)*(N+4) of this year 2023 😊♥😊
COOL CATS CHALLENGES!
How old will Jan be on his birthday next month?
What was Jan’s exact date of birth?
Remove just one letter from the word FEELING and rearrange the remaining letters to make a proper, six-letter English word.
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. M😊’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?
The next brainteaser is going to feature Jan’s three cats…and one of my own… so that we can have a funny tale of pairs of pairs of furry tails!! 😊♥😊♥
7.
Imagine that I write down the ages (which are all different whole numbers of years) of Jan’s three cats and of my cat. The range (or age difference) between the oldest cat and the youngest of the four is exactly double Jan’s favourite number.
For each possible pair of cats among the four, I write down the sum of the two cats’ ages, until the list of sums is complete. Next, for each possible pairof sums from that list, I calculate the sum of the two sums involved! I put the results of those particular calculations in a box. The smallest result in the box is 22, and the largest result in the box is 50. Your fun brainteaser is to figure out Jan’s favourite number ! 😊😊
By the way, where does a cat go if it loses its tail…?!
…It goes to the retail store!!
Did you hear about the secret agent Maths teacher who couldn’t get on the plane?!… …He let the cat out of the bag!! 😊😊
Bright B😊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’.
Black is threatening …Bh4#, but…it’s White to play and force checkmate in just 3 moves! 😊
10. The Cats’ Sky-HighB😊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 😊
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 😊.
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 ♥
With kindest wishes as always,
Paul M😊twani ♥
“You have already been pruned and purified by the message I have given you.”–Bible verse, John 15:3 ♥
Here’s Wishing You a Happy Valentine’s Day in 3 x 3 days from now…😊♥😊
P.S. = Puzzle Solutions (being posted on 17.2.2023)
Jan will be 3 x 4 x 5 = 60 years old on his birthday next month.
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.
Jan’s favourite number is 7. Jan’s favourite colours are Red and Green.
3 x 7 x 9 = 189.
In the Chess puzzle, White forces checkmate with 1 Rh5+! Kxh5 2 Qh7+ Kg5 3 h4#.
Part 1: The maximum possible perimeter is 84 + 1763 + 1765 = 3612 cm; the maximum possible area is 84 x 1763 ÷ 2 = 74046 square centimetres; Part 2: The minimum possible perimeter is 84 + 13 + 85 = 182 cm; the minimum possible area is 84 x 13 ÷ 2 = 546 square centimetres.
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 😊
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!
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!
Jan, Raphaël and I love the breathtaking ‘cascade de surprises’ that one can experience in Mathematics, wonderful Chess studies, and via marvellous compositions in Music, for example. Right after our Chess at lunchtime on Tuesday this week, Jan treated me to his interpretation of one of Beethoven’s masterpieces, which he described eloquently as, “Playing Chess with Sound” ♥
It’s White to play and win! Have fun solving the way! 😊♥😊Happy birthday today, Peter! Have a really wonderful time, and keep enjoying music! ♥
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 CDseach, 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. 😊♥😊
Out of the many millions of books that have ever been written, if I had to pick just 3–my absolute favourite number!–of them to keep enjoying forever, then my top selection would probably be the following:-
The Holy Bible, because it’s a perfect book revealing to us the Word of God, which can be trusted totally and is of supreme importance.
My selected text today is: Jesus said, “I am the way, the truth, and the life. No one comes to the Father except through me.”-John 14:6 ♥
2. After my clear first choice above, it’s not at all easy for me to pick a second book in preference to all other books, but I’m sure that a very strong candidate would be: ’15 Minutes Alone With God’ by Bob Barnes.
Though the book has the subtitle ‘For Men’, every page has wonderful reflections for everyone. Here is a short extract from page 9 of my copy: “Time with your heavenly Father is never wasted. If you spend time alone with God in the morning, you’ll start your day refreshed and ready for whatever comes your way. If you spend time alone with Him in the evening, you’ll go to sleep relaxed, resting in His care, and wake up ready for a new day to serve Him.” ♥
If I fast-forward to pages 185-188 of the book, there’s a four-page article entitled I Didn’t Believe It Anyway, which includes the following powerful poem:
‘Twas thenight before Jesus came and all through the house
Not a creature was praying, not one in the house.
Their Bibles were lain on the shelf without care
In hopes that Jesus would not come there.
The children were dressing to crawl into bed,
Not once ever kneeling or bowing a head.
And Mom in her rocker and baby on her lap
Was watching the Late Show while I took a nap.
When out of the East there arose such a clatter,
I sprang to my feet to see what was the matter.
Away to the window I flew like a flash
Tore open the shutters and threw up the sash!
When what to my wondering eyes should appear
But angels proclaiming that Jesus was here.
With a light like the sun sending forth a bright ray
I knew in a moment this must be THE DAY!
The light of His face made me cover my head.
It was Jesus! Returning just like He had said.
And though I possessed worldly wisdom and wealth,
I cried when I saw Him in spite of myself.
In the Book of Life which He held in His hand
Was written the name of every saved man.
He spoke not a word as He searched for my name;
When He said, “It’s not here,” my head hung in shame.
The people whose names had been written with love
He gathered to take to His Father above.
With those who were ready He rose without a sound
While all the rest were left standing around.
I fell to my knees, but it was too late;
I had waited too long and thus sealed my fate.
I stood and I cried as they rose out of sight;
Oh, if only I had been ready tonight.
In the words of this poem the meaning is clear;
The coming of Jesus is drawing near.
There’s only one life and when comes the last call
We’ll find that the Bible was true after all!
3. No further book is really needed, but still I thank God every day for having let me enjoy many thousands of fascinating puzzles in my life so far. For me, a compilation of all those puzzles, about Chess, Mathematics, Words and more, would certainly be a treat! I believe that the puzzles in store in Heaven will be better and more magical than I can possibly imagine. For the moment, I can only offer what I know right now. So, I would like to share some surprises with you, since God gives us good thoughts to be shared. Here comes fresh puzzle ideas that came yesterday evening and in the morning today…with some extra bonuses this evening!
I would like to specially dedicate the puzzles to my excellent colleague Jens Van Steerteghem, his brother Nick, and their father Eric, as all three gentlemen are passionate about puzzles and have great talent for solving them!
Get ready for a race…but first rearrange the letters of TIME RAN to make a seven-letter female first name, the nice name of Jens & Nick’s mother 😊. A wee clue is that her name begins with MA, and it’s a very popular name in Belgium 😊.
A SNEAKY SPEED BRAINTEASER 😊♥😊
The name ‘Eric’ always makes me think of the famous missionary Eric Liddell–affectionately known as ‘The Flying Scotsman’–who won the Gold Medal in the 400m race at the 1924 Paris Olympics. Fast-forwarding 99 years to the present 2023…imagine that Eric Van Steerteghem runs a long distance from A to B at an average speed of 3 metres per second. On the way back from B to A (following exactly the same route as before, only in the opposite direction, and naturally more tired than before), Eric’s average speed is 2 metres per second.
The brainteaser is to figure out Eric’s average speed for his entire run from A to B to A. (Being an expert in Physics, Mathematics and more, Jens could tell you immediately that the average speed will not be 2.5 metres per second! Can you do like Jens and figure out the correct value?)
CHINESE FOOD WORD PUZZLE 😊♥😊
Thinking of delicious Chinese food… rearrange the letters of PRC CALORIES to make a proper 11-letter English word!
CUBOID BRAINTEASER 😊♥😊
Part 1: As a very quick warm-up before the main Part 2, imagine a cube with its dimensions (equal length, width and height) in centimetres (cm).
If a particular cube’s volume (in cubic centimetres) is numerically equal to its total surface area (in square centimetres), then what must be the cube’s exact dimensions?
Part 2: If the length, width and height of a certain cuboid are all exact whole numbers of centimetres, and if the cuboid’s volume (in cubic centimetres) is numerically equal to its total surface area (in square centimetres), then what is the maximum possible height of the cuboid?
Your super-fun brainteaser challenge is to find nine different positive whole numbers to fill the nine grid boxes (with one number per box) so that the total product (when you multiply all the nine numbers together) will be 1000000000 = 1 billion. Also, the mini-products of the three numbers in any row or in any column or in either of the two main diagonals should always give the same results in each of those eight cases. That is: three row products, three column products & two main diagonal products must all equal each other.
RECIPROCALS BRAINTEASER 😊♥😊
For this puzzle, we need to know that, in Mathematics, the reciprocal of any non-zero number n is 1 ÷ n.
Imagine that a lady on her birthday today said, “The difference between the reciprocal of my new age now and the reciprocal of the age I’ll be in a year from now is equal to the reciprocal of the year when my younger sister was born.”
Your brainteaser is to figure out the lady’s new age now, and figure out the exact year when the lady’s younger sister was born.
You know that I like the number 141, as it features in one of my email addresses, pmotwani141@gmail.com. Here in blog post #146, I should give a special mention to the number 14641, which equals the fourth power of my house number 11 😊
146 BRAINTEASER 😊♥😊
Part 1: In our normal base ten, the number 146 = 6 x 1 + 4 x 10 + 1 x 10 squared.
However, in another base B (not base ten), 222 (base B) = 146 (base 10).
Figure out the value of B.
Part 2: This involves a new base, N. We are told that
222 (base N) = xyz (base 10),
where xyz represents a proper three-digit whole number.
The brainteaser is to figure out the maximum possible value of xyz, and the corresponding value of N.
A Wee Dose of Chess To Finish! 😊♥😊
Part 1: Though White is down on material, it’s White to play and win. Part 2: If it were actually Black to play, what would be the strongest move?
It’s my intention to publish solutions to all the puzzles around the time that blog post #147 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 ♥
P.S. = Puzzle Solutions!
TIME RAN = MARTINE
Eric’s average speed for the entire run was 2.4 metres per second. That can be verified using the formula Average Speed = 2vw ÷ (v+w), in which v=3 and w=2, the respective speeds for the outward and return runs covering equal distances.
PRC CALORIES can turn one’s diet upside down because they make RECIPROCALS !! 😊
Regarding length L, width W and height H, when a cube has L = W = H = 6cm, then its volume = 6 x 6 x 6 = 216 cubic centimetres, and the total surface area of its six faces is 216 square centimetres because each one of the faces has an area of 6 x 6 = 36 square centimetres.
In the cuboid part of the brainteaser, a maximum whole number height of 42cm is achievable when the length and width are 3cm and 7cm in either order.
Given that the volume was numerically equal to the total surface area, I used LWH = 2LW + 2LH + 2WH and then 1 = 2/H + 2/W + 2/L.
Letting L=3 helps to ‘use up’ two thirds of the 1, leaving only one third or 2/6. So W can’t then be 6, but it can be 7, letting us solve directly for the optimal H.
(If the length and width are 4cm and 5cm in either order, we would find that H = 20cm; smaller than our optimal 42cm.)
In the Multiplication Magic Square, we must have the number 10 in the central box, and all the other factors of one hundred can be filled in the rows in (for example) this order (starting from the top-left box):- 20, 1, 50; 25, centre 10, 4; 2, 100, 5.
In the reciprocals brainteaser, the lady is 44 and her younger sister was born in the year 1980.
It makes use of the fact that 1/n – 1/(n+1) = 1/(n(n+1)). In the puzzle, n(n+1) has to be the year when the younger sister was born. The only value for n that gives a suitable value for n(n+1) in the reasonably recent past is n=44, and then n(n+1) = 44 x 45 = 1980.
In the number bases brainteaser, 222 (base N) has the value of 2 + 2 x N + 2 x N squared. If you were to generate an accurate table of different values for N and the corresponding values of 2 + 2 x N + 2 x N squared, it would show, for example, that 2 + 2 x N + 2 x N squared = 146 when N = 8 and 2 + 2 x N + 2 x N squared = 926 when N = 21 and 2 + 2 x N + 2 x N squared = 1014 when N = 22.
So, the answers asked for in the puzzle are:- B = 8; xyz = 926; N = 21.
In the Chess puzzle, 1 Bg5+ Kg8 2 Qh7+ Kf8 3 Qh8# is the fastest win for White.
If it were actually Black’s turn to move, then (though it’s true that 1…Qxg3+ would win easily) the quickest forced win is 1…Rh1+! 2 Kxh1 Qg1#, a key checkmating pattern 😊
My family and I hope that this message finds you keeping well, and we would like to wish you a very happy new year, 2023 ♥
We’ve got some super-fun puzzles for you to enjoy solving! 😊♥😊
Puzzle #1: A Magical New Year Brainteaser!
Michael and Jenny write down two positive whole numbers. They calculate the sum by adding their two numbers together, and they also calculate the product by multiplying their two numbers together.
Guess how many whole numbers there are from their sum up to and including their product as well…there’s exactly 2023 whole numbers !
Your fun challenge is to discover Michael & Jenny’s numbers!
There are three different possible solutions! 😊♥😊
Additional Notes:
1. The purposeful meaning behind the words ‘as well’ above is that the sum and the product are both included in the 2023 whole numbers, with the sum at the very start and the product at the very end. Be extra-careful when considering the difference between the product and the sum…
2. Very sincere thanks to Teun Spaans (on 6 January 2023) for having kindly mentioned this brainteaser on his site justpuzzles.wordpress.com, but please note well point 1, just above.
Some Maths students know that 5! means 1 x 2 x 3 x 4 x 5 = 120.
Here in blog post #145, it’s nice to note that the special number 145 = 1!+4!+5!
Puzzle #2: A Good Book Puzzle ♥
A boy is enjoying currently reading two of the 66 distinct books of The Bible that he has been given as a gift ♥
Your puzzle is this: If you wanted to choose two out of 66 distinct books to read, how many different selections would be possible?
Puzzle #3: A Champion’s Challenge 😊
I recently received a happy message from Cansu, an excellent former Maths student of mine whom I always think of as ‘Champion Cansu’! 😊
In honour of C CANSU, use the numbers 3, 3, 1, 14, 19, 21 (in any order that you want) to make the target number 2023. You may also use parentheses ( ) and any of the operations +, -, x, ÷ as you wish. 😊
Puzzle #4: A Neat Word Puzzle
Rearrange the letters of the word LISTEN to make another proper six-letter English word.
There are four different possible solutions! ♥😊😊♥
Puzzle #5: Dedicated to Elton (a former student of mine from 2012-2013 who likes the Royal Game of Chess) 😊
Black appears to be down on material by more than 2 knights… but there’s an invisible black knight somewhere on the f-file… Can you discover two possible locations for it, so that in either case it will then be Black to move and win by force!?
Puzzle #6: OUR ANGLE Brainteaser 😊
The OUR ANGLE picture is composed of several straight lines (ORGE, URA, LGA, NR, NG). x and y are positive whole numbers. The angles x°, 2x°, y°, 2y° are not drawn to scale.
Your brainteaser is to figure out the maximum possible size (in degrees) for angle RAG that fits correctly with the given information.
It’s my intention to publish solutions to all the puzzles around the time that blog post #146 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 2023, with lots of happiness in everything that you do ♥
With kindest wishes as always,
Paul M😊twani ♥
“You have decided the length of our lives. You know how many months we will live, and we are not given a minute longer. You set the boundary, and no one can cross it.”–Job 14:5
Joke: Why should you keep your left foot still at the beginning of January?
You’ll start off the New Year on the right foot!
P.S. = Puzzle Solutions (being posted now on 6 January 2023)
Magical New Year Brainteaser
If we let Michael & Jenny’s positive whole numbers be x and y, then their sum is x+y and their product is xy, which are respectively the very first and very last of 2023 consecutive whole numbers. Therefore, the difference xy – (x+y) = 2022 (not 2023). However, the puzzle can be solved really neatly by noting that xy – x – y + 1 = 2023, because we can then use factorisation to get that (x-1)(y-1) = 2023. In other words, 2023 is the product of whole numbers (x-1) and (y-1) which must both be factors of 2023 😊😊
2023 = 1 x 2023 or 7 x 289 or 17 x 119 (because 2023 = 7 x 17 x 17), and so we get that x-1 = 1 & y-1 = 2023 or x-1 = 7 & y-1 = 289 or x-1 = 17 & y-1 = 119, leading to
(x, y) = (2, 2024) or (8, 290) or (18, 120). Those are our 3 distinct solutions ♥
Naturally, x & y are interchangeable, but in the context of this number puzzle we wouldn’t count 2024 & 2 (for example) as being a different pair from 2 & 2024.
A Good Book Puzzle
The number of different possible selections of two books from 66 distinct books is 66 x 65 ÷ 2 = 2145, which is nice here in blog post #145 😊
A Champion’s Challenge
We saw in the first puzzle that 2023 = 119 x 17, and having that target in mind can help us to use 3, 3, 1, 14, 19, 21 to easily make 2023 as follows:
((19 + 21) x 3 – 1) x (3 + 14) 😊♥😊
A Neat Word Puzzle
Listen = Silent = Tinsel = Enlist = Inlets ♥
In the chess puzzle, if Black’s invisible knight is on f4, then 1…Ne2# delivers checkmate instantly!
Alternatively, if the invisible knight is on f2, then Black wins beautifully with 1…Rh1+! 2 Bxh1 Nh3+ 3 Kh2 Qe2+! and then either 4 Bg2 Qxg2# or 4 Kxh3 Qh5#, a very pretty checkmate! 😊♥😊
OUR ANGLE Brainteaser
The OUR ANGLE picture is composed of several straight lines (ORGE, URA, LGA, NR, NG). x and y are positive whole numbers. The angles x°, 2x°, y°, 2y° are not drawn to scale.
Note: In all of the following steps of working, the angle sizes are in degrees.
Step 2: Angle RNG = 180 – angle NRG – angle NGR; after now using results from Step 1 above, and simplifying the algebraic terms, we get that angle RNG = 3x + 3y – 180 or, in factorised form, angle RNG = 3(x + y – 60).
Step 3: From the result of Step 2 above, we can conclude that x + y must be greater than 60.
Step 4: Angle ARG = 2x & angle AGR = 2y, so angle RAG = 180 – 2x – 2y, which can also be written as 180 – 2(x + y).
Step 5: Since x + y is greater than 60 (from Step 3 above), angle RAG must be less than 180 – 2 (60); so angle RAG must be less than 60 degrees. That would normally be our final conclusion… However…
Step 6: Since we were given that x and y are positive whole numbers, then the minimum possible value for x + y is 61, to be greater than 60.
Therefore, when x & y are whole numbers,
the maximum possible measure for angle RAG is 180 – 2 (61) = 180 – 122 = 58°.
(It’s also worth noting that, when angle RAG = 58°, angle RNG = 3°.)
Very warm congratulations to everyone who enjoyed trying and solving some or all of the puzzles 😊♥😊
We can see a piece of Heaven in places of peace on Earth ♥
When this day (19 December 2022) began, there were exactly six days = 144 hours until the start of Christmas Day ♥
By midday at the centre or heart of each new day, God has granted us another 144 super-precious 5-minute gifts of time (as 144 x 5→720 minutes = 12 hours), and it’s good to aim to use all of them to thank and honour Him in everything that we do ♥
I found today that ABCDF contains more than one interesting idea. Thinking of the product of the numbers that correspond to the normal positions of the letters, 1 x 2 x 3 x 4 x 6 = 144, right here in Blog Post #144. I count myself and everybody as being the missing ‘E’ for ‘EVERYONE’… I know that we will all go on to G, H, I, J, K, L…with God in Heaven provided that we believe gratefully In Jesus, King of Love ♥
We could note very briefly in passing that L→12, the (square) root of 144, but it’s really much more important to recognise properly that God’s Love for all of us is at the root of our salvation.
“For God so loved the world that He gave His one and only Son, that whoever believes in Him shall not perish but have eternal life”–John 3:16
As many of my students–and my colleagues–like puzzles, I’ll offer several now for everyone’s enjoyment 😊
Word Puzzles
Rearrange the letters of NEAR GIANT to make the name of a 9-letter country.
Rearrange the letters of MR TALENTS IN U to make a proper 11-letter word that all the students of Musica Mundi School use 😊♥😊
Numbers Puzzle
Use all of Cleo’s numbers 1, 6, 3, 6 and any operations that you want from+, -, x, ÷ to get the target number of 144. You may also use parentheses ( ) as you wish.
Special Puzzle in honour of my colleague Jens Van Steerteghem
JENS is a brilliant physicist, chemist & mathematician, too. Suppose that J=10, E=5, N=14 & S=19. Now here comes a super-fun puzzle…Add up the values of any two or any three of the four letters, then multiply by the left-over total, and divide by 4 in honour of JENS.
What is the maximum possible final result?
It’s a delightful wee Christmas present to Jens and everyone who loves mathematical puzzles, and in fact it’s possible to figure out the answer mentally using nice logic 😊♥😊
Puzzle Regarding A Lovely Lady (use A=1, B=2,…,Z=26 in this puzzle)
A former colleague of mine from my previous school in Belgium sent me a lovely Christmas card by email. In the lady’s six-letter first name, there’s no A. The product of the values of a certain four of the letters is 100. The product of the values of a certain five of the letters equals 5 x 144.
What is the exact product of all six letters? 😊
B😊NUS: Can you make a smart conjecture to guess the lady’s well-known first name, that many ladies have had?
A Really Beautiful Brainteaser ♥😊♥ (use A=1, B=2,…,Z=26 in this brainteaser)
Raphaël writes a proper six-letter English word that uses 5 distinct letters, including an R. The product of the values of all the letters in the word uses 4 distinct digits and begins with 5703…
Your mega-fun brainteaser is to figure out Raphaël’s six-letter word 😊
A Memorable Chess Game between Timothée and Raphaël at Musica Mundi School ♥😊♥
Chess Game in honour of my colleague Emile Daems
Emile Daems–a great colleague of mine at Musica Mundi School in Waterloo, Belgium–has been enjoying discussing some famous Chess openings with me. So, I’m including a photo of a recent game for Emile and other fans of The Royal Game ♥
Black’s sixth move, 6…Nd7!?, is rarer than 6…Ne4, but it’s very interesting 😊
A Whisky Puzzle in honour of my friends ‘Happy’ & Mandi!! 😊😊
Many people from Scotland and elsewhere enjoy some whisky! The word ‘whisky’ also brings ‘malt’ to mind, or the adjective ‘malty’. Here’s one description that I came across: “Generally speaking, a malt taste can be described as having a combination of flavours. It tastes sweet and nutty, but is also described as tasting similar to toast, caramel, coffee or fruits like raisins. The reason for its sweet, almost dessert-like taste has to do with how malt is made from barley.”
Here’s the puzzle now: Use all the letters of MALTY + H to make the name of a six-letter town where my friends ‘Happy’ & Mandi live, in the English county of Lancashire 😊
Here’s thinking of you, dear ‘Happy’ & Mandi 😊😊
A Maths Mega-Brainteaser is coming next, in honour of my students and my brilliant colleague Jens & his ingenious brother Nick Van Steerteghem (who wrote a computer program specifically to solve another recent brainteaser!–special congratulations also to Raphaël Murphy who solved the brainteaser directly himself!!)
Part 1
Imagine that I put the numbers 1, 2, 3, 4, 5, 6 & 7 (one of each) in a bag.
Raphaël, Tarik and Wout each take a number out of the bag.
Damla, Sophie and Jens also each take a number out of the bag.
I then announce, “The total sum of Raphaël’s, Tarik’s and Wout’s numbers is exactly equal to the total sum of Damla’s, Sophie’s and Jens’ numbers! Furthermore, the total sum of the squares of Raphaël’s, Tarik’s and Wout’s numbers is exactly equal to the total sum of the squares of Damla’s, Sophie’s and Jens’ numbers!!”
Your fun brainteaser is to figure out exactly which number is still in the bag.
Part 2 (Super-Tricky!!)
A large group of Maths fans goes with me to visit an old Spiritual Maths monk at a monastery on a high hill. The number on the monastery building is a proper three-digit number. When we get to the door and see the number, I tell everyone that the sum of the squares of its digits equals the monk’s age!
The monk’s age is also equal to the sum of the fourth powers of three distinct positive whole numbers (just meaning different from each other).
Your mega-challenge brainteaser is in three parts:-
2.1: How old is the monk?
2.2: What is the smallest possible three-digit number that could be on the monastery building, given the clues above?
2.3: What are the fourteen different possible three-digit numbers that could be on the monastery building, consistent with the clues? (Note: That’s fourteen answers including the answer to part 2.2.)
Note also that the Spiritual Maths monk is not hundreds of years old like Yoda!! The monk is at an age that many other people have reached, too, in human history. Of course, I didn’t strictly need to mention that. Even if the monk’s current age could have been, say, 144–or 22 years more than the oldest person on record to-date 😊!!–the first clue relating his age to the sum of the squares of the digits on the three-digit door number meant that he couldn’t be more than 9 squared + 9 squared + 9 squared, or 243 years old now!! Still, my sincere answer to the question, “What could be better than living to be 144+99 years old?” is: “Living forever in Heaven.”
It is my intention to publish full solutions (God-willing, as always) to Blog Posts #142 & 144 before 9 January 2023, when the next semester’s lessons at Musica Mundi School begin.
In the meantime, dear students, colleagues and other readers, please do feel free to send in your best solutions to any or all of the puzzles, if you like ♥
The school’s cats visit the Maths classroom frequently, and they’re already pondering the fresh puzzles now!! 😊😊😊
My family and I would like to wish you and everyone a very blessed, merry Christmas soon, followed by a wonderful, happy New Year ♥♥♥
With kindest wishes as always,
Paul M😊twani ♥
My colleague Solomon is embarking on a special journey very soon. May God bless you in all that you do, dear friend ♥
Meanwhile, here’s wishing Headmaster Herman a very happy birthday tomorrow!
Enjoy a lovely piece in perfect peace! ♥
Certainly let any nice school cat add the missing ‘E’ (from much earlier in the article), because SCHOOL CAT + E = CHOCOLATES
for everyone…
…and add comes from 144 😊
P.S. = Puzzle Solutions (being posted on 31.12.2022)
NEAR GIANT = ARGENTINA
MR TALENTS IN U = INSTRUMENTAL
Using the numbers on Cleo’s sweater, (6×6)(1+3) = 36 x 4 = 144
In the puzzle about JENS, since J=10, E=5, N=14 & S=19 have a total sum of 48, we can achieve the optimal result with (10+14)(5+19)÷4 = 24 x 24÷4 = 144 again! 😊😊
In the puzzle about the lady’s six-letter name, the product of all her letter values = 3600, the LCM of 720 (= 5 x 144) & 100
Her name is DEBBIE (with 4 x 5 x 2 x 2 x 9 x 5 = 3600) 😊
Raphaël’s six-letter word is MEMORY, for which the product of the letter values is 13 x 5 x 13 x 15 x 18 x 25 = 5703750
(There were some cases to check, but since we were given that the word included an R–with letter value 18–the overall product had to be an even number and also a multiple of 9, and so in this particular puzzle the last digit had to be 0 & the sum of all the digits had to be a multiple of 9, which helped enormously to narrow down the cases for checking 😊)
MALTY + H = LYTHAM, where ‘Happy’ & Mandi live 😊😊
In part 1 of the Maths Mega-Brainteaser, 2+3+7=1+5+6 and, crucially, 2 squared + 3 squared + 7 squared = 1 squared + 5 squared + 6 squared; the unused number 4 is left in the bag 😊
In part 2, we don’t have to consider 4 to the power of 4 because that’s 256, which is a bit too old!! Instead, 1 to the power of 4 + 2 to the power of 4 + 3 to the power of 4 = 1 + 16 + 81 = 98, the monk’s true age ♥
Now we know that 98 = the sum of the squares of the digits of the monastery’s three-digit door number, which could be any of:-
since 1 squared + 4 squared + 9 squared = 3 squared + 5 squared + 8 squared = 7 squared + 0 squared + 7 squared = 98 in those cases or permutations of them; clearly 149 would be the smallest possible proper three-digit door number meeting the monk’s requirements! 😊
Theodore Roosevelt famously said, “A thorough knowledge of The Bible is worth more than a college education”, and I’m certain that he was right. Personally, I’ll always be truly grateful for all my excellent teachers of Mathematics, Physics, Chess, and many other fascinating subjects, but it’s still much more important to come to know God’s Word well through The Bible, which contains all the examples, truths, and commands that God wants us to know, believe, respect, and live by. In prayer, we can ask God to grant us increased understanding, wisdom and good powers of discernment. Mark 12:30 gives a clear, unmistakable message: “Love the Lord your God with all your heart and with all your soul and with all your mind and with all your strength.”
There are many people who want to have faith in God, but may be struggling with doubts. Three lines below here, there follows a free link directly to a really helpful recent article by Rev. John Piper (the founder of http://www.desiringgod.org)→
It’s well worth thinking about that article carefully. The good author has done a superb job of helping to answer questions which have troubled billions of people in the past, right up to the present time.
I am also thankful for every moment I have as the Mathematics teacher at Musica Mundi School (MMS), where I have the privilege and pleasure of working with wonderful students and colleagues. Since many of them enjoy puzzles, I will do my best to offer a bumper-sized gala feast now near Christmas! ♥😊♥
First, rearrange the letters of now King to make a proper seven-letter English word.
Also, rearrange the letters of xy gala to make a proper six-letter English word.
Next, here comes a wee joke… What might someone say if he/she didn’t know the meaning of the word ‘AXON’ in Biology?…”It’s getting on my nerves!!” 😊
Thinking of miracles performed by Jesus, use the letters of GLORY HEAL AXON to make OR + a proper 11-letter Maths-related English adverb. There’s a clue in the lovely photo just below…
Damla & Sophie are talented, young mathematicians, as is Valentin (pictured in the next photo).Valentin’s eraser shows a famous image of Wolfgang Amadeus Mozart. Mozart’s birthday was 27 January, which brings to mind the three-digit number 271. Quick Puzzle: Can you read my mind and say why I’m also thinking of 2 x 71 now!?
Just before we meet more puzzles, I’d like to share a happy, musical memory with you… One evening in the school a couple of years ago, I was sitting doing Maths work in the dining hall, while simultaneously enjoying listening to a beautiful music album. A Turkish student named Idil (a fabulous cellist) came into the hall and immediately asked about the lovely album. She liked it so much that I sent her a link to the full album, right away. I think that it’s perfect for Christmas, and so I’m sharing it now with you, too! 😊♥😊
Duration: Just under 44 minutes. Happy listening!
I would also like to take this opportunity to wish all MMS students lots of enjoyment and success in all the concerts in which they are performing during this wonderful, festive period ♥
Musica Mundi School concerts are a delight to all who experience them ♥
Meanwhile, more puzzles are coming in fast now!
Imagine that a lady once declared on a weekday, “I know that I was working 142 days ago because it was a weekday back then. My birthday is 142 days from now, and that will also be a weekday; not during a weekend.”
Your fun challenge puzzle is to figure out what day of the week the lady’s birthday was going to be on. Also, on what day of the week did the lady make her statement?
Next, can you think of seventy-one consecutive integers which have a total sum of 142 ?
Here comes another interesting puzzle… The total sum of all the prime numbers less than N is a whole multiple of 71. If the prime number N is included too, then the new total sum is also a proper multiple of 71. Exactly what number is N ?
And now I’ll share a very short, true story with you… Just a few days ago, while driving on my way to work, I found myself behind a vehicle with a number plate that included 2 ♥ 7 1, where ♥ stands for a particular digit. Given the value of ♥ that I saw, I realised that the four-digit number 2 ♥ 7 1 must be a proper multiple of 99. Your fun challenge is to figure out the value of ♥. Also, in general, if ABCD is a proper multiple of 99, then what is the special connection between AB and CD ? (One exception to the general rule is 9999, in which AB = 99 = CD.)
BRAINTEASER
Imagine that XYZ stands for a proper three-digit whole number.
Prove that XYZ times 3 can never equal YZX.
Find two solutions to the equation XYZ times 3 equals YZW, where the only restriction is that W does not equal X.
It’s time now for a Chess puzzle to delight Timothée, Raphaël, and other fans of the Royal Game ♥Black would love to queen the g-pawn, but… right now it’s White to play and force checkmate in just three moves! ♥
By the way, let’s call Raphaël’s favourite whole number R. It’s a single-digit number. Here in blog post #142, it’s nice that 1 ÷ R gives the recurring decimal 0.142857142857… Also, 2 ÷ R or 3 ÷ R or 4 ÷ R or 5 ÷ R or 6 ÷ R all produce decimals with exactly the same recurring digits, in different orders. If you’ve already figured out the value of R, then it will only take an extra moment to figure out the date this month of Headmaster Herman’s birthday, coming in R+1 days from now 😊
Here’s sending early good wishes for Headmaster Herman whose birthday is coming in R+1 days from now 😊
It’s my hope and intention (God-willing as always) to publish solutions to the puzzles around the time of the next blog post. In the meantime, please do feel free to send in your best solutions to some/all of the puzzles, if you like 😊
My family and I would like to wish you and everyone a very blessed, merry Christmas and a happy New Year, coming soon ♥😊♥
Jesus said, “In My Father’s house are many rooms. If that were not so, I would have told you, because I am going there to prepare a place for you.” John 14:2
We don’t have to wait until Easter–or even until Christmas– to stop and recognise with gratitude Jesus’ sacrifice to save us all ♥ “For God so loved the world that He gave His one and only Son, that whoever believes in Him shall not perish but have eternal life”– John 3:16.
With love and kindest wishes as always,
Paul M😊twani ♥
P.S. = Puzzle Solutions (being posted on 30.12.2022)
now King = Knowing
xy gala = galaxy
GLORY HEAL AXON = OR + HEXAGONALLY
2 x 71 = 142, the number of this blog post
142 days = 20 weeks + 2 days, so the lady was speaking on a Wednesday, looking forward to her birthday to come on a Friday (20 weeks + 2 days later), while also thinking back to a Monday when she was working, 20 weeks + 2 days earlier
The seventy-one consecutive integers from -33 to 37 have a total sum of 71 x the number at the very middle of the long list of numbers = 71 x 2 = 142, as required 😊
The prime number N = 71
♥ = 8; 2871 = 99 x 29; note that 28 + 71 = 99, and in general AB + CD = 99 when ABCD is a multiple of 99 (except in the case of 9999)
Consideration of the place values of the digits in XYZ shows that XYZ = 100X + 10Y + Z, while YZX = 100Y + 10Z + X;
it’s not possible that 3(100X + 10Y + Z) = 100Y + 10Z + X because that would imply that 300X + 30Y + 3Z = 100Y + 10Z + X, leading to 299X = 70Y + 7Z, which is a clear contradiction since 70Y + 7Z = 7(10Y + Z), a multiple of 7, but 299 is not a multiple of 7 and if X=7 then 299X = 299 x 7 = 2093 which is far too big to equal 70Y + 7Z, given that digits Y & Z can’t be bigger than 9; so the proof is complete 😊
In part 2 of the brainteaser, the requirement boils down to
300X – W = 7(10Y + Z);
if X = 1 & W = 6, then 300X – W = 300 – 6 = 294 = 7 x 42, a multiple of 7, so we’ve found that it works with 3 x 142 = 426;
similarly, if X = 2 & W = 5, then 300X – W = 600 – 5 = 595 = 7 x 85, a multiple of 7, so we’ve also found that it works with 3 x 285 = 855;
there are no higher solutions, because if X = 3 or more, then 300X – W would be far too big to equal 7(10Y + Z), given that digits Y & Z can’t be bigger than 9 😊
In the chess puzzle, White forces checkmate with 1 b6! cxb6 2 a7 followed by 3 a8=Q#, and ‘underpromoting’ to a rook at the end is also perfectly sufficient 😊
Raphaël’s favourite number is R=7
Headmaster Herman’s birthday came on December 20, which was 7+1 days after this blog post was first published (on December 12) ♥
Paul Motwani 