May 2023 – Paul Motwani

Blog Post #157: Nice Prize Puzzles in Honour of Andrée!! 🎁🎂💖😊

Dear All,

Last year, Jenny and I were delighted that our very dear friend, Andrée, was able to come from Luxembourg to celebrate my turning 60. Now we are really looking forward to going in two weeks’ time to join in lots of great fun for Andrée’s upcoming birthday! 😊🎂😊

I decided to offer a fresh collection of prize puzzles in honour of Andrée! 😎

The people who send me the best answers to all of the puzzles will receive a nice gift ✔🎁😍 I would still warmly encourage you to not be shy to send in your solutions even if you might not be sure about some of them.

I look forward hopefully to giving lovely gifts to some people at Andrée’s birthday party, to some students and colleagues at Musica Mundi School, and to other readers who may send in very good solutions 👌

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

Prize Puzzle #1: Anagram starring Harriet & Andrée 😊

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

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

HARRIET SENDS

ANDREE – – – – – –.

Harriet might say, “Why would I send those to Andrée?!” and Jenny could reply jokingly, “We’ve got Paul’s back! Let’s keep the word a secret…Sh….” 😉

Prize Puzzle #2: Happy House Number 😁

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

Prize Puzzle #3: Starring Andrée, Anthony & Zoe 😊💕😊

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

What exactly are the three numbers that Andrée, Anthony & Zoe must have written down? (As an extra, optional challenge for Maths fans, enjoy proving algebraically that the solution is unique. 👍)

Prize Puzzle #4: Thinking of Andrée’s Party 🎶🎂😊

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

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

Prize Puzzle #5: A Puzzle about most of Your Ages!! 😍

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

What exactly is my mystery number, M?

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

With lots of love always

From Jenny, Michael and me xxx

Paul 😊

“A sweet friendship refreshes the soul”–Bible verse, Proverbs 27:9 💖

***Thank You, Andrée, for taking that nice photo at number 11 eleven years ago!!*** 😍

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

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

🎁 Andrée, Anna, Anthony, Chantal & John, Deborah, Hilary & Richard, Susan, Zoe 💖

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

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

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

Blog Post #156: You Are A Lovely, Truly Delightful Miracle!! 💖😊

Dear All,

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

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

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

Nice Word Puzzle 😊

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

There are (at least) three possible solutions!

**Keep a smile on your face and a rainbow in your heart** 💕

I recently promised a friend that this article being offered freely here would include some good, fun number/number-theory puzzles. I wish you lots of enjoyment with them now. 😎✔

Power of Friends Brainteaser 🙌

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

A remarkable detail is the fact that 31 x 5 x (H^A + H^B + H^C + H^D) = H^E – H^A.

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

New 2023 Brainteaser 😎

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

Double All-Play-All Chess Brainteaser (with no Chess moves!! 😃♟🤣)

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

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

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

The brainteaser is to figure out, with proof, three possible values for P, the number of players. 👍👌

A Double High-Five Brainteaser!! 🙌🙌

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

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

Advanced Algebra Triple Brainteaser!!! 😍👌😎

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

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

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

Chess Puzzle (with moves this time!! 😃😂)

It’s my intention to publish solutions to all the puzzles around the time that blog post #157 comes out, God-willing as always.

In the meantime, please do feel free to send me your best solutions to any or all of the puzzles, if you like 😊.

I would like to round off this article by most sincerely wishing you a very blessed month of May and a wonderful weekend now, with lots of happiness in everything that you do ❤.

Beautiful Bonus! 😊🎂💖

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

The final result of…😃 5.5…is to wish you a beautiful birthday now on 5 May 😍🎂💕

With kindest wishes as always,

Paul M😊twani ❤

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

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

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

In the big H brainteaser, we were given that

31 x 5 x (H^A + H^B + H^C + H^D) = H^E – H^A

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

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

and ‘cancelling’ the common factor H^A on both sides of the equation gives

155 (1+H+H²+H³) = (H⁴ – 1)

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

155 (1+H+H²+H³) = (H-1) (1+H+H²+H³)

which simplifies to

155 = H-1

so H = 156, which is nice here in blog post #156 😍

In the ‘2023 Brainteaser’, we were given that

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

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

and so

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

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

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

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

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

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

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

which simplifies to 2 x 10^D-1 👍✔

In the advanced Advanced Algebra Triple Brainteaser, where

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

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

factorisation leads to

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

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

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

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

In the Chess puzzle,

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