Dear All,

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

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

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

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

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

**Prize Puzzle #1**: **Anagram starring Harriet & AndrΓ©e** π

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

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

**HARRIET SENDS**

**ANDREE – – – – – –**.

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

**Prize Puzzle #2:** **Happy House Number** π

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

**Prize Puzzle #3: Starring AndrΓ©e, Anthony & Zoe** πππ

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

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

**Prize Puzzle #4: Thinking of AndrΓ©e’s Party** πΆππ

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

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

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

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

*What exactly is my mystery number, M?*

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

With lots of love always

From Jenny, Michael and me xxx

Paul π

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

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

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

π ** AndrΓ©e, Anna, Anthony, Chantal & John, Deborah, Hilary & Richard, Susan, Zoe **π

During the coming week, I will also give a prize in person at Musica Mundi School to my terrific puzzle-solving colleague, ** Jens Van Steerteghem**, and I’ll certainly include a gift for

**, who loves having fun with all the blog posts π**

*Jens’ father, Eric*It’s time now to share the puzzle solutions:-

- HARRIET SENDS is an anagram of ANDREE
**SHIRTS**π - 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,
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 βπ Here’s a fun extra-bonus challenge π:*AndrΓ©e’s house number 13*π*Not*counting my own house number, 11, or AndrΓ©e’s house number, 13, what is the next smallest whole two-digit house number which can be multiplied by another whole number to produce a result that consists entirely of nines? Exactly what number would you multiply the house number by in order to get the string of nines result? - Suppose that A, B & C are the positive whole numbers that AndrΓ©e, Anthony & Zoe wrote down, put in increasing order from smallest to largest. We were basically given that (A
^{2}+B^{2})/2=C & (B^{2}+C^{2})/2=157. So, A^{2}+B^{2}=2C & B^{2}+C^{2}=314. Noting that B^{2}=2C-A^{2}βC^{2}+2C-A^{2}=314βC^{2}< 314β**C β€ 17**. Also, C^{2}+2C+1-A^{2}=315β C^{2}+2C+1=315+A^{2}β(C+1)^{2}Λ 315βC+1β₯ 18β**Cβ₯17**. Having proved that C β€ 17**and**Cβ₯17, the unique solution has to be**C=17 exactly**. Since A^{2}=(C+1)^{2}-315, we get A^{2}=18^{2}-315=324-315=9β**A=3**. Recalling that B^{2}=2C-A^{2 }β B^{2}=34-9=25β**B=5**. Summary:**A=3, B=5, C=17**is the unique solution. Many congratulations indeed to everyone who found the numbers 3, 5 & 17. That puzzle was really well done πππ - “Thinking ahead to AndrΓ©e’s party, I have a particular English six-letter plural word in mind. If I remove the penultimate letter of the word, the remaining five-letter word is still a proper English word,
*and its last two letters are the same as each other*.” I intended the last part to refer to the remaining five-letter word, but if it would be referring instead to the original six-letter word, then there’s a lovely possibility which Zoe thought of: the word CARESSβCARES. The word that I actually had in mind was, and Anna and Deborah both got it! ππ*GUESTSβGUESS* - My mystery number,
. When whole numbers are divided by 7, the remainder is either 0, 1, 2, 3, 4, 5 or 6. The remainder is zero only when the dividend (the number being divided) is a whole multiple of 7. Otherwise, when the remainder is 1, 2, 3, 4, 5 or 6, the quotient involves the fractions 1/7, 2/7, 3/7, 4/7, 5/7 or 6/7. Their decimal forms are 0.142857…, 0.285714…, 0.428571…, 0.571428…, 0.714285…, 0.857142… which all involve 1, 5, 7 in some order and fit nicely here in Blog Post #157 π AndrΓ©e’s party is starting in 007 days from now…πππ*M = 7*