Today, 25 November, it’s exactly one month until Christmas, and I recently promised some friends that I would do my best to publish a nice, fresh article before then. I had intended to wait a few more days, but I try to always remember that every precious moment counts, and to never make any arrogant assumptions regarding the future. An important reminder of that is, “Do not boast about tomorrow, for you do not know what a day may bring”-Proverbs 27:1.
While it’s not wrong to have good, personal hopes and intentions, we need to know and respect that God’s perfect plans will always prevail, and not always in the way or at the time that any person can predict.
Many of my friends like puzzles very much as I do, and so I will try to offer now some lovely puzzles, starting with one for a lady who has her birthday on day number D x D in December, when her new age will be the two-digit number EF (in which E is smaller than F, and note also that D, E and F are all whole numbers bigger than 1). Here in Blog Post #140, it’s nice that D x D x E x F = 140. Your first fun puzzle here is to figure out the date of the lady’s birthday, and the new age that the lady will be.
Next, I’m remembering Max (a student from my previous school), who liked to remember facts such as: there are 1440 precious minutes in a day. Your second fun puzzle is to figure out (preferably without using a calculator), what is the smallest positive whole number of standard days for which the total number of minutes will be a whole multiple of 140?
As 3 is my absolute favourite number, I hope that you’ll particularly enjoy the third fun puzzle, coming in just a wee M😊ment… There was just one particular year in the 20th century such that its four-digit number could be expressed as the product A3 x B3, where A3 and B3 are both two-digit whole numbers. A few days ago, I discovered that A3 x B3 + 14 x 140 = C3 x D3. The fun brainteaser is to figure out the precise values of A, B, C & D. (Naturally, A & B are interchangeable in this puzzle, as are C & D.)
One of the most precious gifts that we can be granted is wisdom from God. King Solomon used that gift well for a while. ‘Solomon’ is also the name of one of my excellent, current colleagues at Musica Mundi School. I’m almost two months too early in writing this to Solomon, but I would like to wish him a wonderful, happy birthday for the date A3.01.20A3, when he’ll be A3 years old 😊😊😊.
Other colleagues with birthdays coming sooner include:-
Jan V.L. in 3 days’ time; Herman in A + A3 days’ time; Peter in 3 + A3 + A3 days’ time.
Have fun figuring out their birthday dates 😊😊😊.
Clearly, Jan V.L.’s birthday (3 days after 25 November) is 28 November, which brings to mind the number 2811. Here comes a very beautiful puzzle featuring that number… Suppose that J, A, N, V & L stand for certain consecutive whole numbers, in increasing order. The nice challenge is to discover their values such that (J x A x N x V x L) ÷ (J + A + N + V + L) + 3 = 2811 ♥♥♥♥♥
One of my favourite chess puzzles was composed by W.Speckmann in the year 14 x 140 = 1960. I’m sure that FIDE Master Arben Dardha will love it, too. Happy birthday today, dear Ben! ♥
God-willing, I will publish solutions to all the puzzles around the time when I hope to share Blog Post #141 with you. In the meantime, you are more than welcome to send me your solutions to some or all of the puzzles, if you like. I wish you a very blessed, happy weekend now.
Special congratulations to Michael and Annie, two lovely young people who got engaged on Trent Bridge, Nottingham, tonight ♥😊😊♥
Congratulations also to Eric & Marianna (lovely former Maths students of mine) who recently married near Paris ♥♥
With love and kindest wishes as always,
Paul M😊twani ♥
As I don’t want to waste this precious moment, I’ll leave you with even more than an extra couple of couples of nice bonus puzzles!!!! ♥😊😊♥
First, suppose that the letters C, O, U, P, L, E and S each stand for different, positive, whole numbers. Then, what would be the minimum possible value for the sum C squared + O squared + U squared + P squared + L squared + E squared + S squared ?
Next from our huge chocolate box of puzzles, can you rearrange the letters of WAIST to make another proper English word?
Waiting now and looking forward to 25 December has inspired this penultimate wee puzzle today: (25 – W) x (25 – A) x (25 – I) x (25 – T) = 25, where W, A, I & T represent different integers. What is the value of the sum W + A + I + T ?
The final puzzle tonight is in honour of Alicia, the eldest daughter of Nathan Braude, a new colleague at Musica Mundi School. As Alicia’s birthday was on 5 October, I’m thinking of the three-digit number 510. The number 6 will also be starring in this wee puzzle, since the names ‘Alicia’ and ‘Nathan’ each have six letters. Now I’ll tell you that 6 x 5 x the age that Alicia will be on her birthday in (just under) five years from now equals 510. Can you be almost as fast as Alicia and figure out the year when she was born?!
By the way, though I’m now 60 years old, I still love to solve Junior Maths Challenge puzzles! I’m betting that Alicia will enjoy them, too! 😊 So, specially for Alicia and all Maths fans, here is a link to a further feast of wonderful puzzles: https://www.ukmt.org.uk/competitions/solo/junior-mathematical-challenge/archive
For all Chess enthusiasts-including Daniel Mathiesen from Härnösand, Sweden-I’m including a link to two of my favourite sites: http://www.chess.com/ccc & http://www.ideachess.com/chess_tactics_puzzles/checkmate_n/34288
(I couldn’t resist giving everyone yet another b😊nus!!…and so the second link, just above, leads to a delightfully crisp case of ‘White to play and force checkmate in 3 moves’ even though the material is equal there ♥)
P.S. = Puzzle Solutions (being posted on 2 December 2022)
2 x 2 x 5 x 7 = 140; the lady has her birthday on December 4 (= 2 x 2), and she’ll be turning 57.
In the second puzzle, 7 days works, because 1440 x 7 ÷ 140 = 144 x 7 ÷ 14 = 144 ÷ 2 = 72, a whole number.
23 x 83 = 1909, and 1909 + 14 x 140 = 3869 = 53 x 73.
Herman’s birthday is on December 20 (2 + 23 = 25 days after 25 November, when this article was first published).
Peter’s birthday is on January 13 (3 + 23 + 23 = 49 days after 25 November).
Solomon has his next birthday on 23.01.2023, when he’ll be 23 years old!
(9 x 10 x 11 x 12 x 13) ÷ (9 + 10 + 11 + 12 + 13) + 3 = 2811 ♥
In the Chess study, White forces checkmate with 1 Qe6! Bg4 2 Qa2+! Ra4 3 Qg8 b4 4 Qc4#; beautiful play! 😊
The minimum possible value for the sum C squared + O squared + U squared + P squared + L squared + E squared + S squared = 1 squared + 2 squared + 3 squared + 4 squared + 5 squared + 6 squared + 7 squared = 1 + 4 + 9 + 16 + 25 + 36 + 49 = 140, nice for this blog post #140 ♥😊♥
WAIST rearranges to make WAITS.
If (25 – W) x (25 – A) x (25 – I) x (25 – T) = 25 for different integers W, A, I & T, then, in some order, (25 – W), (25 – A), (25 – I) & (25 – T) must equal -1, 1, -5 & 5. So, in some order, W, A, I & T have to equal 26, 24, 30 & 20. Therefore, the sum W + A + I + T must be 20 + 24 + 26 + 30 = 100 exactly ♥
In the puzzle about Alicia, 6 x 5 x 17 = 510; Alicia will be 17 in 2027 (five years from now); she was born in 2010.