Welcome to this ‘Good Lives Global Prize Competition’ with free entry for everyone of every country! The more puzzles you solve, the better for sure, but you may possibly still receive some prize for submitting some good answers even if they’re not all perfect. So, please don’t be shy! Just go ahead and enjoy the puzzles (detailed below), and don’t hesitate to send in your answers by email to firstname.lastname@example.org, preferably by Sunday 17 January; the sooner, the better for you!
I felt moved and inspired when listening to a wonderful, old classic song by the band ‘Hot Chocolate’. You can check it out via this link:
The exact number and type of prizes will depend a bit on the number of entrants, and their ages and locations. Given that I am an International Chess Grandmaster and also a very experienced Mathematics teacher, some people may request a free Maths or Chess lesson (using Zoom, for example) if they win a prize. People whom I can see in person in Belgium, may prefer to receive a signed chess book gift in perfect condition or a box of delicious Belgian chocolates! We’ll see how everything turns out, but, when sending your answers by email, it’s a good idea to also mention your age, country of location, and preference for type of prize in case you are awarded one!
I do most sincerely wish you lots of success and fun in now trying the prize puzzles…
Rearrange the nine letters of GOOD LIVES to make three proper words which together make a beautiful, true statement. (Admittedly, there could be more than one answer to that first puzzle, but I don’t know anything more important than the statement I’m thinking of.)
The time on the clock in the first photo above is 20:37:48. Exactly how many seconds must pass to go from that time to 20:52:00? (I particularly like this puzzle because I was born at 8.52pm on June 13, 1962.)
Assume that the time on the clock in the second photo above is 20:38:12. Exactly how many DEGREES did the seconds hand have to turn through to go from the time in the first photo to the time in the second photo?
PUZZLE #4: THE SPECIAL ‘EYES’ BRAINTEASER!
(For this puzzle, we need to know what a PALINDROME is. MUM, DAD, EYE, 99 and 101 are all examples of palindromes because they each have their characters in the same order from left to right as from right to left.)
Look again at your numerical answer to Puzzle #2. Let’s represent that number you got by the letter S.
Now suppose that E x Y + E = S
where E and Y are both positive whole numbers and Y is actually a palindrome with two or more digits.
What is the smallest possible palindrome solution for Y so that
E x Y + E = S?
Also, what is the largest possible palindrome solution for Y so that
E x Y + E = S?
It is my intention that
winners will be notified by email by next weekend, and possibly before then.
God-willing as always,
solutions to the puzzles will also be published around the same time.
Let’s conclude today’s blog post with a wee chess puzzle, just below. I reckon that it will be pretty easy even for moderately experienced players. You don’t have to be Beth Harmon to solve this one!
If you think that you’ve figured out the chess solution, include it an email to me with your other puzzle solutions, and I’d love to be congratulating you very soon!
PUZZLE #5: DOUBLE INVISIBILITY, DOUBLE MATE!
BLACK has TWO different invisible pieces located somewhere on the board such that:-
- If it’s White’s turn to move now, White can force checkmate in two moves.
- If it’s Black’s turn to move now, Black can force checkmate in two moves.
Happy solving, and have a lovely weekend now!
With kindest wishes as always,
Paul Motwani xxx
COMPETITION PUZZLE SOLUTIONS (being published at night now on 16 January, and the competition is now officially completed and closed).
As 22-year-old Niklas of Helsinki University in Finland is one of the main winners, I will publish his solutions to the mathematical puzzles.
You can read lots of other fun details in Blog Post #81: Top of the Class, Nice Niklas!!
Puzzle 1: GOOD LIVES makes GOD IS LOVE.
Puzzle 2: 852 = 14 x 60 + 12.
Puzzle 3: By counting how many seconds it has to turn (24 seconds), we can get the following equation: 24/60 = x/360˚. From there, we can rearrange it as follows: x=360˚ x 24 / 60 = 144˚ which is the answer.
I believe there could be a much easier way to come to the solutions, but I had to figure this out the difficult way! I believe the answers are as follows…
For calculating what is the smallest possible palindrome Y so that the other requirements are true, I tried with inserting Y=11 (11 being the smallest palindrome with two or more digits). Solving the equations, there exists a positive whole number for the value of E so that the equation E x Y + E = S =852 is true, which is E=72.
Thus, my answer for the first part of Puzzle 4 is Y = 11.
For the second part of that puzzle, I figured that the lower our number E is, the higher our palindrome number Y is going to be.
Rearranging E x Y + E = S => Y=(S – E)/E => Y = S/E – 1 (with S=852, of course).
By trial and error (a Mathematics teacher’s shock for sure!), with the value E = 4 (since 1, 2 or 3 for E do not lead to palindrome results for Y) we get the value Y = 212, which is therefore the highest palindrome with two digits or more that fits all the requirements such that E x Y + E = S is true!
Congratulations to Niklas!
He was always an excellent young student of mine way back in 2007-2008, and in Maths he’s clearly ‘gone from strength to strength’, as the good saying goes!
There was one more brainteaser…Puzzle 5 on Chess…
MR. MO’S CHESS BRAINTEASER SOLUTION
The required invisible pieces for Black were a queen on e3 and a bishop on h7.
With White to move, 1 Rd8+ Qe8 2 Rxe8# is checkmate, whereas with Black to move the finish would be 1…Be4+ 2 Kh3 Qxh6#!