Blog Post #163: Precious Time❤️

Dear Friends,

We’re already 50 days into 2024, but I have been so busy trying to do my best with many commitments that I’m only now finally finding time to publish my first blog post of this year. As I met up with lots of Chess friends in recent days during the school holiday, it’s a pleasure to now share many happy moments via nice photos 😊❤️😊

Time for a quick, wee Chess joke…!!

What’s the first name of a brilliant Belgian Chess player who can be as good as a Woman International Master!?

Answer: Wim!

P.S. If you see me with a black/blue bruise soon, you’ll know my joke backfired!! 😂

To do well in Mathematics or Chess it’s essential to practise a lot, and that requires precious time and great dedication.

I will now share with you an original Maths brainteaser which I created to celebrate the birthday of a friend named Eric who was turning 66 earlier this month…

I sent this on 9 February:

“Suppose that the Quadratic Equation x² + bx + c = 0

has two positive whole number solutions for x.

In honour of Eric’s birthday on 12 February, we’ll say that c – b = 122.

Your Prize Puzzle Challenge now is to solve the equation x² + bx + c = 0.

Please feel free to send me your two solutions for x by 12 February 👍😊

Wishing you lots of fun and happy solving,

Paul/Mr. M😊.”

Eric used EXCEL to solve the puzzle, which was OK as it was for his birthday, after all!! 😂

In general, though, I always hope that some people will send in a beautiful analytical solution worked out by themselves.

Warm congratulations to 14-year-old Wonderful Wout Callens who did solve the puzzle and sent me this, for which I will be delighted to present him with a nice prize in school this week:

“Dear Mr. Mo,

These are the steps I used to solve the problem:

c-b = 122 so c = b+122.

As the solutions are required to be positive whole numbers, we can derive from the quadratic formula that b<0 and that √(b² – 4ac) must be a whole number.

We know that a=1 and that c=b+122 and substituting that into the square root and rearranging the expression, we get that √((b-2)² -492) must be a whole number.

Then I tried out values for b for which √((b-2)² -492) was equal to a whole number and found the values b=-42 or 46, but as b<0, b =-42.

Hence c=122-42=80.

Substituting a=1, b=-42 and c=80 into the quadratic equation, we get that x equals either 2 or 40.

Enjoy your holidays 😊

Wout.”

Thank you and congratulations again, dear Wout! 👍😊

I wrote personally to Eric on his birthday to say,

Dear Eric,

Happy birthday!! Have a really wonderful time!

Congratulations on finding that x = 2 or 40 in the puzzle I sent on Friday.

In this New Year ’24, add up 24 + 2 + 40 to get a nice total for your birthday!!!

With kindest wishes as always to your whole family,

Paul 😊.

Now, keen young or older fans of Mathematics could be interested to learn that there actually exists a super-fast method for solving the puzzle in a matter of seconds without even needing a calculator, and without even needing to figure out the individual values of b or c!! First, I will present a quick piece of theory which helps to crack the puzzle speedily👌

Suppose that the solutions to the quadratic equation x² + bx + c = 0 are x = p or x = q. Then (x-p)(x-q) = 0, which leads us to x²-(p+q)x + pq = 0. Comparing coefficients in the quadratic equations can help us to realise that p+q = -b and pq=c. (Note: Those results can also be proved using the Quadratic Formula.) So, c-b = pq+p+q. Next comes a highly noteworthy mathematical trick… c-b+1 = pq+p+q+1 which factorises to (p+1)(q+1).

In the specific case we had where c-b = 122, we get that c – b + 1 = 123, and so (p+1)(q+1) = 123 = 3 x 41 (or 1 x 123). Therefore, since the puzzle required that p & q had to be positive whole numbers, it follows that p+1 = 3 & q+1 = 41 (or vice-versa), and p=2 & q=40 (or vice-versa). 💕

Those positive solutions are unique, but having a different value for c-b such as 2 would mean that c-b+1 = 3 which is 1 x 3 for (p+1)(q+1), and then we wouldn’t be able to get positive solutions for both p & q. Also, a different value for c-b such as 23 would mean that c-b+1 = 24 which is (1 x 24 or) 2 x 12 or 3 x 8 or 4 x 6 for (p+1)(q+1), and we would consequently get several possible pairs of values for p & q (unless we were to pin down specific individual values for b & c rather than just saying what value c-b has).

Quick Word Puzzle:

Rearrange the letters of RETRIED to make another proper seven-letter word also beginning with R.

Valère De Buck was rightly very pleased to discover a beautiful finish as White (to play and win) in a lovely game of his in Wachtebeke.

Valère calculated 1 fxe4 fxe4 2 Qg4+ and then 2…Kf6 3 Rf1+ or 2…Qf5 3 Rxe4+ Kf6 4 Rf7+ or 2…Kxd6 3 Bf4+! gxf4 4 Qxg6+ or 3…Kc6 4 Rc7#, checkmate! ❤️

Solution to Word Puzzle (given earlier, above):

RETIRED

I will round off this article with the following message:

With kindest wishes as always,

Paul M😊twani ❤️