Dear All,

I’m not just 3 or 4 days early with surprises here…it’s actually 3^{4} + 4 days in advance of the 12 July birthday of Jens Van Steerteghem, a brilliant colleague who loves solving fun brainteasers, as do many of our students at Musica Mundi School πΆπ

Just yesterday evening, a special idea which involves “Jens’ 12/7″π came to me, and so I decided happily to share this surprise π super-early now!! π It was tempting to wait until tomorrow and then be precisely 12 x 7 days before 12 July…but instead we’re (12 – 7 + 12) x (12 – 7) days early! π

OK, here we go… There’s literally an infinite choice for pairs of numbers a and b such that a^{2} β b^{2} = 12/7. Imagine that I pick such a pair, and I whisper the numbers to Eric and Martine (Jens’ lovely parents) ππ.

To turn up the sneaky-level button π, Eric then calculates a^{2} + 2ab – 3b^{2} and Martine calculates 2a^{2} – 3ab + b^{2}.

Next, Nick (Jens’ brother) * multiplies Eric’s and Martine’s results together*.

The Advanced Algebra Birthday Brainteaser for Jens (and other great solvers!) to figure out is this: ** what is the most extreme final result that Nick could get**?

In other words, the ‘most extreme’ possibility is either a minimum value or a maximum value. The fun challenge is to determine which of those two is correct, and also to figure out the actual extreme value.

To make the challenge all the more interesting and surprising, it can be completely solved ** without** using Calculus at all! So, it really is an Advanced Algebra brainteaser coming far in advance of Jens’ 12/7 birthday!

Compared to that one, a very wee snack bonus puzzle, here in Blog Post #155, is to **use all six of the numbers 3, 4, 3, 4, 7 & 12 to make 155**. Parentheses ( ) and operations involving +, -, x, Γ· can be used freely, as you wish.

Another quick one is: **how can the digits within 12 and 7 combine** *in a somewhat different type of calculation* **to get the result 127**?

And here’s quite a nice puzzle: **how many whole numbers from 1 to 1000 inclusive do not contain Jens’ favourite digit 7**?

The following neat brainteaser gives me happy memories… π ** How old was I when I could then truthfully say** (

*all in relation to that time*), “The product of my age n years ago and my age n years from now is 1207”? π

Joke Time! π

With apologies to lovely lady Martine, why should men be good at solving the earlier algebra brainteaser?!

Answer: **M**artine **E**ric & **N**ick (MEN) were starring in it! π€£

To make up for that one…I’ve got another algebra brainteaser for Jens’ family and everyone to enjoy!

It’s my intention to publish solutions to all the puzzles around the time that blog post #156 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 π.

I would like to round off this article now by most sincerely wishing you a very blessed week, with lots of happiness in everything that you do β€.

With kindest wishes as always,

Paul Mπtwani β€

βI am the vine; you are the branches. If you remain in me and I in you, you will bear much fruit; apart from me you can do nothing.”–Bible verse, John 15:5

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

Let a^{2} + 2ab – 3b^{2} = p and 2a^{2} – 3ab + b^{2} = q. Then 3p + 2q = 3(a^{2} + 2ab – 3b^{2}) + 2(2a^{2} – 3ab + b^{2}), which simplifies to 7a^{2} – 7b^{2} or 7(a^{2} – b^{2}). We were given that a^{2} – b^{2} = 12/7, and so 7(a^{2} – b^{2}) = 7*12/7, which is 12. We’ve proved that 3p + 2q must have the constant value 12. That is, ** 3p + 2q = 12** is absolutely forced, given that a

^{2}– b

^{2}= 12/7.

So, q = (12 – 3p)/2, and the product pq = p(12-3p)/2 or -1.5p^{2} + 6p. Using standard facts regarding quadratic expressions, the graph of the function f(p) = -1.5p^{2} + 6p would be a parabola with a maximum value when p = -6/(2*-1.5)βp = 2. The actual maximum value of the function is then -1.5*2^{2} + 6*2β-6 + 12β** 6 is the maximum-possible value** π

One may ask, how can we be sure that it’s really possible to have p = 2 and q = 3 in order to achieve pq = 6, the claimed maximum-possible value of pq when a^{2} – b^{2} = 12/7 ? Do suitable values exist for ‘a’ and for ‘b’ which would get us to p = 2 and q = 3 ? Well, if you check out the solutions to the next blog post (#156), you’ll see a proof showing that a = (p+3q)b/(2p-q). Requiring p = 2 & q = 3 also then requires that a = 11b. So, (11b)^{2} – b^{2} = 12/7β121b^{2} – b^{2} = 12/7β120b^{2} = 12/7βb^{2} = 1/70β** b = 1/β(70) or -1/β(70)** and, since a = 11b, we find that

**πβ**

*a = 11/β(70) or -11/β(70)*Here in Blog Post #155, we were invited to **use all six of the numbers 3, 4, 3, 4, 7 & 12 to make 155**. Parentheses ( ) and operations involving +, -, x, Γ· could be used freely, wherever we wish. One solution is (3+3+7) x 12 – (4Γ·4) = 155. Note that it’s also perfectly sufficient to write **(3+3+7) x 12 – 4Γ·4 = 155** if we’re familiar with the standard, internationally-accepted rules regarding ‘Order of Operations’ πβ

In the next puzzle, **2 ^{7} – 1 = 127** π

Just as there are one thousand whole numbers from 1 up to and including 1000, there’s also one thousand such possibilities from 000 to 999. However, if we exclude numbers containing any 7, then for each of the three – – – place-value positions, we only have nine (instead of ten) options (namely 0, 1, 2, 3, 4, 5, 6, 8, 9 but not 7), and so the total number of such possibilities is 9 x 9 x 9β** 729** π

** How old was I when I could then truthfully say** (

*all in relation to that time*), “The product of my age n years ago and my age n years from now is 1207”? Note that 1207 = 17 x 71, and the number mid-way between 17 & 71 is (17+71)/2 which is 44. Note also that 44 – 27 = 17 & 44 + 27 =71. So,

**when I could truthfully say, “The product of my age 27 years ago and my age 27 years from now is 1207.” π**

*I was 44*We use the facts that each interior angle of a regular hexagon, regular pentagon and a square is 120Β°, 108Β°, 90Β° respectively. We also make repeated use of the fact that the sum of the angles in a triangle is 180Β°. So, the angles immediately to the left and to the right of the top vertex of the pentagon are (60-a)Β° and (60-b)Β° respectively, with a 108Β° angle between them. Therefore, (60-a)Β° + (60-b)Β° + 108Β° = 180Β°β**a + b = 48**.

Similarly, the angles just above the top edge of the square are (18+c)Β° & (18+d)Β°. So, (18+c)Β° + (18+d)Β° + 108Β° = 180Β°β**c + d = 36**.

Therefore, ** a + b + c + d = 48 + 36 β84**.

Also, ** the maximum-possible value for the product a*b*c*d is 24*24*18*18β186624** π

White wins beautifully with **1 Qh6+!! Kxh6 2 Nf5++ Kg6 3 Rh6#** **or** via the similar **1…Kh8 2 Ng6+ & 3 Qxh7#** **or 2 Qxh7+ Kxh7 3 Nf5+ Kg6 4 Rh6#** π