Dear Readers,

I very much hope that you will all enjoy this article, which is especially dedicated to Christopher (one of my first students at my former school in Belgium, nearly 15 years ago) and to Kristina, an outstanding student at the beautiful Music school where I now work as the Mathematics teacher.

Christopher’s family have made positive use of social media to publish good posts which help to raise people’s awareness of Mental Health matters. That is an important and complex subject which can potentially impact everyone. Here on this site, all the puzzles I post are a small personal contribution that I hope will bring enjoyment and healthy mental stimulation to all readers.

__CHRISTOPHER’S BIRTHDAY BONUS!__

Christopher already celebrated his birthday this month. You can quickly figure out when it was, if I tell you that Christopher’s day-number within November is equal to the square root of the number of days remaining in November **after** Christopher’s birthday.

Kristina is a world-class violinist who loves getting to play an almost magical instrument from the year 1737. A couple of days ago, the number 1737 starred in a fun Maths SMS that I sent to a colleague and to Kristina. SMS is a palindrome (which reads the same forwards as backwards), and 1736^{2}+1737^{2}+1738^{2}=9051509, a palindromic number.

Some High School Maths students would easily manage to prove that, though the sum of any three consecutive whole numbers is always a multiple of 3, the sum of the **squares** of any three consecutive whole numbers is **never** a multiple of 3. Just very occasionally, though, the sum of the squares does give a palindromic result, like 9051509.

Here are other such cases:-

4^{2}+5^{2}+6^{2}=77

11^{2}+12^{2}+13^{2}=434

37^{2}+38^{2}+39^{2}=4334

566^{2}+567^{2}+568^{2}=964469

(1736^{2}+1737^{2}+1738^{2}=9051509, already given)

8339^{2}+8340^{2}+8341^{2}=208666802

16084^{2}+16085^{2}+16086^{2}=776181677

17552^{2}+17553^{2}+17554^{2}=924323429

17932^{2}+17933^{2}+17934^{2}=964777469, and I reckon that’s the last such case where the result is under 1 billion.

__ONE BILLION TRICK CHALLENGE!__

Your “one billion trick challenge” is to determine exactly how many magic triples (like Kristina’s 1736, 1737, 1738) there are for which the sum of the squares gives a palindromic result between one billion and two billion !

Solutions to the puzzles will be given at the time of blogpost #63 (not #62).

As you’ll be considering “magic triples”, and since 3 is my absolute favourite number, my wife and my son and I would like to wish you all a very happy November now.

__CHECKMATE BONUS!!__

Consider the chess position shown below.

Imagine that White and Black **each** have one extra, invisible pawn somewhere on the board. You get to decide exactly where the two invisible pawns will be, but not anywhere on the 7^{th} rank (e.g. NOT on d7 beside Black’s king). You also get to decide whose turn it will be to move. The challenge is to place the two invisible pawns so that one side can then force checkmate in as few moves as possible.

Wishing you lots of enjoyment,

Paul Motwani.

**SOLUTIONS TO PUZZLES**** **(being posted now on 30 December 2019)

Christopher’s birthday is **5 November**.

In the chess puzzle, if White has a pawn on f3 and black has a pawn on f5 or d5, then after **1…f4+ or 1…d4+ 2 Ke4, 2…Re2#** is a pretty checkmate!

In the “One Billion Trick Challenge”, there are **no solutions at all** between one billion and two billion.

Proof

Note that, if n-1, n and n+1 are three consecutive whole numbers, then the sum of their squares can be simplified to 3 x n squared + 2. If that were to start and finish with a digit 1 (which any palindrome between one billion and two billion would have to), then 3 x n squared would have to end with a 9, and n squared would have to end with a 3. Since no square numbers end with a 3, there are no solutions.