Blog Post #21: Everyone Is Invited

Dear Readers,

Though I may or may not have the pleasure of getting to meet you in person during this short life, we are all invited to a glorious and everlasting celebration afterwards, and to be sure of arriving there, we only have to gratefully accept God’s invitation and happily let everyone know that the Good News is for them too.

Whenever I see myriads of dancing autumn leaves, or fields full of beautiful flowers, and other miraculous sights which shouldn’t be taken for granted, I picture many millions of people dancing in Heaven, glowing with happiness, knowing that they are there because God is perfect love.

Even before we get to Heaven, we can learn to wisely put all our cares in God’s hands, and then genuinely become…


Of course, we must still use well our God-given gifts, and so I invite you to enjoy tussling over the following brainteaser…


What is special about the sentence THE FIVE JUMPING WIZARDS BOX QUICKLY?

SPECIAL CONGRATULATIONS if you realised that THE FIVE JUMPING WIZARDS BOX QUICKLY is a pangram containing every letter of the English alphabet, A-Z (once or more).


This recently-published Batsford book (see also ISBN 978-1-84994-446-5 & by George Huczek, a Canadian Correspondence Chess Association Senior Candidate Master, is a wonderful reference guide to lots of key tactical terms from Attraction to Zwischenzug. The exercises and illustrative games featuring masters and grandmasters make this book a very useful work for students, and even for more experienced players looking to refresh and sharpen their chess skills.

One of hundreds of great puzzle positions in the book is from the game Alekhine-Supico, Lisbon 1941.

It is White to play and force checkmate in at most 4 moves.


Why are cats really hard to beat at chess?

It’s because they have nine lives!


Yesterday, I logged on to Facebook to wish a very happy 59th birthday to Andrew, a Scottish friend born in the year ’58, who was turning from being 58 to 5×9+5+9 now! Then I saw that another fine ANDREW–whose first name is an anagram of WARNED–had posted part of a 1939 chess study by C. de Feijter. It’s beautiful but tricky!

It’s White (moving up the board) to play and draw.


  1. How many trailing zeroes would there be at the right-hand end of the enormous number we’d get by multiplying together all the 158 numbers that we see printed on the blocks in Incredible Isabelle’s Multiplication Maze? Do not use a calculator!!
  2. What would be the first non-zero digit, immediately to the left of all the trailing zeroes?



White wins really quickly with 1 Qg6!! fxg6 (or 1…Rg8 2 Qxh7+ Kxh7 3 Rh3#) 2 Nxg6+ hxg6 3 Rh3+ Qh4 4 Rxh4#, checkmate.


White draws with 1 Kb7 a5 2 Kc7 Kc5 (or 2…a4 3 f5) 3 Kd7 Kd5 4 Ke7 Ke4 5 Ke6! (intending 6 f5) 5…Kxf4 6 Kd5, and the white king will catch the black a-pawn.


  1. The only way that we’ll get zeroes at the end of the product of all the numbers is from 5s multiplying 2s (or any even number) to produce many 10s. As there are eighteen 5s in the maze, and more than enough even numbers to pair up with them, the entire product will end with eighteen trailing zeroes on the right.

Though it’s not actually required, the total product is 

  1. Imagine that the blocks 8×2 and 7×4 were not missing from the maze. Then, the left-hand numbers in each of the nine columns and the right-hand numbers in each of the nine rows would contribute 1x3x4x6x7x8x9x(2×5)=36288 x 10 towards the overall product. That reconfirms why the product ends with nine plus nine zeroes, or eighteen zeroes.

Now consider 8×8=64, ending with a 4; (8×8)x(8×8)=64×64 ends with a 6 because 4×4=16; (8x8x8x8)x(8x8x8x8) still ends with a 6 because 6×6=36; (8x8x8x8x8x8x8x8)x(8x8x8x8x8x8x8x8)=8^16 also ends with a 6; but 8x8x8x8x8x8x8x8x8x8x8x8x8x8x8x8x8x8=8^18 ends with a 4. However, the blocks for 8×2 and 7×4 are missing from the maze. So, the overall product must be divided by 8x2x7x4=448. Then 448 x “the reduced product” must end with a 4 and eighteen zeroes. That can only happen (remembering 8×8=64 ending with a 4) if the reduced product ends with an 8 and eighteen zeroes; not a 3 and eighteen zeroes (though 8×3=24 ending in a 4) because 3 is an odd number, but the reduced product should be even, even without all its zeroes!

As it’s less than 10 weeks until Christmas, I’ll finish with this thought, picture & link: 7mins 7secs.

Author: Paul A. Motwani

My name is Paul Motwani, but my colleagues, my students and their parents mostly call me "Mr. Mo"! My middle initial, A, stands for Anthony, because I was born on the official feast day of St. Anthony of Padua, the patron saint of miracles and of lost souls. I love teaching Mathematics and Chess, and giving fun-packed talks and shows in schools and clubs. The popular ingredients of Math, Chess, Mystery and Magic are my "Fantastic Four", and I give prizes too! I am an International Chess Grandmaster, and (loooooong ago!) I was the World Under-17 Champion. I am the author of five published chess books and hundreds of newspaper articles. I live with my wonderful wife and son in Belgium. I also love music, movies and puzzles. I blog at My e-mail address is You can find me on Facebook, too.

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