Within the past few days, lots of people from numerous countries were captivated yet again by wonderful concerts seen live inside the Bach Concert Hall at Musica Mundi School, Waterloo, Belgium, or enjoyed online via top-quality broadcasts. My personal appreciation of the beauty of music, and its profound power to touch and heal, continues to grow.
I am especially grateful to Leonid Kerbel and Hagit Hassid-Kerbel, Musica Mundi School’s founders, who hired me three years ago to be the Mathematics teacher in their magical environment.
On Tuesday evening this week, the staff, the students, and many special friends and great supporters of the school were treated to a fabulous end-of-school-year dinner BBQ. Looking back at the following photo, I would like to say…
I do have a nice warm-up puzzle for us now…Assume that there are two long tables with an equal number of places on the left side and on the right side of both of the tables (the same at both tables). The cutlery includes a knife and fork at each place. If the total number of knives and forks is at least 100, what is the exact minimum possible total?
If the number of places is the same on the left and on the right of both tables, then the total number of places must be a multiple of 4. Since there is a knife AND a fork (2 items) at each place, the total number of knives and forks must in fact be a multiple of 8. The smallest multiple of 8-meeting the other condition that the total must also be at least 100-is 104 (which equals 8 x 13).
Though that correct puzzle answer of 104 does fit perfectly here in Blog Post #104, I should confess that the actual total needed at the BBQ was more, because the Musica Mundi School Family is big (and is likely to keep on growing!) 😊.
The following fun brainteasers are dedicated to the school’s graduating students, some of whom were pictured above.
Imagine that each graduate in the photos picks a different whole number, chosen from 1 to 8.
No two people pick the same number, but clearly some numbers will end up not being picked since only five of the graduates were pictured in the photos.
(Still, I will pick the number 9 for my own number since I’m older!)
Now imagine that, in each photo, we’ll multiply together the numbers chosen by all the people in that photo, and we’ll note the results.
Suppose that the products obtained from the three photos are P1, P2 and P3 respectively.
SOLUTIONS (being posted now on 14 August 2021)
Brainteaser Part 1: Also suppose that P2/P1 (meaning P2 divided by P1) turns out to be a whole number. Can you figure out then exactly what number P2/P1 would have to be? What numbers would Aude and Taïlin have to pick to make that happen?
Solution 1: When calculating P2/P1, we don’t have to worry much about the numbers chosen by Kristina and Quintijn. That’s because they star in all the photos, and so their numbers in P2 are effectively ‘cancelled out’ by their same numbers in P1. That can help us to realise that P2/P1 will only produce a whole number result when Aude’s number x Taïlin’s number in P2 is a multiple of my number 9 in P1. That can only happen if Aude and Taïlin choose 3 and 6 (or vice-versa). Then, since 3 x 6/9 = 2, two is the only possible WHOLE number result for P2/P1.
Brainteaser Part 2: Still assuming that P2/P1 is a whole number, what is then the largest whole number answer that we could get for P3/P1? What number would Thaïs have to pick to make that happen?
Solution 2: To maximise the value of P3/P1, in addition to Aude and Taïlin still choosing 3 and 6 as above, we want Thaïs to choose the largest available remaining number, which is 8. Then, P3/P1 will be 3 x 6 x 8/9 = 16.
Mindreader Fun Word Puzzle 😊
All the students are special stars in their own ways, but for this puzzle let’s pick…
Which one other mystery letter am I thinking of so that my mystery letter, together with STAR AUDE, can be rearranged to make a proper 9-letter word? (The word has featured in today’s blog post.)
Solution: G STAR AUDE = GRADUATES !
Solutions will be posted here by the time of the next blog post, if not before. Please do feel free to send me your own solutions by email when you crack the brainteasers and/or the word puzzle and/or the following bonus treat…
Yes, let’s conclude with a delightful chess puzzle (by A. Silvestre, 1899) for Kristina and Quintijn…
Solution: White plays 1 Rh8! followed by a winning discovered check(mate) with the bishop according to which move Black plays after 1 Rh8.
Wishing everyone a happy new month of July now,
Paul Motwani xxx