Group, Math Riddles, Murderous Maths, Notes on everyday life

MathsJam 18 June 2019

This was the last MathsJam this academic year and a very social one. Simon has enjoyed the outside setting and mingled with the other mathematicians.
Simon had also contributed two problems (numbers 3 and 5). Unfortunately we haven’t seen anyone solve them (the problems come from the Russian math olympiad published in the magazine Kvantik).
Simon had recognized two problems he knew the answers to. Here he is explaining the persistences problem (number 13).
The smallest number with persistence 5 is 679
Simon trying to prove that the solution to problem number 2 is e.
Math Riddles, Murderous Maths, Notes on everyday life, Simon's sketch book

Trinagular birthday probabilities

“What is the chance that two people in a group of, say, 30 people would have their birthday on the same day?” I asked Simon as we were sitting on a bench by the river Schelde late last night, waiting for his Dad and sister to arrive by boat. The reason for this question was that one of the professors at Simon’s MathsJam club turned out to have celebrated his birthday exactly on the same day as I the week before. Besides I was afraid of Simon getting bored just sitting there, “enjoying the warm evening”. At first, I thought he didn’t hear my question and repeated myself a couple of times. Then I noticed he was so silent simply because he was completely immersed in the birthday problem.

Eventually, at that time already on Antwerp’s central square, Simon screamed with joy as he told me the formula he came up with involved triangle numbers! “It’s one minus 364/365 to the power of the 29th triangle number!” he shouted. “It’s a binomial coefficient, the choose function!”

Simon’s solution defining the probability of two people having the same birthday in a group of n people. The highlighted diagonal in the Pascal triangle are the triangle numbers. For example, 15 is the 5th triangle number. So in a group of 6 people, the probability would be 1 minus 364/365 tothe power of 15.
A few days later Simon told me his previous formula wouldn’t work for a group of 366 people and quickly came up with a simpler formula, without any triangle numbers.
Crafty, Geometry Joys, Group, Logic, Milestones, Murderous Maths, Notes on everyday life, Together with sis

Vladimir Krasnoukhov at MathsJam Antwerp!

all these beautiful puzzles we have received from Vladimir Krasnoukhov

When we arrived at the MathsJam last Tuesday, we heard a couple of people speak Russian. One of them turned out to be a well known Russian puzzle inventor Vladimir Krasnoukhov, who presented us with one colorful puzzle after another, seemingly producing them out of thin air. What a feast! Simon got extremely excited about several puzzles, especially one elegant three-piece figure (that turned out to have no possible solution, and that’s what Simon found particularly appealing) and a cube that required graph theory to solve it (Simon has tried solving the latter in Wolfram Mathematica after we got home, but hasn’t succeeded so far).

Vladimir told us he had been making puzzles for over 30 years and had more than 4 thousand puzzles at home. Humble and electricized with childlike enthusiasm, he explained every puzzle he gave to Simon, but without imposing questions or overbearing instructions. He didn’t even want a thank-you for all his generosity!

Vladimir Krasnoukhov and Simon

Vladimir also gave us two issues of the Russian kids science magazine Kvantik, with his articles published in them. One of the articles was an April fools joke about trying to construct a Penrose impossible triangle and asked to spot the step where the mistake was hidden:

Simon was very enthusiastic about trying to actually physically follow the steps, even though he realized it would get impossible at some point:

Simon and Neva constructing the shape that “allows” to convert it into Penrose impossible triangle (as seen in the optical illusion in the Kvantik magazine)
the next step was already impossible

Simon’s also working on other math problems from the magazine, so more blog posts about Kvantik will follow. We’re very happy to have discovered the website https://kvantik.com

You can find out more about Vladimir Krasnoukhov’s puzzles on planetagolovolomok.ru

Group, Milestones, Murderous Maths, Simon teaching, Simon's sketch book

MathsJam Antwerp 23 October 2018

Euphoric fun at MathsJam Antwerp @MathsJamAntwerp last night, where Simon solved two 2×2 Rubik’s Cube puzzles and one tricky maths problem, and simply enjoyed socialising with like-minded folks. In the video below he explains how he solved the Rubik’s Cube puzzles:
1. Solve the cube so that on every face the 4 colors are all different;
2. Solve the cube so that not only the 4 colors on every face are different but also every face has a different color combination.

After we stopped filming, Simon added that a cube like this has 8! times 3^8 possibilities in total, because the cube has 8 corners and every corner has three orientations.

Simon also talks about the Choose function and symmetries:

Simon showing his solution to university maths students.

Simon also solved a tough problem (one of several tough problems) that asked to sum up the digits in x, if x equals 1111…1111 (number with 100 ones) minus 222…222 (number with 50 twos):

Simon spent the rest of the time trying to prove the ‘cosine rule’, an equation similar to Pythagoras’s theorem, that defines the side c of any triangle if it’s opposite to angle C: c² = a² + b² – 2abcosC. He got stuck with the proof, but luckily, with so many university professors walking around, he got great help from one of them, who came up with an alternative proof using vectors!