This blog is about Simon, a young gifted mathematician and programmer, who had to move from Amsterdam to Antwerp to be able to study at the level that fits his talent, i.e. homeschool. Visit https://simontiger.com
Simon has come up with an equation to solve the Too many Twos, the puzzle mode of the Add ‘Em Up game:
x is the number of twos I used to clear out just a single two at a time
y is the number of twos I used to clear out six twos at once.
We have two pieces of information. At the beginning, the twos are arranged in a pattern with 40 twos in it. And the number of twos I can use to clear out the whole grid is 25.
x + 6y = 40
x + y = 25
We thought we solved it, but no! The reason why is because of the way the twos are arranged there were spots where there were exactly 6 twos neighbouring an empty cell. And there was only one spot where there wee more than 6. Our equation says that there must be 3 of those. the way I solved this problem was by considering a third variable, z = the number of twos that I place without clearing any twos in the grid. So now our two equations look like this:
x + 6y – z = 40
x + y + z = 25
With a little bit of cleverness though, we know that these are all integers. You don’t have 2.7 twos! That doesn’t exist! Which means that we can use some number theory to narrow it doen. After solving these equations we get: x = 25 – y – z and y = 3 + 2z/5
We’ve got a fraction. We need to carefully choose the z for this to result in an integer! This is only true if z is divisible by 5.
I don’t want to check infinitely many solutions. Luckily, we know one more quite obvious thing: all of our variables must be positive. So if z gets too large, x will become negative. How large? Let’s just be lazy and use trial and error. Let’s draw a table. In our table we now only have four solutions that we need to check. The first one, with 0 z‘s, clearly doesn’t work.
This has been one of Simon’s most ambitious (successful) projects so far and a beautiful grand finale of 2019, also marking his channel reaching 1K subscribers. The project – approximating Euler’s number (e) in a very weird way – is based upon a Putnam exam puzzle that Simon managed to prove:
Inspired by the Card Flipping Proof by Numberphile, Simon created his own version of this proof. He made a solitaire game and proved why it would be impossible to solve with an even number of orange-side-up circles. He drew all the shapes in Microsoft Paint, printed them out and spent something like two hours cutting them out, but it was worth it!
If there’s an odd number of orange circles in the middle, then the end pieces are the same, both orange or both white. In both cases the total number of orange circles will also be odd. If there’s an even number of orange circles in the middle, then the ends have to be different (one orange, one white).
In the case of odd number of orange pieces, the ends have to match. In the case of an even number of orange pieces, you would have pieces that point the same way at both ends. “Now we’ve proven that to make this puzzle possible it has to have an odd number of orange pieces”, Simon says.
Why? Imagine a stick figure that always walks to the right, but always faces in the direction of the arrow (as in it can’t go backwards). It would flip every time it reaches an orange circle. Focusing on everything except the ends, if there are an odd number of orange circles, the puzzle pieces would face the other way. Which means that the end pieces are the same, and therefore the end circles are the same. If there are an even number of orange circles in the middle, the puzzle pieces would face the same way. Which means that the end pieces are different, and therefore the end circles are different.
Simon finds this sort of proof easy, but I felt like my brains are going to boil and dripple through my ears and nostrils. He patently exlained it to me several times and types the above explanation, too.
Today, Simon returned to a problem he first encountered at a MathsJam in summer: “Pick random numbers between 0 and 1, until the sum exceeds 1. What is the expected number of numbers you’ll pick?” Back in June, Simon already knew the answer was e, but his attempt to prove it didn’t quite work back then. Today, he managed to prove his answer!
The same proof in a more concise way:
At MathsJam last night, Simon was really eager to show his proof to Rudi Penne, a professor from the University of Antwerp who was sitting next to Simon last time he gave it a go back in June. Rudi kept Simon’s notes and told me he really admired the way Simon’s reasoning spans borders between subjects (the way Simon can start with combinatorics and jump to geometry), something that many students nurtured within the structured subject system are incapable of doing, Rudi said. Who needs borders?
Later the same evening, Simon had a blast demonstrating the proof to a similar problem to a larger grateful and patient audience, including Professor David Eelbode. The first proof was Simon’s own, the second problem (puzzle with a shrinking bullseye) and proof came from Grant Sanderson (3Blue1Brown) on Numberphile.
“Don’t allow any constraints to dull his excitement and motivation!” Rudi told me as Simon was waiting for us to leave. “That’s a huge responsibility you’ve got there, in front of the world”.
During Chinese lesson yesterday, Simon came up with what he calls his “Cycle formula” to calculate all the permutations of placing n numbers in a cyclical order (like on a clock face). He also proved the formula. Wait, Chinese lesson? Yes, I know, this guy manages to squeeze some math everywhere. His Chinese tutor loved it by the way. “Well, we’ve both learned something!” Simon exclaimed delightfully.
Take any real number and call it x. Then plug it into the equation f(x) = 1 + 1/x and keep doing it many times in a row, plugging the result back into the equation.
At some point you will see that you arrive at a value that will become stable and not change anymore. And that value will be… φ, the golden ratio!
But this equation also has another answer, -1/φ. If you plug that value into the equation, it will be the same, too. The real magic happens once you have rounded the -1/φ down (or up), i.e. once what you plug into the equation is no longer exactly -1/φ. What happens is that, if you keep going, you will eventually reach… φ as your answer!
While in Southern France, Simon really enjoyed solving this puzzle (he originally saw in a Brilliant.org vid). He was so happy with his solution he kept drawing it out on paper and in digital apps, and later shared the puzzle on Twitter. This sparked quite a few reactions from fellow math lovers, encouraged Brilliant to tweet new puzzles and now Brilliant follows Simon on Twitter, how cool is that!
Simon was showing Dad a graph of how technology is developing exponentially, y = a^x. Dad asked for a specific value of a, and Simon said: “All exponentials are stretched out or squished versions of the same thing.” He then quickly came up with the proof (“a few lines of relatively simple algebra”). “If all exponentials are pretty much the same, that means that all exponentials have proportionately the same derivative.”