# Funding available for doctoral programme in Oxford

Studentships are still available for the 4-year Life Sciences Interface Doctoral Training Centre in Oxford, which I went through starting in 2005. This is a very strong programme dedicated to students with a degree in physical or mathematical sciences who want to get involved in applications to the Life Sciences. It’s a great opportunity to meet edge-cutting researchers and to have a better grasp of various application fields before starting a DPhil.

Studentships are available for UK and EU students and the application deadline is 4 June.

# Pandigital approximation to e

I spent some time this week-end trying to find a mathematical puzzle whose solution is 2718, for the “Mathematics on a shelf” competition, and the first trail was to look into properties of Euler’s number $e$. The following result is not useful in any way, but it is amazing: an approximation of $e$ using all the digits from $1$ to $9$ exactly once, and which is correct to $18457734525360901453873570$ decimal digits (that’s more than $10^{26}$ digits!):

$e \approx \left( 1+9^{-4^{7-6}}\right)^{3^{2^{85}}}.$

# Candy branching process

Christian Robert has basically forced me to post my solution to last week’s Le Monde problem:

Two kids are given three boxes of chocolates with a total of 32 pieces. Rather than sharing evenly, they play the following game: Each in turn, they pick one of the three boxes, empty its contents in a jar and pick some chocolates from one of the remaining boxes so that no box stays empty. The game ends with the current player’s loss when this is no longer possible. What is the optimal strategy?

Although we don’t know the number of chocolates in each box, the kids do. A kid loses when at his turn, he is faced with the configuration (1,1,1): he has to leave one empty, thus losing. Note that the total number of chocolates decreases at each turn.

Because the initial number of chocolates is a power of 2, the first kid (A) is guaranteed to win. Let us look at the parity of the number of chocolates in each box. The strategy of player A is to give to the other kid (B) a configuration of the form (odd, odd, odd). Then B will necessarily return (odd, odd, even) to A. Kid A can empty a box with an odd number of chocolates, split the even number into two odd numbers, and B is faced again with (odd,odd,odd). Eventually, B will end up with (1,1,1) and lose.

Note also that any (odd, even, even) configuration wins: empty one of the even boxes, and split the other even number into two odd numbers.

The only remaining question is what to do with a configuration (even, even, even). Per the above, the first player to introduce a box with an odd number of chocolates loses. Kid A’s strategy is now to give B a configuration of the form (4a+2, 4b+2, 4c+2). To avoid odd numbers, B must give A a configuration of the form (4a+2, 4d+2, 4e), where the number of chocolates in box 3 is a multiple of 4. Kid A can then empty one of the first two boxes, giving B (4a+2, 4f+2, 4g+2) again, and so on until B is faced with (2, 2, 2), at which point he has lost.

This can be easily generalized: let a configuration be of the form $\left(2^n a, 2^n b, 2^n c\right)$, with n as large as possible (so at least one of a, b, c is odd). If a, b and c are all odd, the configuration loses; otherwise, it can be transformed into a configuration where a, b and c are all odd, and the other player loses.

Because the initial total number of chocolates is a power of 2, the initial configuration is necessarily a winning configuration.

# “Mathematics on a shelf”

A nice new mathematical puzzles competition (in French) is open until 16 May: “Mathematics on a shelf“, a collection of 19 puzzles. Many are quite classical and all can be solved by high school students. The hardest is probably the 11th: create a puzzle, the solution of which must be 2718. The most elegant puzzle wins.

The most obvious property of 2718 is that it is the floor of $1000 \times e$, but that doesn’t make an elegant puzzle yet!