# CNRS interview

My (short) CNRS interview was just held. Unsurprisingly, given the number of candidates, there was trouble having it on time, but the jury was very friendly.

I had trained myself to present my work and my research project in 120 seconds, which is a challenge. Thankfully, the members of this jury seemed to be statisticians, so I didn’t need to spend time to explain what MCMC is, for example.

The 5 minutes of questions did not leave time to discuss my research project a lot. I was asked what I was most proud of, a question on my PhD and which lab I would like to work in, and that was it. It must be very frustrating for the jury, who needs to make a decision without having time to look at the details of each candidate’s work (I assume they make a shortlist and look a bit more at the work of a small number of candidates).

I have no idea whatsoever on how well or badly the interview went, nor on how important it is. I heard that a few years ago, the Mathematics section of the CNRS only held interviews because they were legally obliged to, and that the interview lasted only a few seconds: the time to sign a sheet of paper. Hopefully, the interviews have at least some weight nowadays.

The results should be (at least unofficially) available in a few days.

# Le Monde problem: “Nationale 7”

It seems that solving mathematical puzzles from Le Monde is becoming the main focus of this blog. This week’s problem is about a road with 100 trees: 50 elms and 50 plane trees, in a random order. We are asked to show that whatever the order of the trees, there exists a sequence of 50 consecutive trees with exactly 25 elms and 25 plane trees.

Let $v_k (k=1,\ldots 51)$ be the number of elms between positions $k$ and $k+49$; we are asked to prove that there exists a $k$ such that $v_k=25$. Note that $v_1+v_{51}=50$, since that includes all the trees exactly once. By symmetry, we can assume that $v_1\leq 25$ and $v_{51}\geq 25$. Note also that the sets considered for $v_k$ and $v_{k+1}$ differ by only one tree, hence $| v_k - v_{k+1} | \leq 1$. Looking at the sequence $(v_k)$ we need to go from an integer below $25$ to an integer above $25$ by taking steps of size $0$ or $1$. At some point, we will necessarily go through $25$.

The exact same proof holds with only 98 trees (49 of each species). However, going down to 96 trees (48 of each), the following arrangement has no set of 50 consecutive trees with half of each: 24 elms, then 48 plane trees, then 24 elms.

# Le Monde problem: “The prediction”

This week’s Le Monde mathematical problem is:

The scene is set in 2010. Young $N$, an integer looking for his soul-mate, consults a fortune-teller, who says: “Yes, you shall encounter the beloved number [call it $X$], a number strictly greater than you, but later! That year [call it $Y$], you will be the only [my emphasis]  two positive integers whose sum is equal to the year, and whose product is a multiple of the year.” When (at the earliest) will $N$ meet his soul -mate?

This is the kind of question which could be asked at the FFJM semi-finals in a couple of weeks (which my schedule doesn’t allow me to attend, unfortunately). The word only is key. (Christian solves the problem without that word.)

We have, for some integer $k$, $X>N\\Y>2010\\X+N=Y\\XN=kY$

and we want to minimize $Y$.

Write $Y=ab^2$, with $a$ and $b$ integers and $b$ as large as possible. Then it is clear that both $X$ and $N$ are multiples of $ab$. Write $X=xab$ and $N=nab$. Suppose that we have a solution to the problem, with $(x,n)\neq (1,2)$; then the couple $(x+1,n-1)$ would give another solution. Hence $X=ab$, $N=2ab$ and $ab^2=Y=X+N=3ab$, so $b=3$.

The problem is now to find the smallest $Y=9a>2010$ such that no perfect square divides $a$ (otherwise $b$ would not be as large as possible). $2016=9\times 4^2 \times 14$ and $2025=9\times 15^2$ don’t work, so we move to $2034=9\times 2\times 113$, which does work.

The solution is then that in year $Y=2034$, young $N=678$ will meet $X=1356$ and they will live happily ever after.