Problem A4. Consider a rectangular region, where and are integers such that . The region is to be tiled using tiles of the two types shown below
(The dotted lines divide the tiles into squares.) The tiles may be rotated and reflected, as long as their sides are parallel to the sides of the rectangular region. They must all fit within the region, and they must cover it completely without overlapping.
What is the minimum number of tiles required to tile the region?
Here are a few of the problems of the Putnam 2016 contest. I choose to only list problems which I managed to solve. Most of them are pretty straightforward, so maybe the solutions posted here may be very similar to the official ones. You can find a complete list of the problems on other sites, for example here.
A1. Find the smallest integer such that for every polynomial with integer coefficients and every integer , the number
that is the -th derivative of evaluated at , is divisible by .
Hints. Successive derivatives give rise to terms containing products of consecutive numbers. The product of consecutive numbers is divisible by . Find the smallest number such that . Prove that does not work by choosing . Prove that works by working only on monomials…
This post is not about math… It’s about a watch which is elegant, smart and a good activity motivator. The Withings Activité is all the above and more. Apparently this watch is on a market for quite a while now, but I didn’t hear about it until recently (via a Facebook add, the irony… I usually despise ads). What caught me was the nice quality design and the promise that it can do more than just tell time.
So what exactly can this watch do?
- tells time via an analog display
- has a silent vibrating alarm
- counts your steps
- shows progress towards the daily goal
- monitors sleep
- can monitor running and swimming
- battery lasts 8 months (that’s forever compared to other fancy smartwatches out there)
I recently wrote a brief introduction to FreeFem++ in this post. FreeFem is a software designed for the numerical study of partial differential equations. It has the advantage of being able to easily define the geometry of the domain, construct and modify meshes, finite element spaces and solve problems on these meshes.
I use Matlab very often for numerical computations. Most of the numerical stuff I’ve done (take a look here if you want) was based on finite differences methods, fundamental solutions and other classical techniques different from finite elements methods. Once I started using finite elements I quickly realized that Matlab is not that easy to work with if you want some automated quality meshing. PDEtool is good, but defining the geometry is not easy. There is also a simple tool: distmesh which performs a simple mesh construction for simple to state geometries. Nevertheless, once you have to define finite element spaces and solve problems things are not easy…
This brings us to the topic of this post: is it possible to interface Matlab and FreeFem? First, why would someone like to do this? Matlab is easier to code and use than FreeFem (for one who’s not a C expert…), but FreeFem deals better with meshes and solving PDE with finite elements. Since FreeFem can be called using system commands, it is possible to call a static program from Matlab. FreeFem can save meshes and variables to files. Let’s see how can we recover them in Matlab.
There is a tool called “FreeFem to Matlab” developed by Julien Dambrine (link on Mathworks). There’s also a longer explanation in this blog post. I recently tried to use the tool and I quickly found that it is not appropriate for large meshes. It probably scans the mesh file line by line which makes the loading process lengthy for high quality meshes. Fortunately there’s a way to speed up things and I present it below. I will not cover the import of the data (other than meshes) since the function importdata from the FreeFem to Matlab tool is fast enough for this.
Problem 8. Let be a positive integer and denote by the ring of integers modulo . Suppose that there exists a function satisfying the following three properties:
- (i) ,
- (ii) ,
- (iii) for all .
Prove that modulo .
Problem 7. Today, Ivan the Confessor prefers continuous functions satisfying for all . Fin the minimum of over all preferred functions.