### Thinking Out Loud – EGMO 2013 Problem 2

**EGMO 2013, problem 2**. Determine all integers \(m\) for which the \(m\times m\) square can be dissected into five rectangles, the side lengths of which are the integers \(1, 2, 3, \dotsc, 10\) in some order.

EGMO 2018 will be held in Florence, Italy.

Contestants are given three problems to solve in each of the two competition days.

April 9th - 15th, 2018.

**EGMO 2013, problem 2**. Determine all integers \(m\) for which the \(m\times m\) square can be dissected into five rectangles, the side lengths of which are the integers \(1, 2, 3, \dotsc, 10\) in some order.

by/di **Emanuele Tron**

In this series of posts we would like to talk about women in mathematics who are not as well known as, say, Hypatia or Emmy Noether, and are at the same time quite different from the usual mathematicians' stereotypes. Our mathematician of the day is Claire Voisin, who managed to balance having a family with five children and doing high level research in Hodge theory (earning a CNRS Gold Medal in the process).

*In questa serie di post vorremmo parlare di matematiche meno famose di, ad esempio, Ipazia o Emmy Noether e che, allo stesso tempo, non corrispondono ai consueti stereotipi sui matematici. La matematica di oggi è Claire Voisin,
che è riuscita a bilanciare le esigenze di una famiglia con cinque figli e la ricerca nell'ambito della teoria di Hodge (vincendo tra l'altro una medaglia d'oro del CNRS).*

**EGMO 2013, problem 3.** Let \(n\) be a positive integer.

- Prove that there exists a set \(S\) of \(6n\) pairwise different positive integers, such that the least common multiple of any two elements of \(S\) is no larger than \(32n^2\).
- Prove that every set \(T\) of \(6n\) pairwise different positive integers contains two elements the least common multiple of which is larger than \(9n^2\).