Netherlands win gold medal at anniversary edition Benelux Mathematical Olympiad

April 2018

In the last weekend of April, the tenth edition of the Benelux Mathematical Olympiad took place. The Dutch delegation won a total of 8 medals. With a grand total of 160 points, the Netherlands managed to beat the other two Benelux countries.

Anniversary edition

Since 2009, high school students from Belgium, the Netherlands and Luxembourg have challenged each other at the Benelux Mathematical Olympiad. For this anniversary edition, also teams from France, Ireland and Switzerland were invited. Each team consisted of ten students. The Dutch team was selected from over 10,000 participants in four preliminary rounds. The best half of all participants won a medal (bronze, silver or gold).

The results of the individual members of the Dutch team are as follows:
  • GOLD (24 points) – Nils van de Berg, 6 vwo, Sint-Oelbertgymnasium Oosterhout (NB)
  • SILVER (22 points) – Jovan Gerbscheid, 4 vwo, St. Ignatiusgymnasium Amsterdam
  • SILVER (22 points) – Szaby Buzogany, 5 vwo, Corderius College Amersfoort
  • SILVER (21 points) – Lammert Westerdijk, 6 vwo, Stedelijk Gymnasium Leeuwarden
  • SILVER (20 points) – Thomas Chen, 6 vwo, Gymnasium Haganum Den Haag
  • BRONZE (17 points) – Tim Vogels, 5 vwo, RSG Pantarijn Wageningen
  • BRONZE (15 points) – Richard Wols, 5 vwo, CSG Dingstede Meppel
  • BRONZE (12 points) – Floris Venselaar, 6 vwo, Christelijk College de Populier Den Haag
  • 5 points – Hanne Snijders, 4 vwo, Marnix Gymnasium Rotterdam
  • 2 points – Philippe van Elderen, 6 vwo, Atheneum College Hageveld Heemstede

We congratulate the Dutch team for their outstanding efforts!

The participants fought a battle in four challenging math assignments. For this, they had a total of four and a half hours. For the math enthusiasts among you, here's the hardest of the four problems.

Problem 2

In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.

(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)

➢ Prove that the tourist can purchase the book if the denomination of each banknote is smaller than 49.

➢ Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is 49.

Transtrend as partner

For young talented students, the road towards such a contest is a once in a lifetime experience. As partner of the Olympiad we support paving that road, among others by teaching math classes at our offices.

Dutch Mathematical Olypiad

Go to DMOchevron-right
More information

More information on the Benelux Mathematical Olympiad, including all statistics and assignments, can be found here:

Go to BxMO