What is so hard about this IMO problem?

Over 90% of International Mathematical Olympiad (IMO) participants scored zero marks for this question. But the solution is beautifully simple.

An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer
from 1 to 1 + 2 + 3 + 4 = 10 occurs exactly once:

Is it possible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to (1 + 2 + 3 + … + 2018) exactly once?

You will probably need to read it a few times to get your head around it. But as far as mathematical problems go, it’s relatively easy to understand what this one is asking. Solving it, on the other hand, is not easy at all.

When I first read this problem, I was intrigued by its apparent simplicity. So I gave it a try….

• I recognised why an equilateral triangle with n rows must contain
  1 + 2 + 3 + … + n numbers.
• I was able to come up with other suitable anti-Pascal pyramids with 4 rows.
• I tried to find one with 5 rows, but could not.