In Project Euler There are loads of problems that end up with a number theoretic solution. Problem 139 is no exception to that. The problem readsLet (a, b, c) represent the three sides of a right angle triangle with integral length sides. It is possible to place four such triangles together to form a square with length c.For example, (3, 4, 5) triangles can be placed together to form a 5 by 5 square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5 square can be tiled with twenty-five 1 by 1 squares.Given that the perimeter of the right triangle is less than one-hundred million, how many Pythagorean triangles would allow such a tiling to take place?I will give you two different approaches to solving it.

Continue reading...Saturday, March 2, 2013

Problem 138 of Project Euler readsFind ∑ L for the twelve smallest isosceles triangles for which h = b ± 1 and b, L are positive integers.The key to solving this problem is definitively to only consider the rightangled part of the triangle, such that you get a triangle consisting of one of the sides with length L, h and half the base (which I will denote x).I started out by figuring out that solutions would be primitive pythagorean triplets which we have worked with in Problem 9 among other places. So I tried to build a solution where I check to see if the triplets fulfill the condition that 2x ± 1 = h. However, I quickly ran into the fact that the solution would take forever to complete. So I had to take a step back and try another approach.

Continue reading...Saturday, April 28, 2012

I can realy feel that the problems become harder and harder. Problem 94 of Project Euler took me a good while to figure out. The problem is easy to understand and readsWe shall define an almost equilateral triangle to be a triangle for which two sides are equal and the third differs by no more than one unit.Find the sum of the perimeters of all almost equilateral triangles with integral side lengths and area and whose perimeters do not exceed one billion (1,000,000,000).We are dealing with a triangle with two sides of length a and one side of length b = a-1 or b = a+1. I will treat these as two different cases for the next paragraf or two.

Continue reading...Saturday, October 22, 2011

Based on the problem description for Problem 66 of Project Euler I thought we had left the continued fractions for a while. Once I started reading up on the maths behind it and trying to solve the problem I got quite a lot wiser. But let's start from the beginning and look at the problem description which readsConsider quadratic Diophantine equations of the form: x2 – Dy2 = 1 Find the value of D 1000 in minimal solutions of x for which the largest value of x is obtained.

Continue reading...This site uses cookies.

Saturday, March 9, 2013

6 Comments