Read the whole post at Project Euler 267: Billionaire

]]>Read the whole post at Project Euler 146: Investigating a Prime Pattern

]]>**Some positive integers n have the property that the sum [ n + reverse(n) ] consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible**

**How many reversible numbers are there below one-billion (109)?**

This one is insanely easy to write a brute force method and that is the first thing I did. However, as we shall see there is a more analytic approach to the problem as well.

Read the whole post at Project Euler 145: How many reversible numbers are there below one-billion?

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Read the whole post at UVa Online Judge New Platform

]]>**In laser physics, a "white cell" is a mirror system that acts as a delay line for the laser beam. The beam enters the cell, bounces around on the mirrors, and eventually works its way back out.**

The specific white cell we will be considering is an ellipse with the equation 4x2 + y2 = 100

The section corresponding to -0.01 ≤ x ≤ +0.01 at the top is missing, allowing the light to enter and exit through the hole.

The light beam in this problem starts at the point (0.0,10.1) just outside the white cell, and the beam first impacts the mirror at (1.4,-9.6).

**How many times does the beam hit the internal surface of the white cell before exiting?**

We will simply brute force our way through this problem, by calculating the laser beams path through the cell, and check if it hits the exit. In order to do that, we need to calculate how the laserbeam reflects. Once we know the angle of the reflecting beam, we can calculate the corresponding line, since we have the point of reflection. Once we have a line parameterization of the reflecting line, it is simply a matter of finding out where the line and the ellipse intersect. This will be the next point out laser beam hits. Confused yet? Don't be. I will elaborate on it. Lets start by finding the slope of the reflecting beam.

Read the whole post at Project Euler 144: Investigating multiple reflections of a laser beam.

]]>**Let ABC be a triangle with all interior angles being less than 120 degrees. Let X be any point inside the triangle and let XA = p, XB = q, and XC = r.**

If the sum is minimised and a, b, c, p, q and r are all positive integers we shall call triangle ABC a Torricelli triangle. For example, a = 399, b = 455, c = 511 is an example of a Torricelli triangle, with p + q + r = 784.

**Find the sum of all distinct values of p + q + r ≤ 120000 for Torricelli triangles.**

After solving it, I can see why there are so few other people who have solved it. Because it was really difficult, and took a whole lot of research for me.

Read the whole post at Project Euler 143: Investigating the Torricelli point of a triangle

]]>**Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x - y, x + z, x - z, y + z, y - z are all perfect squares.**

I don't think we can manage to iterate over all possible values of x, y and z. So let us see if we can use the relations that has to be squares to something.

Read the whole post at Project Euler 142: Perfect Square Collection

]]>**A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.**

Some progressive numbers, such as 9 and 10404 = 1022, happen to also be perfect squares.

**Find the sum of all progressive perfect squares below one trillion (10 ^{12}).**

I ended up getting the right idea when I was working out. I guess some times it really does help to do something else. In this problem it comes down to some really basic properties and insights so lets start with those

Read the whole post at Project Euler 141:Investigating progressive numbers, n, which are also square.

]]>**Consider the infinite polynomial series AG(x) = xG1 + x2G2 + x3G3 + ..., where Gk is the kth term of the second order recurrence relation Gk = Gk-1 + Gk-2, G1 = 1 and G2 = 4; that is, 1, 4, 5, 9, 14, 23, ... .**

**We shall call AG(x) a golden nugget if x is rational. Find the sum of the first thirty golden nuggets.**

In Problem 137 I mentioned in the end that the problem could be solved using a Diophantine equation. This is exactly the way I will go for this problem.

Read the whole post at Project Euler 140: Modified Fibonacci golden nuggets

]]>**Let (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.

Read the whole post at Project Euler 139: Pythagorean tiles

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