The Project Euler thread

[benguino]

Re: The Project Euler thread

Decided to start doing this again. Going to start from the beginning because I know when I first started my solutions were crap. I used my favorite programming language for this one:

Some actual code:

/*
* We use the fact that 3+6+9+12+... = 3(1+2+3+..) from which we can
* use the n(n+1)/2 formula for the sum of the first n positve integers
*/
       int a = 999/3; //number of multiples of 3 less than 1000
       int b = 999/5; //number of multiples of 5 less than 1000
       int c = 999/15; //number of multiples of 15 less than 1000
       System.out.println(3*(a)*(a+1)/2+5*(b)*(b+1)/2-15*(c)*(c+1)/2);

[benguino]

Re: The Project Euler thread

Ok, screw the whole going in order thing, I'm just going to do problems I haven't done yet haha. This was a nice paper-n-pencil-n-calculator one as well :) Level 1 get.

[rushyrulz]

Re: The Project Euler thread

I tend to just brute force everything :/

[stargroup100]

Re: The Project Euler thread

I tend to just brute force everything :/
*solution to problem nine*

if you haven't done these kinds of problems with pencil/paper you should try it

the first time you realize how to do it it's quite enlightening and exciting

[leonid]

Re: The Project Euler thread

pencil&paper will only get you so far

[Reincarnate]

Re: The Project Euler thread

Not every problem can be done with pencil and paper.

[leonid]

Re: The Project Euler thread



Rubix sent me to solve this

A bit of paperwork and then lots of wolframalpha

[Reincarnate]

Re: The Project Euler thread

Nicely done

[stargroup100]

Re: The Project Euler thread

sorry I should've been more clear

I was referring to problem nine specifically, involving pythagorean triples

[rushyrulz]

Re: The Project Euler thread

brute forcing 12 and it's not going so well. I'm almost an hour in and still no solution.

The sequence of triangle numbers is generated by adding the natural numbers. 
So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

 1: 1
 3: 1,3
 6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

In hindsight, I probably would have used a much more efficient factor breakdown algorithm had I known it would take this long.

[Reincarnate]

Re: The Project Euler thread

using brute force = not very interesting

usually more fun to try for the faster approaches

[rushyrulz]

Re: The Project Euler thread

I'm sure there's a very elegant mathematical solution for all of these, but I'm just trying to get as far as I can :P

[stargroup100]

Re: The Project Euler thread

I did the first 30 (excluding 12) and even a couple of the 300+ ones before I figured out 12. I don't know why but that one was hard for me

[rushyrulz]

Re: The Project Euler thread

Probably since there are so many different things you need to account for in 12. Not only do you need a gigantic 500+ factor number, but it also needs to be a summation of n number which is tricky to check for.

[Guest15937]

Re: The Project Euler thread

My brute force of 12 took 15 seconds to run.