The Project Euler thread

[klimtkiller]

yay

[Reincarnate]

Oh how those green checkmarks make you feel warm and fuzzy inside

[Zapmeister]

so today i remembered i had an account on this thing with 26 solved that i hadn't touched in over three years.

up to 60 now, running out of easy ones really fast. let's see how this goes. even if nothing happens it'll probably be a more productive use of my spare time than cookie clicker or messing around with bitcoin faucets

i was the 30000th person to do #38 yay

[evanescence_death4ever]

I think...I'm going to start this. O_O Time to fail.

[lilylily608]

I will be going in linear order from start to finish!

[rushyrulz]

Originally Posted by lilylily608

I will be going in linear order from start to finish!

you won't finish sorry.

[sweet2kill210]

hmmm, I managed to do the first 5 in python a while back. BUT I will try to get through more later because next semester I'm taking a college class for C++ which I've been wanting to learn for a long time. :)

[leonid]

who knows.

[Reincarnate]

leonid you have not solved a problem in... almost 900 days.

tear

[Guest15937]

I've done about 70 of them, going in order from most solved to least solved.

[reuben_tate]

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);

[reuben_tate]

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]

I tend to just brute force everything :/

[stargroup100]

Originally Posted by rushyrulz

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]

pencil&paper will only get you so far

[Reincarnate]

Not every problem can be done with pencil and paper.

[leonid]

Rubix sent me to solve this

A bit of paperwork and then lots of wolframalpha

[Reincarnate]

Nicely done

[stargroup100]

sorry I should've been more clear

I was referring to problem nine specifically, involving pythagorean triples

[rushyrulz]

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.