The Project Euler thread

[kaiten123]

just did 96 which was cool since i never really used recursion in a way that was non-trivial before
no idea how you guys do some of the higher numbered ones, highest one i did was 243, which for some reason is many times easier than everything around it, didnt even write any code.

edit: got to lvl 3 :)

[kaiten123]

brute forced 407 lol
never saw that box at the bottom right before, is that new? or does it pop up when you solve something recent?

[Reincarnate]

tis new

[leonid]

i should get back to solve rest of the problems

[Reincarnate]

Managed to squeak onto the 408 list; hard one

[Arkuski]

This is pretty awesome I'm giving some of these a go

[Reincarnate]

I think 411 will be on the easier side this week... I hope.

[Reincarnate]

411 wasn't too bad -- pretty much an exercise in programming

[Reincarnate]

414: Got this one somewhat quickly but it took me forever to bring it down to under a minute, lol. In the end got it to 10 seconds.

[Reincarnate]

415... only 3 solvers after 7 and a half hours.

Yeah I can pretty much kiss my 100% goodbye

It's been fun, PE :(

[MracY]

Originally Posted by Reincarnate

2 mod 3 is 2

(x mod y is the remainder of x / y)

PS use Python instead, it's cooler

sumTotal = 0
num = 1
while num < 1000:
if (num % 3 == 0) or (num % 5 == 0):
sumTotal = sumTotal + num
num = num + 1
print sumTotal

Damn thing doesn't save formatting but w/e

Just as a note to people who want to join Project Euler: this solution is not necessarily representative of what you will be doing.

Here are some of my crude notes for problem 1, that used logic in stead of programming:

vertical axis is m
horizontal axis is n


Sum of the series of 5 is 0.5nm+0.5m
Sum of the series of 3 is 0.5pq+0.5q

1000/5 = 200.
The last value here is 1000, so n = 199
m = 1000-5 = 995

1000/3 = 333 with 1 remainder, so p = 333
q = 1000-1 = 999

Sum5 = 0.5*199*995+0.5*995
= 99500

Sum3 = 0.5*333*999+0.5*999
= 167166

99500+166833=266333



There is overlap between the sums of multiples of 15.
1000/15 is 66 with 10 remainder, so x = 66
y = 1000-10 = 990

Sum of this series is 0.5xy+0.5y
= 0.5*66*990+0.5*990 = 33165

So the answer is: 99500+166833-33165 = 233168

[leonid]

@Reincarnate use code tag

while 1:
  print "INDENT"
  if "INDENT":
    print "MORE INDENT"
print "NEVER REACHES HERE"

[reuben_tate]

Maybe ill try to play around and do some more problems this weekend :) (I haven't touched PE in awhile) . gotta start on my take-home test and other school stuff first though xP

[Reincarnate]

415 is a worthy challenge -- very fun but hard. Not many solvers on the first week. Whoever sk7 is, he's insane.

also damn didn't know code tag existed on ffr

[Reincarnate]

Got my program for 416 working but mothereffing thing is taking eons

[reuben_tate]

Started doing this in my free time again :3 Just got problem 34 this morning. Problem and code below; any tips that can make me a better programmer or mathematician is greatly appreciated :P

problem:

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Note: as 1! = 1 and 2! = 2 are not sums they are not included.

solution: (in Java)

//this will be our final answer
		int probSum = 0;
		
		//at some point, the sum of the factorials of the digits won't be large enough to add up to the number
		//so choose our digits to be 9 and find when we have too many digits
		//our upper bound for curious numbers becomes k
		int i = 1;
		int nFact = factorial(9);
		int k = 9;
		while(i*nFact > k)
		{
				k += 9*Math.pow(10, i);
				i++;
		}
		
		for(int j=10; j <=k ; j++)
		{
			String[] digits = (j+"").split("");
			
			int sum = 0;
			for(int l = 1; l < digits.length; l++)
			{
				sum += factorial(Integer.parseInt(digits[l]));
			}
			
			if(j==sum)
			{
				probSum += j;
			}
		}
		
		System.out.println(probSum);

EDIT: cool, got another :3


Problem:

Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 53 = 49714.

What is the total of all the name scores in the file?

Solution:

int totalScore = 0;
		
		File file = new File("names.txt");
		Scanner scan = null;
		try 
		{
			scan = new Scanner(file);
		} 
		catch (FileNotFoundException e) {}
		String nameString = scan.next();
		String[] names = nameString.split("\",\"");
		ArrayList<String> names2 = new ArrayList<String>();
		for(int i = 0; i < names.length; i++)
		{
			names2.add(names[i]);
		}
		
		Collections.sort(names2);
		for(int i = 0; i < names2.size(); i++)
		{
			char[] chars = names2.get(i).toCharArray();
			int charSum = 0;
			for(int j = 0; j < chars.length; j++)
			{
				charSum += chars[j] - 'A' + 1;
			}
			totalScore += charSum*(i+1);
		}
		System.out.println(totalScore);

[Arkuski]

Learned some jython, and these are all much easier

[Reincarnate]

Anyone else try their hand at 415? It's a beast :X

[Reincarnate]

pumped for saturday ;O