Project Euler...

[Shah-Cuber]

5 days of struggling with #246 ... I hate geometry problems >_>
Just calculate a given point from the center of the ellipse ... 'nough said ... Used Wolfram Alpha for most of the calculations =P

[bbi5291]

Instead of using Wolfram Alpha, why not download the engine it uses for its calculations - Wolfram Mathematica? http://***********
lol, the domain is censored, but the link still works

[Dan]

bbi5291 @ 7th July 2009, 5:30 pm wrote:

lol, the domain is censored, but the link still works

Theres a reason for that, no links to warez.

[Shah-Cuber]

Finally level 6 ...
I have a headache right now, but it took me a week to solve #248, which is numbers for which Euler's totient function equals 13!, a lot of little details to deal with primes, ahhg! ... it was hardcoded ...
Well, at least I can rest now ...
They changed the problem a while back from 100,000 to 150,000 for the number to find.
... great ... i still have to finish Coresilience and Ambiguous Numbers, this is torture. I'll sleep now ...

[A.J]

I am kind of bored now that there aren't any more PE questions to do....it has been about 2 weeks w/out PE..

so I am suggesting new questions to them

and I got addicted to SC....
(I am in a MathCamp at washington....it is soooooooo fun!)

[matt271]

to anybody who cares... i think i have the best solution to Problem 40

Quote:

An irrational decimal fraction is created by concatenating the positive integers:

0.123456789101112131415161718192021...

It can be seen that the 12^(th) digit of the fractional part is 1.

If d_(n) represents the n^(th) digit of the fractional part, find the value of the following expression.

d_(1) ? d_(10) ? d_(100) ? d_(1000) ? d_(10000) ? d_(100000) ? d_(1000000)

my solution in c99 is: (i hope this hides for non-spoilers)

c:

#include <stdio.h> #include <stdlib.h> int a[] = {1, 10, 190, 2890, 38890, 488890, 5888890}; int b[] = {1, 10, 100, 1000, 10000, 100000, 1000000}; int getindex(int, int*); int matt(int); inline int digit(int, int); int main(int argc, char** argv) {     int i, p = 1;     for (i = 1; i < 1000001; i *= 10)         p *= matt(i);     printf("%i\n", p);     return (EXIT_SUCCESS); } int getindex(int x, int a[]) {     int i;     for (i = 0; i < 7; ++i)         if (a[i] > x)             break;     return --i; } int matt(int x) {     int i = getindex(x, a);     x -= a[i];     int q = x / (i + 1);     int r = x % (i + 1);     q += b[i];     return digit(q, r); } inline int digit(int q, int r) {     return q / b[getindex(q, b) - r] % 10; }

the c code itself is not that impressive, but the math i used i think is. i have (what i like to think is) a brilliant method to derive the numbers u see in array a[]

does anybody care? anybodys thoughts? i wanted to post this on the thing for it, but its locked for archive.

[Cyril]

Haha, I suppose naming a function after yourself is the best alternative to becoming famous enough as to have others do it for you. Anyways, nice, clean, morally justified solution- out of laziness, I just used an ostringstream and bashed it.

[A.J]

Very nice solution Matt.

I like the clean code (you don't see that very often nowadays). However, I do not believe that you have the best solution.

Your method is certainly a good method, but I think there's a better one. However, I might be mistaken.

Once again, I like the cleanliness in the code.

[Cyril]

http://www.research.att.com/~njas/sequences/A033307

Oh wow, there's a closed-form formula. But this is impractical, as there would be precision errors with computing the Lambert W-function. Interesting though.

[Shah-Cuber]

Weeeeee - It's 3 am
Renewing post after 9 months ...
Coming back for new problems & more headaches
Enjoy!

[A.J]

Well, the latest problem, p289, is giving me quite some trouble. However the latest few, < 289, were pretty easy. Maybe Project Euler wants < 50 people to have solved all the problems