Project Euler Forum

Posted: **Sat Jan 10, 2015 6:47 am**

Is the answer <50,000?

Posted: **Sat Jan 10, 2015 9:18 am**

Posted: **Sat Jan 10, 2015 4:32 pm**

didn't think so T.T ...
I got 62444, but it was wrong

Could I pm someone my code? I think I'm just not getting all of them...

Posted: **Sat Jan 10, 2015 8:09 pm**

Perhaps you can study the pdf for problem 9 first.

Posted: **Sat Jan 10, 2015 9:36 pm**

hk wrote:Perhaps you can study the pdf for problem 9 first.

I don't understand the paragraph after eq 9.3
I implemented those conditions, but i'm getting that m<k<2m is always false :\

Posted: **Sun Jan 11, 2015 1:46 am**

Just some observations:

m > n > 0

m < m + n < m + m = 2m

Since k = m + n,

m < k < 2m

And a question:

If you are "running" loops, are you keeping n < m?

Posted: **Sun Jan 11, 2015 4:02 am**

I define n as n = k - m, like it says in the pdf, so yeah, it should be <m
but before that I pick m from a list of divisors of the perimeter/2, like it says, and then k from a list of odd divisors of p/(2m).
And it so happens that for p = 100, k is never between m and 2m >.> ...

Posted: **Sun Jan 11, 2015 9:21 am**

Eh, Problem 9 (View Problem) says a+b+c=1000.

Posted: **Sun Jan 11, 2015 1:39 pm**

hk wrote:Eh, Problem 9 (View Problem) says a+b+c=1000.

Is the method not generalizable to any p?

Posted: **Sun Jan 11, 2015 1:57 pm**

solarmew wrote:
hk wrote:Eh, Problem 9 (View Problem) says a+b+c=1000.
Is the method not generalizable to any p?

Yes, but there are not always unique integral solutions. There can be multiple solutions or no integral solutions, at all.
Problem 75 states this.
In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.

Your result indicates that there are no right triangles with perimeter = 100, whose sides are all integers.

Tom

Posted: **Tue Feb 24, 2015 8:27 pm**

EnDorphin wrote:Alright, I figured out what I did wrong for the one before and I fixed it, but still not correct.
So I wanted to check my new values.

If we let X(n) be the function which gives the number of solutions for the problem when L <= n,
then for L = 50, 100, 500 and 2000, I get these values:

X(50) = 6
X(100) = 11
X(500) = 56
X(2000) = 210

If these are still wrong, could I may be PM anyone the L values for when the max length is 2000?
If these values are all correct, then could someone take a look at my code? It's in Java

I get the same for 50 and 100. But I get 59 for X(500) and 222 for X(2000)

I'm wondering if at least one of us is right with their numbers...

Posted: **Tue Feb 24, 2015 8:44 pm**

leghorn wrote:I'm wondering if at least one of us is right with their numbers...

You are right.

Posted: **Tue Feb 24, 2015 9:03 pm**

So far so good, but my result of xx446 won't work. Maybe someone could give a yes or no for X(135,000) = 14774?

Posted: **Tue Feb 24, 2015 9:11 pm**

I give a "Yes".

Posted: **Tue Feb 24, 2015 9:34 pm**

Vielen Dank. My code is going wrong somewhere. Now I've got to find out where.

Posted: **Sat Oct 28, 2017 1:27 am**

I agree with Oliver 1978.

X(50) = 6
X(100) = 11
X(500) = 59
X(2000) = 222

Posted: **Sat Oct 23, 2021 5:22 pm**

*sigh* this is the one problem, which is considered easy, but still too long lasting for me

Unfortunate, getting the same numbers up to X(2000) -.-

EDIT:
oh, the reason being... 150 000 used as a limit instead of 1 500 000

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