Problem 075
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As your posts will be visible to the general public you are requested to be thoughtful in not posting anything that might explicitly give away how to solve a particular problem.
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Don't start begging others to give partial answers to problems
Don't ask for hints how to solve a problem
Don't start a new topic for a problem if there already exists one
See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
- solarmew
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- hk
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- solarmew
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Re: Problem 075
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...
I got 62444, but it was wrong
Could I pm someone my code? I think I'm just not getting all of them...

366541_5799d51e95657e1a227f2cb86bd181de
- hk
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Re: Problem 075
Perhaps you can study the pdf for problem 9 first.

War ruins the life and health of untold numbers of innocent children.
- solarmew
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Re: Problem 075
I don't understand the paragraph after eq 9.3hk wrote:Perhaps you can study the pdf for problem 9 first.
I implemented those conditions, but i'm getting that m<k<2m is always false :\

366541_5799d51e95657e1a227f2cb86bd181de
- dawghaus4
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Re: Problem 075
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?
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?
- solarmew
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Re: Problem 075
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 >.> ...
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 >.> ...

366541_5799d51e95657e1a227f2cb86bd181de
- hk
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Re: Problem 075
Eh, Problem 9 (View Problem) says a+b+c=1000.

War ruins the life and health of untold numbers of innocent children.
- solarmew
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Re: Problem 075
Is the method not generalizable to any p?hk wrote:Eh, Problem 9 (View Problem) says a+b+c=1000.

366541_5799d51e95657e1a227f2cb86bd181de
- dawghaus4
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Re: Problem 075
Yes, but there are not always unique integral solutions. There can be multiple solutions or no integral solutions, at all.solarmew wrote:Is the method not generalizable to any p?hk wrote:Eh, Problem 9 (View Problem) says a+b+c=1000.
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
- Oliver1978
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Re: Problem 075
I get the same for 50 and 100. But I get 59 for X(500) and 222 for X(2000)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
49.157.5694.1125
- Georg
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Re: Problem 075
You are right.leghorn wrote:I'm wondering if at least one of us is right with their numbers...
- Oliver1978
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Re: Problem 075
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?
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- Georg
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- Oliver1978
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Re: Problem 075
Vielen Dank. My code is going wrong somewhere. Now I've got to find out where.
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MadPat
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Kamiccolo
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Re: Problem 075
*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
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