Problem 001

[hk]

Perhaps you noted this icon

icon_pdf.gif

next to the link to the forum.
This icon gives access to an overview for the problem , with a reasonable efficient solution at the end of it.

[Programmist]

Right - got the PDF. Thanks for the tip. I used a repeating 7-element pattern to solve it. I don't want to get any more specific so I don't give it away to anyone who has not solved the problem. Mine was not the most efficient solution, but might be unique.

@Ikcelaks: I'm not sure what you mean by "The question doesn't ask you to find the numbers. It asks you to find how many numbers there are." My reading of the question says it asks for a sum of numbers.

[rayfil]

My reading of the question says it asks for a sum of numbers.

You are absolutely correct on that point. However, there are means of arriving at the solution without the need of identifying any of the numbers.

[quilan]

Paging Dr Gauss to this thread...

[hk]

Right - got the PDF. Thanks for the tip. I used a repeating 7-element pattern to solve it. I don't want to get any more specific so I don't give it away to anyone who has not solved the problem. Mine was not the most efficient solution, but might be unique.

@Ikcelaks: I'm not sure what you mean by "The question doesn't ask you to find the numbers. It asks you to find how many numbers there are." My reading of the question says it asks for a sum of numbers.

Given a cool head and what you learned in the pdf, you can transform your repeating 7-element pattern also into a O(1) algoritm.

[Ikcelaks]

My reading of the question says it asks for a sum of numbers.

Oops, I should have reread the problem, because I must have forgotten it. On the other hand, you could take my slip-up as a decent hint.

[AlbertoEAF]

I've tried this over and over and doesn't work, I'm doing it in python and in the attachment you have my program's source code.

Could someone send me the solution (result)? Or tell me what is wrong with my program? Any of those will do

EDIT: I've done it, it was probably a problem in the server because my answer didn't change but in the day after it worked.

[Keybounce]

Perhaps you noted this icon

icon_pdf.gif

next to the link to the forum.
This icon gives access to an overview for the problem , with a reasonable efficient solution at the end of it.

Nope. I cannot find this. I also did not see a link to the problem.

Lets see if this works: Problem 1 (View Problem)
Yep

Keybounce
New solver.

[hk]

Perhaps you noted this icon

icon_pdf.gif

next to the link to the forum.
This icon gives access to an overview for the problem , with a reasonable efficient solution at the end of it.

Nope. I cannot find this. I also did not see a link to the problem.

Lets see if this works: Problem 1 (View Problem)
Yep

Keybounce
New solver.

Of course you must solve the problem first before the mentioned links are shown.

[h2o4444]

Hello, I'm a newbie at Python so I'm trying to learn by doing. I have no idea how to even do the first problem on ProjectEuler, can anyone help me?

The first problem is:
"If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000."

Thanks.

[triglios]

there is 2 parts in this problem , find all the number from 3 to 1000 multiple of 3 or 5 , and sum them. the divisibility test is a%b in python (it returns the remainder of a/b), i think you have all you need here.

[MekeorMelire]

I HATE LANGUAGES!!! - they are very inaccurate!

or means in my eyes "Logical disjunction".
the truth table of a "Logical disjunction" can you see here:http://en.wikipedia.org/wiki/Logical_di ... ruth_table.

but in Problem 1, there's no "either" so actually it should be a "Logical disjunction", but it isn't!
PLEASE correct the problem to:

If we list all the natural numbers below 10 that are multiples of either 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of either 3 or 5 below 1000.

[elendiastarman]

Ah, I beg to differ. The correct includes 15, 30, 45, etc... (although not twice), all of which are a multiple of 3 and 5. So yes, it is a logical disjunction.

[gipsy86147]

Edit by hk:spoiler removed.

[hk]

First of all: don't post solutions for the problems in this forum. The problems are meant to be solved by the members themselves.
Secondly: if you'd read the overview for this problem you'd known that this solution is far from optimal.

[Slaunger]

...
Secondly: if you'd read the overview for this problem you'd known that this solution is far from optimal.
(that's why I didn't snip your code).

Albeit I agree, it is not an optimal solution, it is a solution which is fine for the beginning PE solver. Thus, I think it is a spoiler, and I would urge it to be either hidden as a spoiler or (preferably) removed. If I were a beginning PE user coming to the forum, I would be a bit dissapointed to see an actual solution, as it would not be what I was after. OK, problem 1 is a bit special, as it is the simplest and widely publicly known. E.g., a slightly better Python version (using a sum over an iterator instead of a list) is available on Wikipedia in the article describing Project Euler:

http://en.wikipedia.org/wiki/Project_Euler

But out of principle I do not think we should have any complete published solutions in this forum visible to the eye when scrolling thorugh the posts for a specific problem.

[hk]

Spoiler removed.

[Slaunger]

Thank you, hk

[BeulerBeuler]

Quilan's advice above is so good that I would go farther. If you take his advice, you can solve this problem in your head without a computer program. (Who else did this?)

[rae3012]

Hello, I am a very new programmer, and I was told about the Euler site by a friend. Today is my first day to use Python, and I need some help regarding the code that I have. It seems like it would work from a logical standpoint, but I cannot get it to execute. If someone would be so kind as to offer their assistance via PM, I would greatly appreciate it.