Project Euler Forum

Thanks for fixing the captchas. I appreciate the new looks of the site. I have to mention it though: where has the 'remind me' checkbox at login gone?

milhous wrote:Greetings from California.

I wanted to let you know that the CAPTCHA authentication is broken, no image will appear when trying to enter the answer. Thanks!

Reproduced this. I was using FF 6.0.2. Maybe the image calls for the wrong location of the captcha script?

It appears to work (as far as I see superficially). I have just created a PE project having no main class, but a 'problems' package and a 'utils' package; all problems have their own file with a 'main' method. In my opinion, such is easier than your suggested method.

I apologise for posting blatant lies. I should read the problem description more carefully the next time somebody asks a question.

But do you have to choose your own f? Or is the given f already 1/4? Or is the question what is the best f you can choose? And i assume this is random, how can we know the random generator of a specific language is the same as any other one? All throws have a random 50% chance of winning or losing,...

It is a correct proof if I look at it superficially. Also, (and therefore :) ) it is a true statement. Let me give another (somewhat shorter) proof: Let $p,q$ be prime factors of $a$ such that $pq\mid a$ (to avoid trivialities), and $p\geq q>\sqrt{a}$. Then $pq>a$, contradicting the assumption. Henc...

RichardBuckalew wrote:'remove leading zeros before performing the addition'

Could you explain what difference this could possibly make? Last time I checked, I could calculate 120+21=141=120+021.

Any decent compiler/IDE should warn you about the variable x possibly being not initialized; you should consider one. As you are new to this forum, I will attend you to the fact that this particular section is reserved solely for discussing possible misunderstandings of PE problem wordings. The righ...

I would say it is flawed. The only obvious thing that might work and occurred to me would be to change condition 1 in the following way: 1. The bounding box for the one segment intersects the other segment, and vice versa. So not the bounding boxes, but a stronger condition. It sounds like it could ...

142857 is also a LEFT rotation of 714285. What is your point? There is only *one* left-rotation, and only one right-rotation (this can in fact be deduced from the problem's wording). It consists of transferring the last digit to the first place. So 142857 is (the right-rotation of)^5 714285, and al...

How about the nice 'search...' box at the top right? Works brilliantly for me

Someone (on another thread) mentioned that there are combinatoric solutions for "almost all of the questions up to 37" and 24 in particular is shouting "combinatorics" at me but I can't see how this is possible and again why would you write a program to do it this way? Sorry if ...

Schwein wrote:the only thing i don't understand, is why that does not give me all the different combinations makeable...

How about combining larger, 'composite' capacitors in series or parallel? (This is what hk meant above as well)
Think about $1 = \frac1{1+1}+\frac1{1+1}$.

I still have an ingrained dislike for definitions. I work in an industry where there are tons of brand new terms. Often there are three or four terms that mean the same thing, or a term that different people claim to mean quite different things. I often find myself translating. Certification exams ...

Heh, I wasn't expecting something slick like that. My instinct for combinatorial problems always seems to be to resort to something awkard and laborious... As with most PE problems, thinking up the correct algorithm is a lot more important than choosing the exact implementation. I can advise you to...

Suppose you have an array of prime factors, say of the form {{2, 2}, {5, 7}, ...} which says the number is $2^2\cdot 5^7\cdot\ldots$. Then you could iterate over the array and then each component as follows (in pseudo-code): factors = {{2, 2}, {5, 7}, ...}; result = {1}; for each factor in factors: ...

Yeah.....the thing is wont we have to make a list of 2 million numbers to apply that pseudocode ?????? I mean isn't there some other efficient way In the end, there isn't really. One could use the fact that 2 is the only even prime to bring it down to 1 million. Then, as the primeQ flag is just a b...

I just solved the problem but my algorithm takes WAAAYY to much time. It take like 3 or 5 minutes, which is way to much. I did it using the hint from wikipedia of trivial division wich says: It consists in testing whether n is a multiple of any integer between 2 and √n. So that implied that all pri...

Could someone give me one (hopefully) small hint?? is the answer greater than 775146.09922452678 or smaller than it, ie the square root of 600851475143??? Yes Your answer should have been No,Yes. To that end, I think it is amusing that you state that I should have asked everybody falling under the ...

chicharito14 wrote:Could someone give me one (hopefully) small hint?? is the answer greater than 775146.09922452678 or smaller than it, ie the square root of 600851475143???

Yes