Project Euler Forum

Hibernatus34 wrote: I have yet to understand why so many people want to post their useless code.
I wish there was a page dedicated to that, separate from the forums, so everyone would be satisfied.

Cf. viewtopic.php?f=5&t=1421&p=26961#p26961

What about n = 24? 12 has more divisors than 25 (6 vs. 3). Still, you are not too far off. Try to think in prime factorisations.

Hey, can someone help me out in this? I'm not an experienced programmer and I don't know how to put the the digits of 2^1000 into an array. I mean, I know arrays and how to get digits and everything, but how do you put this specific number into the form of an array? Iterate the following: Take modu...

This question confuses me. What's the difference between the 3X2X1 with 2 layers cuboid, and a 3X2X2 cuboid. I don't seem to be grasping how a 3X2X1 cuboid with 1 layer requires 22 cubes, but that same cuboid with 2 layers requires 46 and that same cuboid at 3 layers requires 78. How are these cubo...

Even though for example 1/3 is not an integer, you can still divide by 3 modulo 29. This is because 3*10==30==1 (29), so 10 acts just like 1/3 in that if you multiply it by 3 you get 1. Could you then say $\frac{1}{3} \equiv 10 \pmod{29}$ In some sense yes, but I would always prefer to write $3^{-1...

I'm sure this is not what you are asking, but why not take a single row of n rectangular polyominoes? (or is there a fixed rectangle? if so, why are the two example rectangles not identical?)

For completeness' sake, I think you find it useful to know that $a\pmod{b}$ is invertible (has a solution $c$ to $ac\equiv 1\pmod{b}$) precisely when $\gcd(a,b)=1$. For determining (existence of) square roots, there is a useful tool called the Legendre symbol, or more generally the Jacobi symbol. Fo...

+1 from me for this initiative. I think however that it might be easiest to implement if just every (locked) problem gets such a forum. A range of 10 problems would be hard to manage as to whom gets to see what.

Yes, entirely correct. Usually I skip to calculating all the cubes (or other expressions) for 0..n-1 (mod n) as you did. Note that for squares, you would only have to check half of them, as $(-x)^2 = x^2$

Having recently followed a course that briefly touched the incompleteness theorems, let me try to explain them. The first theorem says, that there exist sentences (propositions, if you wish) which are true, precisely when there is no proof that they are true (the notion of proof can be precisely def...

Oh, I'm awfullyy sorry. I will never do that again. How stupid of me. It is not really stupid. It is comparable to someone asking how to make $5000 in 4 days because he has to make a tax payment by then, and you tell him how to request extension of the payment deadline. There's nothing wrong with s...

The problem says nothing about being part of more than one prime triplet . A prime quadruple is just two triplets that aren't disjoint.

I'm still not sure I understand! I think I understand type casting, but what is the point of adding 0.5? How does (int)( 0.5 + Math.sqrt(inn) ) differ from (int)( Math.sqrt(inn) ) ? And isn't it necessary to use (double)inn as the argument in the Math.sqrt-function? Will Math.sqrt(16) (or any other...

jaap's code works around the Java 'rounding' which is actually truncating the decimal part. That's where the 0.5 is coming from. The (int) signifies a type cast from double to int. Make sure you look up type casting and understand it, as it is important in almost every programming language. As for d...

In the Java code, you add 'susu' to sum as opposed to 'su' in the MatLab code. The latter makes more sense because otherwise, you could substitute 'i' for 'susu'. Unfortunately, this forum being publicly visible is not meant to post solutions to PE problems (not even without mentioning the number). ...

You are dividing integers, meaning that the resulting expression will be typecast to an integer internally (or, simpler, that it will be rounded to return an integer). The fact that the result displayed is similar to the actual result tells you that the size of the number is not a problem. I have su...

Quite generally, when one gets to a certain amount of solved problems, there come available more and more solution PDFs and forums. These can be treasures of information if read thoroughly. Most general techniques and algorithms also have a decent Wikipedia page. From there, usually there are also l...

Hmmm...okay, here's the thing... This problem is identical in nature to problem 16 (same deal with a huge number that you need to store and add up digit by digit), yet while my code worked for problem 16, after it spit out the answer for problem 20, the checker rejected it. Now I know it's against ...

Many people already have suggested this. However, I think this feature will take it's time before being implemented. I can even imagine the solution forum being completely reworked somewhere in the future. But that's more a decision for euler c.s..

... The left child simply takes in the same list but starting at the next row. The right child takes in the same list but with the first element of each row removed. It's an ugly loop. Can anyone suggest a better approach? I read this as if you are duplicating elements since they mostly occur both ...