Project Euler Forum

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There is a separate forum specifically for discussing the problems. So please post your queries there. However, it asks that you not post code as it may give the problem away.

As addition and multiplication are commutative and associative, the permutations don't matter. It is just the collection of digits in each number. I found these two by hand by looking for pairs of digits that have the same product, then combining them to produce the same sum. So 2*3=1*6 but 2+3 is 2...

How about 232319 and 161633?

Problem 158 yields an algebraic answer to pencil and paper, though calculating the final answer takes a high-precision calculator

If object_1 is acted on by a force from object_2 that is inversely proportional to the square of the objects distance (think gravity), is there away to determine the velocity of object_1 at a given distance from object_2? So the force between the two things is: F=-1/r^2 where r is the distance betw...

I strongly suspect that one can make an infinite string with three characters. This would answer both questions. An approach, which I have not fully proved, is provided by the rules: start with 0. Each step, take the existing string and replace 0 with 02, 1 with 012, and 2 with 010212. So we have a ...

That would be a better strategy than what I had in mind, so I am very interested to hear it! The best I could achieve with my simple strategy was P(1000,5)=333: Identify 9 products with every 3 parcels we buy, as follows: Calling the nine products a,b,c,d,e,f,g,h and i, we buy the parcels addgh, be...

How about all numbers of the form sqrt(p1*p2*p3*...) for the p's distinct primes? It is certainly explicit and uncountable. I think if you find a set that is linearly dependent over Q then you can prove some sqrt(product of primes) to be rational This set is countable. There's a one-to-one mapping ...

How about all numbers of the form sqrt(p1*p2*p3*...) for the p's distinct primes? It is certainly explicit and uncountable. I think if you find a set that is linearly dependent over Q then you can prove some sqrt(product of primes) to be rational

That is all primes except 2 and 5.

The proofs are not difficult: For the first part, in a fair game the expected value to A cannot change. As his only outcomes are 0 and A+B, his probability of A+B must be a/(A+B) http://en.wikipedia.org/wiki/Gambler%27s_ruin has the result for an unfair coin with an indication of how to get it For t...

So accepting that maxindx₁₃=1.127 x 10²⁰⁰-1, maxindx₁₄ will end with 1.4 x 10¹⁹⁸ 9s. But as your example shows, the number can be a thousand times larger than this. And the sum of digits of maxindx₁₅ will be about 1/9 the actual value of maxindx₁₄, so of order 10^(10¹⁹⁸)

In that case, I think the easiest is to move the squares to the center peg, the triangles to the left, and the squares to the right, using the standard solution. You have to clear the largest square from the left peg, and putting it on the triangles is in the way. So for four of each we need 3*(2^4-...

If the rule is that any square can go on any triangle, we can just view this as a stage of the original game with the number of disks equal to the sum of the squares and triangles. So for 4 squares and 4 triangles we have eight disks with a linear order among them. If we started with all the disks o...

Here is the start of the sheet. Based on the preview, you will have to work on the spacing. The column headings should be clear, and the powers of 2 are the sum of the various Qty columns. Qty 1 is the average number of each level at round 1 = 64/9. Pwin1 is the probability that a character of that ...

My complete results are 1 wins 0.000104453, 2 wins 0.002418018, 3 wins 0.012064299, 4 wins 0.033487702, 5 wins 0.068734128, 6 wins 0.117823111, 7 wins 0.179552404, 8 wins 0.252146623, 9 wins 0.333669262. Note that the 8 probability is within 1/3 of a single win of yours. I think these agree pretty w...

I make it 0.333669262 that a level 9 will win and 0.000104453 that a level 1 will win. This was done in Excel so there will be rounding error. The sheet played each level against an even mix of levels to calculate the winning chance on the first round. For each round after that, it took the level di...

A good example. You are right that things are more subtle than I thought.

If the cells are more complicated and irregular, you may be able to derive the points (I'm not sure how to interpret rmillika 's algorithm, I think it's more complicated than stated). I don't think so for any of the bounded regions. Each bounded region is a convex polygon where each side is a porti...