Project Euler Forum

Which of the two new problems is the hard one?

Wait and see.

I kinda expected this answer

Way back problems used to come in batches of 2 to 4. I used to scan them: which one seems easiest to me. I presume you used that technique too, when fighting to become a 100%-er. So next Saturday and in the weekend of 5/6 September you can use that technique again....

Problem 201 will be very difficult. I've come to this conclusion because the number 201 is difficult. That's because 201 is a Blum integer, and they can with some satisfaction be selected as good primes in the RSA cryptographic algorithm. Surely breaking cryptography is a difficult problem, and thus problem 201 must be tricky indeed.

Problem 202 will also be quite an undertaking. I have thought about this thoroughly, and 202 is a bad, bad integer. That is because 202 can be written as 2x101, meaning the problem can only be solved in one way once you've dealt with the trivial case.

Problem 60 is the oldest problem solved by less than 1000 members, and problem 88 is the oldest solved by less than 500. Another difficult problems is 198. Curiously, all of these problems can be factored into four (not necessarily unique) factors. As 201 and 202 only have two factors, I think they will be less difficult than the ones mentioned. Taking into consideration the argument above, I do however belive both 201 and 202 will be quite tough.

Excellent analysis, Björn. I think we've published the integer angled quadrilaterals seven weeks too early, then.

25 minutes to go!

You sure were correct Björn. I don't have any idea on where to start with neither

I'll wait for the five easy ones in september then

daniel.is.fischer wrote:Wait and see.

still can't say which one the medium one was..

The one with more solutions, of course
Actually, that corresponds to our estimations, though it was closer than expected.