A place to air possible concerns or difficulties in understanding ProjectEuler problems. This forum is not meant to publish solutions. This forum is NOT meant to discuss solution methods or giving hints how a problem can be solved.
Forum rules
As your posts will be visible to the general public you are requested to be thoughtful in not posting anything that might explicitly give away how to solve a particular problem.
This forum is NOT meant to discuss solution methods for a problem.
In particular don't post any code fragments or results.
Don't start begging others to give partial answers to problems
Don't ask for hints how to solve a problem
Don't start a new topic for a problem if there already exists one
I have a very simple question that confuses me. I am stuck on the problem 808 and i really don't know why. I think it is because of the way the problem is posed. Below the problem:
Both 169 and 961 are the square of a prime. 169 is the reverse of 961.
We call a number a reversible prime square if:
It is not a palindrome, and
It is the square of a prime, and
Its reverse is also the square of a prime.
169 and 961 are not palindromes, so both are reversible prime squares.
Find the sum of the first 50 reversible prime squares.
So which number is needed to be added/take in count into the sum once we have found that it is a reversible prime square? Only 169? Or 169 and 961? Or only 13 (169=13*13)? Or 31 (31*31=961)?...
Problem 808 (View Problem) asks to find the sum of reversible prime squares, which means that you do NOT include the primes themselves (or any other number that fails any of the 3 criteria for being a reversible prime square:
Hebrew wrote: Fri Dec 23, 2022 4:47 pmSo which number is needed to be added/take in count into the sum once we have found that it is a reversible prime square? Only 169? Or 169 and 961?
Either of those could be correct, depending on their position in the list of ordered reversible prime squares.