Project Euler Forum

I am having a hard time understanding why 15 is not on the list 15 = 3 * 5 and it seems like those are consecutive primes.


EDIT:
sorry it is a late night.

"An even positive integer N will be called admissible, if ...."

Got me on this one too

Somehow consecutive primes suggest that 3*5 is a possibility.
Fortunate numbers are numbers that are the product of all primes under a certain limit.
I would prefer the phrasing that pseudo-Fortunate means divisible by all primes under a certain limit, and not divisible by primes above or at that limit.
Now it is suggested that a product of only factors 2 is an exception, while it is not.
After I got my answer rejected and I discovered that 15 was missing from the list, it took me a long time before I found that i missed the word even in the description.

Groetjes Albert

To be honest, if you didn't read a word in the description then that's not really the problem's fault.

And powers of 2 are a special case, as the word 'consecutive' only applies to two or more numbers. Other definitions like the one you proposed or the one in the problem's forum are rather clumsy wording.

While there is no need to change the statement at all, perhaps it would be best to change the first sentence as follows:

A positive integer N will be called admissible, if it is a power of 2 or its distinct prime factors are consecutive primes, starting from 2.

Pushed... So speaking of "factors are consecutive primes" means always counting from 2?

leghorn wrote:Pushed... So speaking of "factors are consecutive primes" means always counting from 2?

Since an admissible number must be even, 2 must be a prime factor. Thus, the consecutive primes must start at 2.

Regardless, I agree with TripleM's comment; the wording could be better.

Tom

Well, I think I've got what I needed. But, like so many times, the final result is wrong. Which brings me to this: Consider the numbers, 2, 4, 8 and 16. The PFN for each number is 3. So does the sum make 1, because I have only one distinct PFN, or 3, because that one PFN has the value 3?

Also, is the sum for N < 1000 equal to 39?

[edit] Clarification Update [/edit]

To clarify things myself (there was a typo in my code):
1. The sum for 2, 4, 8 and 16 would be 3.
2. The sum for N < 1000 is not 39.

Differing opinions are appreciated.

leghorn wrote:Consider the numbers, 2, 4, 8 and 16. The PFN for each number is 3. So does the sum make 1, because I have only one distinct PFN, or 3, because that one PFN has the value 3?

It doesn't say to count the distinct PFNs, but to sum them, i.e. add them together. Therefore from the results for 2,4,8,16 you would add 3 to the sum it asks you to calculate. As it asks for the sum of the distinct PFNs, you include the 3 only once in the sum, even though it occurs as a PFN many times.

leghorn wrote:Also, is the sum for N < 1000 equal to 39?

No. Try finding out what the PFN for 512 is.

Edit: I wrote this before/during the edit in the previous post.

Can someone verify the following info?

Sum of distinct PFN for n below 10 ** 2: 17
Sum of distinct PFN for n below 10 ** 3: 50
Sum of distinct PFN for n below 10 ** 4: 109

Are the admissable numbers supposed to be less than 10 ** 9, or is the problem asking for the sum of all different PF numbers below 10 ** 9?

EDIT: I solved it. It was an issue related to how I stored the admissable numbers.

Circling wrote: Fri Apr 30, 2021 3:00 pm Are the admissable numbers supposed to be less than 10 ** 9, or is the problem asking for the sum of all different PF numbers below 10 ** 9?

The former. The admissible numbers have to be less than 10^9. The question is the sum of all distinct pseudo-Fortunate numbers for those numbers.