First Post on Primes

Tinacity · October 31, 2020

Effectively deleted this post as I have to rethink a bit......Sorry for any confusion.

Edited October 31, 2020 by Tinacity
Noticed an error

StringJunky · October 31, 2020

3 hours ago, Tinacity said:

Effectively deleted this post as I have to rethink a bit......Sorry for any confusion.

But your original post has been a success for you, as you now probably know where you might go from here.

Edited October 31, 2020 by StringJunky

joigus · October 31, 2020

23 hours ago, joigus said:

that are not divided by 1, 2, 3, 5, ..., 23 (the 1st 23 primes).

1 is not a prime and those are not the first 23 primes. Don't trust me with numbers. :doh:

8 hours ago, studiot said:

The point is that prime no prime number greater than 23 is divisible by any of the first 9 primes.

The first 9 primes, I should have said. Thank you @studiot.

Charles 3781 · November 1, 2020

9 hours ago, Sensei said:

Yesterday, it was too late. OK. Now, I made program in C/C++:


#include <stdio.h>

//int primes[] = { 2, 3, 5, -1 };
int primes[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, -1 };

bool check(int a, int b) {
	for (int i = 0; primes[i] > 0; i++) {
		if ((a % primes[i]) == 0) return(false);
		if ((b % primes[i]) == 0) return(false);
	}
	return(true);
}

int main(int argc, int *argv[]) {
	int count = 0;
	int end = 1;
	printf("Primes: ");
	for (int i = 0; primes[i] > 0; i++) {
		printf("%d ", primes[i]);
		end *= primes[i];
	}
	printf("\n");
	printf("End %d\n", end);
	int half = end / 2;
	for (int i = 0; i < half; i++) {
		int a = i;
		int b = end - a;
		if (check(a, b)) {
			//printf("Found %d %d\n", a, b);
			count++;
		}
	}
	printf("Count %d\n", count);
	return(0);
}

For test case with 3 primes it gave:

Which looks good.

For OP problem with 9 primes it gave too many results to show them all (so had to disable printf()):

18 mln is over twice more than OP value.

Could you translate your program into BASIC, please, as that's the only programming language I understand?

Sensei · November 1, 2020

6 hours ago, Charles 3781 said:

Could you translate your program into BASIC, please, as that's the only programming language I understand?

If code takes 5-7 seconds on Core i7, in program written in C/C++, in BASIC it will take hours for the same job..

Learn C/C++ on https://en.cppreference.com/w/

Which line of code you don't understand? Tell me which line, I will tell you what it does (even though everything is pretty plain basic, and understandable without words)..

Maybe this?

if ((a % primes[i]) == 0)

% is modulo operator. https://www.google.com/search?q=modulo+operator+c

Quote

end *= primes[ i ];

It is equivalent to:

end = end * primes[ i ];

Edited November 1, 2020 by Sensei

Tinacity · November 1, 2020

I have realised that I formulated my question incorrectly, let's say, by a factor of two, as it were.

And can only apologise humbly for wasting good folks time on here, attempting to advise me on the wrong question.

I am not used to attempting to describe my space of interest and I was so close to it, that I miss_described a basic aspect of my problem space.

Instead of asking folk to "consider 2*3*5*7*11*13*17*19*23 = 223092870 ( the product of the first 9 primes).. Call this number: N"

I should have doubled that.....Because it is the first 223092870 PAIRS of natural numbers summing to N I am interested in examining.....

Thus N in this case should be 4*3*5*7*11*13*17*19*23 = 446185740 so that I can form 223092870 pairs which sum to 446185740 from that.

And hoping therefore.......that I can then assume that the number of pairs (x,y) which sum to N, but where neither x or y has any of the first 9 primes as an exact factor, is given by the product: (3-2)*(5-2)*(7-2)*(11-2)*(13-2)*(17-2)*(19-2)*(23-2) ?

And can also assume that the number of pairs (x,y) which sum to N, but where x and y are even but neither x or y has any of the first 8 odd primes as an exact factor, is also given by the product (3-2)*(5-2)*(7-2)*(11-2)*(13-2)*(17-2)*(19-2)*(23-2) ?

I think it better if I at least establish this first....before attempting any expansion of my questions from there.

I hope I haven't lost everyone with my amateurishness.....!!

studiot · November 1, 2020

1 hour ago, Tinacity said:

any of the first 8 odd primes

Interesting way to put it,

I don't know any even primes other than two

Tinacity · November 1, 2020

I guess I am attempting to respond to gentle criticism of my informality......

I do recognise that a lack of formalism results in confusions which are unfair to anyone giving up their time to consider my posts.

wtf · November 1, 2020

On 10/28/2020 at 8:26 AM, Tinacity said:

Consider, if you would.....the number 2*3*5*7*11*13*17*19*23 = 223092870 ( the product of the first 9 primes).. Call this number: N

Firstly. Am I correct in assuming that there are 3*5*9*11*15*17*21 pairs of numbers which can be summed to make N, where neither is divisible by any of the primes which make its product? Call that X

Haven't followed the thread but this seems a little ambiguous. Let's take a simpler example, N = 2 x 3 x 5 x 7 = 210.

Now you want to consider the pairs of numbers that sum to 210, such as (1, 209), (2,208), ..., (209, 1). Do you care about order? Is (1,209) the same as (209,1) or different? Not a big issue, just a factor of 2, but nice to know what is the intended intepretation.

Now "... where neither is divisible by any of the primes which make its product?" was confusing to me. What is "its" in this context? Do you mean that since 209 = 11x19, and neither 11 nor 19 is one of 2, 3, 5, or 7, we count (1,209) and (209,1) as satisfying your condition?

And you want to count the number of such pairs? Just want to make sure I'm understanding the question. Apologies to all if this has already been covered in the thread.

Edited November 1, 2020 by wtf

Tinacity · November 1, 2020

More ambiguous than ever, since I just recently amended the question by a factor or 2, as it were.

But yes.....Your simpler example does encompass my intentions.

And (1,209) and (209,1) are two answers.

Yes.....the pairs with 209 = 11*19 are relative prime pairs (if we count 1 as a prime that is; and it wouldn't detract from the quality of answer I am hoping for anyway)

I can see the confusion in the phrase "... where neither is divisible by any of the primes which make its product?"

Possibly because I didn't refer to each pair of numbers (x,y) which sum to N . Neither of those should be divisible by any of the factors of N.

@wtf

Here is my attempted amendment to my original question.