I moved this from "Suggestions etc" to this Forum.
Our interpretation is that it is a dream a lot of engineers share.
Problem 263
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Comments, questions and clarifications about PE problems.
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
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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
See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
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hk
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earlbellinger
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Error in problem 263?
"A pair of consecutive prime numbers with a difference of six is called a sexy pair (since "sex" is the Latin word for "six"). The first sexy pair is (23, 29)."
I think this is an error. There are many sexy pairs less than (23, 29): (5,11), (7,13), (11,17), (13,19), (17,23).
I think this is an error. There are many sexy pairs less than (23, 29): (5,11), (7,13), (11,17), (13,19), (17,23).
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sivakd
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Re: Error in problem 263?
Your examples are not consecutive.
puzzle is a euphemism for lack of clarity
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earlbellinger
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endagorion
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Re: Problem 263
Ah. That's the first time I post on this forum, I'm totally confused.
I've "proved" that there are no "engineers' paradises". I may assume that my proof is flawed, but I fail to find the bug. I would appreciate any help.
Now, here we have a number n > (say) 30, which is EP. It's practical, so we have to represent 1, 2, 3, 4, ..., 8 as sums of its distinct divisors. 2 cannot be represented other way than itself, so n is even. 3 = 1 + 2, everything's good. 4 = 1 + 3, but if n is divisible by 3, then n + 3, n + 9 are divisible by 3 and therefore not prime, fail. So n has to be divisible by 4.
Next, 5 = 1 + 4, 6 = 2 + 4, 7 = 1 + 2 + 4. Next, 8. If n is not divisible by 5, 7 or 8 (it cannot be divisible by 6, as earlier), we can't represent 8, and that's a fail.
Assume n is divisible by 8. Then n + 4 and n - 4 are not, but as they're practical, they must be divisible by 5 or 7. They cannot be both divisible by 5 or 7, since their difference doesn't, so one of them is divisible by 7, say, n + 4. But then n - 3 is divisible by 7 and is not equal to 7, therefore not prime. Same for n - 4 and n + 3. Fail, n is not divisible by 8.
If not, n + 8 and n - 8 are also not divisible by 8, similarily one of them is divisible by 5, say, n + 8. But then n + 3 is also divisible by 5, therefore not prime, same for n - 8 and n - 3. We've come to a contradiction.
Where am I wrong? %)
Sorry if I give away something important.
I've "proved" that there are no "engineers' paradises". I may assume that my proof is flawed, but I fail to find the bug. I would appreciate any help.
Now, here we have a number n > (say) 30, which is EP. It's practical, so we have to represent 1, 2, 3, 4, ..., 8 as sums of its distinct divisors. 2 cannot be represented other way than itself, so n is even. 3 = 1 + 2, everything's good. 4 = 1 + 3, but if n is divisible by 3, then n + 3, n + 9 are divisible by 3 and therefore not prime, fail. So n has to be divisible by 4.
Next, 5 = 1 + 4, 6 = 2 + 4, 7 = 1 + 2 + 4. Next, 8. If n is not divisible by 5, 7 or 8 (it cannot be divisible by 6, as earlier), we can't represent 8, and that's a fail.
Assume n is divisible by 8. Then n + 4 and n - 4 are not, but as they're practical, they must be divisible by 5 or 7. They cannot be both divisible by 5 or 7, since their difference doesn't, so one of them is divisible by 7, say, n + 4. But then n - 3 is divisible by 7 and is not equal to 7, therefore not prime. Same for n - 4 and n + 3. Fail, n is not divisible by 8.
If not, n + 8 and n - 8 are also not divisible by 8, similarily one of them is divisible by 5, say, n + 8. But then n + 3 is also divisible by 5, therefore not prime, same for n - 8 and n - 3. We've come to a contradiction.
Where am I wrong? %)
Sorry if I give away something important.
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TripleM
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Re: Problem 263
n may not be divisible by 3, but that doesn't mean other practical numbers can't be.
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endagorion
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Oliver1978
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Re: Problem 263
Can I safely assume there are 28,388 sexy triple-pairs (n-9,n-3)-(n-3,n+3)-(n+3,n+9) below 109?
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