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Old 2018-12-31, 00:02   #12
a1call
 
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Quote:
Originally Posted by SmartMersenne View Post
1 is NOT a prime by definition. The smallest prime is 2.

So primes are NOT a product of distinct primes, because a product is defined by at least 2 terms.
I agree with you, but like I said there are those who consider a product with no factors a valid function just because the infallible Wikipedia has an article about it:

https://en.wikipedia.org/wiki/Empty_product
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Old 2018-12-31, 00:10   #13
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Quote:
Originally Posted by a1call View Post
I agree with you, but like I said there are those who consider a product with no factors a valid function just because the infallible Wikipedia has an article about it:

https://en.wikipedia.org/wiki/Empty_product
"If it is on the internet, it must be true"

Yes but we are talking about at least one term here.

Last fiddled with by SmartMersenne on 2018-12-31 at 00:11
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Old 2018-12-31, 00:38   #14
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Quote:
Originally Posted by SmartMersenne View Post
"If it is on the internet, it must be true"

Yes but we are talking about at least one term here.
It is not much of a stretch to go from a product with no factors being a valid function to one with one factor being valid as well.
Please see post number 6 above.
Regardless of if we accept the convention to be valid or not, the ambiguity of the challenge at not clarifying the concept of product-of-distinct-prime makes the challenge problematic.
We are left to guessing what the "Puzzle-Master" considers a valid convention.
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Old 2018-12-31, 01:54   #15
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Between the example and the requirement that a play actually does something, there seems to be no difficulty understanding which divisors are allowed. It could have been stated more precisely, but the cavils of the willfully obtuse are probably not a major concern of the folks writing these problems.

There does however appear to be a grammatical oopsadaisy. I believe the last comma in the statement of the problem is misplaced. It should be moved one word to the right, from after "numbers" to after "each."
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Old 2018-12-31, 02:08   #16
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Quote:
Originally Posted by Dr Sardonicus View Post
Between the example and the requirement that a play actually does something, there seems to be no difficulty understanding which divisors are allowed. It could have been stated more precisely, but the cavils of the willfully obtuse are probably not a major concern of the folks writing these problems.

There does however appear to be a grammatical oopsadaisy. I believe the last comma in the statement of the problem is misplaced. It should be moved one word to the right, from after "numbers" to after "each."
Right, some of us are just glass is Half-Full type of people by nature. Aren't we now?

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Old 2018-12-31, 02:21   #17
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Quote:
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It is not much of a stretch to go from a product with no factors being a valid function to one with one factor being valid as well.
Please see post number 6 above.
Regardless of if we accept the convention to be valid or not, the ambiguity of the challenge at not clarifying the concept of product-of-distinct-prime makes the challenge problematic.
We are left to guessing what the "Puzzle-Master" considers a valid convention.
The solution won't change in either case. Players can use a single prime.

In the case of allowing a product with no terms: that would get the game into an infinite loop. And if people can't see it, their program will get into an infinite loop, so I would like to see how they will solve it.

Last fiddled with by SmartMersenne on 2018-12-31 at 02:40
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Old 2018-12-31, 15:29   #18
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The set B in the example (but not the set A) can easily be expanded to four positive integers. Not that this is of any use in solving the problem, of course...
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Old 2018-12-31, 22:08   #19
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Quote:
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The set B in the example (but not the set A) can easily be expanded to four positive integers. Not that this is of any use in solving the problem, of course...
Care to elaborate?
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Old 2019-01-01, 03:33   #20
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Quote:
Originally Posted by Dr Sardonicus View Post
The set B in the example (but not the set A) can easily be expanded to four positive integers. Not that this is of any use in solving the problem, of course...
neither does the fact that all sums must not be squarefree. BTW a1call here's the puzzlemasters reply:

Quote:
I meant one or more primes; will update the site.

Last fiddled with by science_man_88 on 2019-01-01 at 03:35
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Old 2019-01-01, 15:03   #21
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Quote:
Originally Posted by SmartMersenne View Post
Quote:
Originally Posted by Dr Sardonicus View Post
The set B in the example (but not the set A) can easily be expanded to four positive integers. Not that this is of any use in solving the problem, of course...
Care to elaborate?
A = {3,99}, B = (1,22,97,526}
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Old 2019-01-01, 17:59   #22
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It is quite an interesting challenge.
I have tens of glasses which are 15/16 full but I am quite peptomestic about filling the last 1/16 of any of them.
One interesting aspect is that the square root of the sums seem to form in interesting and different patterns.
Many are symmetrical around a 45° diagonal.
Others form a row of primes then primesx2 then semi-primes then semi-primesx2 and other patterns as well.
And of course my posts would not be complete without a few painfully annoying nags to some:
The challenge does not define any rules regarding repeating terms within/between the two sets A & B. Without such clarifications many trivial solutions can be found.

Last fiddled with by a1call on 2019-01-01 at 18:01
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