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Xyzzy 2018-12-30 14:09

January 2019
 
[url]http://www.research.ibm.com/haifa/ponderthis/challenges/January2019.html[/url]

science_man_88 2018-12-30 14:48

[QUOTE=Xyzzy;504358][url]http://www.research.ibm.com/haifa/ponderthis/challenges/January2019.html[/url][/QUOTE]

I understand why the original numbers work but I don't see too much in the way of extending the problem. edit: okay I see one flaw with the original numbers.

a1call 2018-12-30 17:47

Here we go again with empty sets and semi empty sets.
Different conventions, conflictingly different solutions.
Makes the challenge flawed in my opinion.

Are primes products of distinct primes?
I would say not, but know of Wikipedia worshippers which would disagree.

science_man_88 2018-12-30 17:58

[QUOTE=a1call;504390]Here we go again with empty sets and semi empty sets.
Different conventions, conflictingly different solutions.
Makes the challenge flawed in my opinion.

Are primes products of distinct primes?
I would say not, but know of Wikipedia worshippers which would disagree.[/QUOTE]

take the individual sums if you look at them you'll find your answer.

a1call 2018-12-30 18:14

[QUOTE=science_man_88;504392]take the individual sums if you look at them you'll find your answer.[/QUOTE]

I disagree.

If Primes are [B]products [/B]of distinct primes then:

4 > 2 > 1
25 > 5 > 1
100 > 20 > 4 > 2 > 1
121 > 11 > 1

Bob wins.

Else if Primes [U]are-not[/U] [B]products [/B]of distinct primes then:

4, 25 and 121 have no winners so Alice will have the only choice of choosing 100 and then:

100 > 10 > 1

Bob wins.:smile:

science_man_88 2018-12-30 18:59

[QUOTE=a1call;504393]I disagree.

If Primes are [B]products [/B]of distinct primes then:

4 > 2 > 1
25 > 5 > 1
100 > 20 > 4 > 2 > 1
121 > 11 > 1

Bob wins.

Else if Primes [U]are-not[/U] [B]products [/B]of distinct primes then:

4, 25 and 121 have no winners so Alice will have the only choice of choosing 100 and then:

100 > 10 > 1

Bob wins.:smile:[/QUOTE]

false if the sum is a prime square, the only way not to repeat primes is for Alice to take the sqrt and Bob to divide by the sqrt a second time.

a1call 2018-12-30 19:19

[QUOTE=science_man_88;504398]false if the sum is a prime square, the only way not to repeat primes is for Alice to take the sqrt and Bob to divide by the sqrt a second time.[/QUOTE]
What exactly are you referring to when you say false?
Please be more specific. Thanks.

science_man_88 2018-12-30 19:25

[QUOTE=a1call;504402]What exactly are you referring to when you say false?
Please be more specific. Thanks.[/QUOTE]

you're just arguing that because you can assume a bunch of cases don't work that it's ambiguous. it's not. you can complain to the puzzlemaster you know.

a1call 2018-12-30 19:29

[QUOTE=science_man_88;504405]you're just arguing that because you can assume a bunch of cases don't work that it's ambiguous. it's not. you can complain to the puzzlemaster you know.[/QUOTE]

I am saying that two different conventions will both work and that's why the challenge is vague.
And frankly I don't need you to let me know if I can or can not raise the issue with anyone.

science_man_88 2018-12-30 19:38

[QUOTE=a1call;504407]I am saying that two different conventions will both work and that's why the challenge is vague.
And frankly I don't need you to let me know if I can or can not raise the issue with anyone.[/QUOTE]

many different conventions work in code as well, you don't complain about that. my point is if you want it to be unvague you can get an explanation from the puzzlemaster not here.

SmartMersenne 2018-12-30 23:41

[QUOTE=a1call;504390]Here we go again with empty sets and semi empty sets.
Different conventions, conflictingly different solutions.
Makes the challenge flawed in my opinion.

Are primes products of distinct primes?
I would say not, but know of Wikipedia worshippers which would disagree.[/QUOTE]

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.


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