20181230, 14:09  #1 
"Mike"
Aug 2002
8,167 Posts 
January 2019

20181230, 14:48  #2  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:
Last fiddled with by science_man_88 on 20181230 at 15:00 

20181230, 17:47  #3 
"Rashid Naimi"
Oct 2015
Remote to Here/There
7E8_{16} Posts 
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. 
20181230, 17:58  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20181230, 18:14  #5  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}×11×23 Posts 
Quote:
If Primes are products of distinct primes then: 4 > 2 > 1 25 > 5 > 1 100 > 20 > 4 > 2 > 1 121 > 11 > 1 Bob wins. Else if Primes arenot products 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. Last fiddled with by a1call on 20181230 at 18:16 

20181230, 18:59  #6  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20181230 at 19:00 

20181230, 19:19  #7 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·11·23 Posts 

20181230, 19:25  #8 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
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.
Last fiddled with by science_man_88 on 20181230 at 19:25 
20181230, 19:29  #9  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·11·23 Posts 
Quote:
And frankly I don't need you to let me know if I can or can not raise the issue with anyone. 

20181230, 19:38  #10 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
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.

20181230, 23:41  #11  
Sep 2017
2^{2}×5^{2} Posts 
Quote:
So primes are NOT a product of distinct primes, because a product is defined by at least 2 terms. 

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