20170529, 16:42  #1 
"Robert Gerbicz"
Oct 2005
Hungary
2656_{8} Posts 
June 2017
There is a new Ibm puzzle:
https://www.research.ibm.com/haifa/p.../June2017.html 
20170529, 20:23  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
5·1,877 Posts 
Nice one! I think I have a "*" solution... (at least two)
Last fiddled with by Batalov on 20170529 at 20:32 Reason: there is more than one 
20170605, 21:37  #3 
"Robert Gerbicz"
Oct 2005
Hungary
5AE_{16} Posts 
There is an update:
"Update (5/6): Division by zero is not allowed, but you can use non integers and unary minus (like the example above); and you get a '*' for every prime solution." 
20170605, 22:15  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
5·1,877 Posts 
Quote:
Or else we can do \(e^{i \pi}\) or something like \(\sqrt 4\). Next month I am going to set up a webcrawler to watch the 'July2017.html' page ...so that I will not lose a couple hours to Robert ;) (This time I only learned of the problem launch from this thread.) 

20170606, 00:53  #5 
Jan 2017
2^{3}·11 Posts 
Not really. You can't insert arbitrary constants or operators into the expression (not integers or rationals either). The text only means that using the listed basic operators is allowed to create noninteger intermediate results (things like (1/3)*9 are allowed); allowing nonrational ones or not makes no difference as there are no operators which could create them.

20170606, 01:38  #6  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20170606, 05:41  #7 
Romulan Interpreter
Jun 2011
Thailand
9358_{10} Posts 
grrr... another silly problem... the formulation is unclear, where and whenever you can/may/must place signs, and what's happening if you don't...
"while other sevendigit numbers simply cannot be solved, such as 0314157" 0+31+41+57=100 skip. 
20170606, 05:46  #8 
Jan 2017
2^{3}×11 Posts 
Nope. There is no limit to integers or any other particular class of intermediate results. Complex numbers would be allowed if the list of operators included any that produced them. Them being allowed just doesn't matter because there's no way to produce them.

20170606, 05:49  #9 
Jan 2017
2^{3}×11 Posts 
I don't think it was unclear. Your "41 + 57" is quite clearly forbidden by the problem statement: "Without changing the order or concatenating two or more digits into a larger number".

20170606, 05:53  #10  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9385_{10} Posts 
Quote:
Quote:


20170606, 06:01  #11 
Romulan Interpreter
Jun 2011
Thailand
10010010001110_{2} Posts 
Whoooops...
Too fast a reading... the concatenated solution just jumped on my eyes from between the rows during I was skimming the text... Haha... Ok, I will look for a solution and send one... to make it up. It may take a while, our algorithms are always slower than yours... 
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