20151029, 18:47  #1 
"Robert Gerbicz"
Oct 2005
Hungary
10110001001_{2} Posts 
November 2015
The Ibm (word) puzzle is out: https://www.research.ibm.com/haifa/p...ember2015.html

20151030, 07:46  #2  
∂^{2}ω=0
Sep 2002
República de California
2^{3}·1,229 Posts 
Quote:
w,a,t,s,o,n = 3,2,3,4,4,9 wow = 343 now = 943 wow*now = 323449 = watson However, this eqn is not uniquely solvable, as it has three nontrivial solutions, all degenerate (excluding the trivial allzeros one): w,a,t,s,o,n = 3,2,3,4,4,9 w,a,t,s,o,n = 3,4,9,5,6,9 w,a,t,s,o,n = 3,7,6,4,8,9 Question: Is a case considered uniquely solvable if it has just one nondegenerate solution, along with some degenerate ones? For example, won*now = watson has the nondegenerate solution: w,a,t,s,o,n = 1,0,7,4,8,5, along with 2 degenerate ones. Last fiddled with by ewmayer on 20151030 at 07:49 

20151030, 11:17  #3  
"Robert Gerbicz"
Oct 2005
Hungary
1417_{10} Posts 
Quote:
Wikipedia has got also a good wording: "Traditionally, each letter should represent a different digit, and (as in ordinary arithmetic notation) the leading digit of a multidigit number must not be zero." So there could be a leading zero only if you use a one letter word. 

20151201, 17:48  #4 
"Mike"
Aug 2002
2^{7}·61 Posts 

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