November 2015
 

The Ibm (word) puzzle is out: [url]https://www.research.ibm.com/haifa/ponderthis/challenges/November2015.html[/url]

[QUOTE=R. Gerbicz;414217]The Ibm (word) puzzle is out: [url]https://www.research.ibm.com/haifa/ponderthis/challenges/November2015.html[/url][/QUOTE]

Comment: If we allow degenerate (> 1 letter maps to same 0-9-valued digit) solutions, their example WOW*NOW = WATSON is soluble, we have e.g.

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 all-zeros 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 non-degenerate solution, along with some degenerate ones? For example,
won*now = watson has the non-degenerate solution: w,a,t,s,o,n = 1,0,7,4,8,5, along with 2 degenerate ones.

[QUOTE=ewmayer;414280]
Question:
Is a case considered uniquely solvable if it has just one non-degenerate solution, along with some degenerate ones? [/QUOTE]

Yes, that is a good interpretation of the problem.

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 multi-digit number must not be zero."

So there could be a leading zero only if you use a one letter word.

[url]https://www.research.ibm.com/haifa/ponderthis/solutions/November2015.html[/url]