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 2015-10-29, 18:47 #1 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 146710 Posts November 2015 The Ibm (word) puzzle is out: https://www.research.ibm.com/haifa/p...ember2015.html
2015-10-30, 07:46   #2
ewmayer
2ω=0

Sep 2002
República de California

3×3,877 Posts

Quote:
 Originally Posted by R. Gerbicz The Ibm (word) puzzle is out: https://www.research.ibm.com/haifa/p...ember2015.html
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.

Last fiddled with by ewmayer on 2015-10-30 at 07:49

2015-10-30, 11:17   #3
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

101101110112 Posts

Quote:
 Originally Posted by ewmayer Question: Is a case considered uniquely solvable if it has just one non-degenerate solution, along with some degenerate ones?
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.

 2015-12-01, 17:48 #4 Xyzzy     "Mike" Aug 2002 2·3·1,361 Posts

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